help-39

[bright apex]

ik but I want to break the fraction

[mortal flower]

wait whats the goal here

is it just to simplify

[bright apex]

no

to break it into more fractions

[mortal flower]

you can factor out (1/(n+1)^2)

and then break it up more

[bright apex]

?

[mortal flower]

both terms have 1/(n+1)^2, no?

[bright apex]

yea but i want to get 2 fractions it is a part of a bigger problem

[mortal flower]

whats the whole problem

[mortal flower]

see what happens if you look at a_n+2

[bright apex]

?

[mortal flower]

they give you a formula for a_n+1

so look at the next term in the sequence

[bright apex]

how can we find

[opaque void]

you here?

[bright apex]

yrs

@opaque void

[opaque void]

you can call what's inside the sum b_n
you want to find sum of b_n from n=1 to inf
by simplifying you'll find that b_(n+1)=1/2 * b_n which is a geometric serie
by calculating b_1 you'll conclude that b_n=(-1)(1/2)^(n-1)

and that makes it easy to calculate sum of b_n (your question)

@bright apex got it!?

[bright apex]

i sec

$you can call what's inside the sum b_n
you want to find sum of b_n from n=1 to inf
by simplifying you'll find that b_(n+1)=1/2 * b_n which is a geometric serie
by calculating b_1 you'll conclude that b_n=(-1)(1/2)^(n-1)$

[jolly parrot]

ArsH

[bright apex]

no I dont

[opaque void]

ok let's do it step by step
you wrote bn?

[bright apex]

what's bn

[opaque void]

bn=what's inside your sum

[bright apex]

k

so sum of bn

[opaque void]

yes exactly

[bright apex]

k

[bright apex]

... so it's easy to find

[opaque void]

yep

you'll find that b(n+1)=(1/2)bn

which is a geometric serie

[bright apex]

how

[opaque void]

just simpfly b(n+1)

[bright apex]

i want to find how we got the geometric series

[opaque void]

b(n+1)=a(n+1)-2/(n+1)²
= (1/2)an + (n²-2n-1)/n²(n+1)² - 2/(n+1)²
= (1/2)an + (n²-2n-1-2n²)/n²(n+1)²
= (1/2)an - 1/n²
=(1/2)(an - 2/n²)
=(1/2)bn
so
b(n+1)=(1/2)bn

focus carefully

then
b(1)=a(1) - 2/1² = 1-2 = -1
so by definition of geometric series:
b(n)=(-1)(1/2)^(n-1)

[bright apex]

can u write and show pls

...

i did till here

[opaque void]

there's no need to split the fraction

[opaque void]

you can start from here assuming what's inside the sum is bn
then go from here

[bright apex]

i want to learn that

i have not learned gp and sum of all terms of gp

[opaque void]

you want to learn splitting fractions?

[opaque void]

what do you mean by gp?

[bright apex]

yes

[bright apex]

geometric progression

[opaque void]

it goes like this
(n²-2n-1)/n²(n+1)² = a/n + b/n² + c/n+1 + d/(n+1)²

but in that question we don't need to split the fraction

[opaque void]

we call bn is a geometric progression we see:
b(n+1)=r*bn where r is the common ration

in our case we found b(n+1)=(1/2)bn so bn is a gp with r=1/2

then we can conclude that
bn=b1*(1/2)^(n-1)

[bright apex]

can u explain this

[opaque void]

b1=(-1)

[opaque void]

you always split fractions like this

for example
1/(n+1)(n+3)² = a/(n+1) + b/(n+3) + c/(n+3)²

now you need to find those constents ( a b c d ..)

[bright apex]

d/n

[opaque void]

here
to find a:
. multiply both sides by n+1 then make n=-1 you'll directly find a

[bright apex]

wait imma try and come

if i got it I may forget

i understood ig

ty

[opaque void]

okey mention me if you need me

[pearl pond]

@bright apex Has your question been resolved?

[bright apex]

Yes

[pearl pond]

Channel closed

Closed by @bright apex

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[scarlet glade]

idk at all how to tell, i don't know when i was supposed to learn this

[pearl pond]

@scarlet glade Has your question been resolved?

[feral sedge]

It's based on the signs of the terms

Well there's a few things it's based on but

[scarlet glade]

so like is the 4rth a hyperbola?

[feral sedge]

If there's one linear term then it'll be a paraboloid

linear means no ^2 sign

[scarlet glade]

ohhh that helps a lot

[feral sedge]

Either elliptic or hyperbolic paraboloid

The axis will match the linear term

[scarlet glade]

the hyperbolic paraboloid doesn't have an axis tho

[feral sedge]

Yeah ig they don't include the "axis" for that one

[scarlet glade]

i did it ty... but that one i dont get still

just got it by poe

[feral sedge]

The elliptical paraboloids, if you look at their graph, very blatantly point in a direction

Then the difference between elliptical and hyperbolic

elliptical the quadratic terms have the same sign

Like z = x²+y²

But hyperbolic paraboloid they're opposite

Like z = x²-y²

[scarlet glade]

ty, imma look at some graphs now lol

.close

[pearl pond]

Channel closed

Closed by @scarlet glade

Use .reopen if this was a mistake.

[feral sedge]

no problem

[scarlet glade]

just noticed i literally could have guessed abcde 😭

[feral sedge]

Oh lmao

thats crazy

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fading ledge]

how did they choose n>=m=1

1?

[glass meadow]

Presumably the indexing starts at 1

They're writing n >= m because that's how convergence is shown (for all n past a certain point, which we can call m)

They could have picked any natural number for m

[toxic lichen]

if you dont like choosing your m as 1 then you can also artificially choose m = 42069

[glass meadow]

Careful someone might pick a certain other number

[fading ledge]

Ohh it starts from natural

Again

Thanks

.close

[pearl pond]

Channel closed

Closed by @fading ledge

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dawn arrow]

ok i need help with this again so i did the cross product of the normal vector of the 2 intersecting planes and then the cross product of that vector and the normal vector of the plane that is perpendicular to the plane we are looking for but how do we find a point to make an equation for the plane

[dawn arrow]

<@&286206848099549185>

[pearl pond]

@dawn arrow Has your question been resolved?

[dawn arrow]

cani get some helkp

[fierce sky]

Your working looks right except it should be <1, -4, 1> instead of <1, -5, 1> but that seems to be a typo, and you also need to find a point on the plane so you can gather an equation

You can get any point that satisfies x - z = 2 and y + 4z = 1 since the line of intersection is always on the plane

[dawn arrow]

but why are we getting the point from those 2 planes that intersect

they intersect at one line

and they are separate planes

[fierce sky]

Because we want to find a point on the plane whose equation we are trying to find

And since we know the plane passes through the line of intersection, we can use any point from that line

[dawn arrow]

right

[dawn arrow]

doesnt this imply that we are using any point thats on these 2 planes and not just the line of intersection

this is so confusing 💔

[fierce sky]

Any point that is common with the intersection of those 2 planes yeah, but that's just entirely the line of intersection

The only(?) way to find a point characterizing the plane is to utilize the line of intersection

[dawn arrow]

the line of intersection

is the cross product i got using the normal vectors

?

😭

[fierce sky]

yep

at least the directional vector of the line of intersection

but you don't need to use that to find a point on the line

you just want one ordered triple that satisfies x - z = 2 and y + 4z = 1, you could for example set z = 0 to get x = 2 and y = 1

[dawn arrow]

i guess my confusion lies in the fact that the 2 planes that intersect are their own planes and i can not grasp the idea that we can just grab any point from these 2 planes because they only intersect at one line and so grabbing just any old point from either of these 2 planes could also be at a point where the 2 planes do not intersect so how would that work

im sorry 😭

[fierce sky]

we're grabbing a point from the intersection of the two planes which will always be on the line of intersection, i.e. simultaneously solving x - z = 2 and y + 4z = 1

because the requirement for our plane is that it passes through the intersection of the two planes

[dawn arrow]

i get that but how do we know that the point that we pick is specifically from the line of intersection and not a random point on either of the planes 💔

[fierce sky]

Because the intersection of the planes x - z = 2 and y + 4z = 1 is the line of intersection

The intersection of two non-parallel planes is always a line

https://www.desmos.com/3d/hboyrbxqpy

[dawn arrow]

omg it just clicked thank you

.close

[pearl pond]

Channel closed

Closed by @dawn arrow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[swift spindle]

I realised that if I didn’t get rid of sqrt(x^2), I have +-, but otherwise I have only + if I got rid of it earlier

How do I make sure I always have options?

[atomic flare]

its because its not a positive once you take away the sqrt over x^2

sqrtx^2 = |x|

in which now |x|= 10, so x=10 & x=-10

[swift spindle]

ohhh Tyty

[spare lark]

or

[swift spindle]

can’t be both?

or is it

.close

[pearl pond]

Channel closed

Closed by @swift spindle

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vast berry]

does musical isomorphism basically turn vector fields into k forms ?

[pearl pond]

@vast berry Has your question been resolved?

[spare lark]

Not at the same time

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[green aurora]

In an urn there are 2 red balls and 6 white ones. Before one is drawn, one is removed from the urn without seeing it. Calculate the probability that the first ball drawn is red.

[green aurora]

I considered 2 events A="probability of drawing a red ball" B="ball removed"

so $P(A|B)=\frac{P(A\cap B}{P(B)}=\frac{P(A)P(B|A)}{P(B)}=\frac{6/8\cdot 1/7}{1/8}=6/7$

[jolly parrot]

Goofy Joe

[green aurora]

but it's wrong

[thorn agate]

B = which ball removed ?

[green aurora]

a random ball

[thorn agate]

the probability red is drawn depends on which ball is removed

it changes for red removed and non red removed

[green aurora]

so should the 2 cases be treated separately?

[thorn agate]

yeah

[green aurora]

2/7+1/7

3/7

[thorn agate]

not quite

2/7 *2/8 +1/7*6/8

[green aurora]

?

[thorn agate]

P(A) = P(B_r)P(A|B_r) + P(B_w)P(A|B_w)

[green aurora]

so 1/4?

[thorn agate]

yeah

[green aurora]

But what difference does it make if I see the ball or not? That is, why in this case it is as if the ball was not removed?

[thorn agate]

if you know which ball you had removed you would just do P(A|ball)

[green aurora]

but why in this case the final probability comes out as if I had not removed the ball, that is 2/8?

[thorn agate]

because the cases become balanced likely ig idk how to phrase it

um sorry but i don't have a good prob knowledge but many do

they will explain it better

[green aurora]

but the event "a ball was removed" is independent of the event "a ball is red", why?

[green aurora]

ok , anyway thanks

<@&286206848099549185>

[thorn agate]

more i think of this it becomes that when you fix a ball being removed the sample space gets reduced but yh
there is a more theoretic explanation

[green aurora]

but even if I remove 2 it doesn't change, why?

[pearl pond]

@green aurora Has your question been resolved?

[green aurora]

<@&268886789983436800>

[pearl pond]

@green aurora Has your question been resolved?

[fierce sky]

why not try and generalize it and see

Given there are r red balls and w white balls, what's the probability that the first ball drawn is red, given that one ball is removed from the urn beforehand without seeing it?

,texsp $$\frac{r}{r+w}\cdot\frac{r-1}{r+w-1}+\frac{w}{r+w}\cdot\frac{r}{r+w-1}=\frac{r\left(r+w-1\right)}{\left(r+w\right)\left(r+w-1\right)}=\frac{r}{r+w}$$

[jolly parrot]

Roy

[green aurora]

Thanks

.close

[pearl pond]

Channel closed

Closed by @green aurora

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[torn lodge]

how do i simplify the fraction to look like the second image as efficiently as possible, I've tried to expand and simplify but I haven't been able to simplify it to look it

[calm wing]

multiply top and bottom by (x-2)

[dense jasper]

Multiply the numerator and denominator by $(x-2)$

[calm wing]

lol the sync

[torn lodge]

i'll try that

got it thanks!

.close

[pearl pond]

Channel closed

Closed by @torn lodge

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[torpid basin]

Hi I'm reviewing my old practice test and don't understand the process of doing this problem it's been a while since we've done it 😭

[plush bramble]

Example 5.3.1
https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/05%3A_Integration/5.03%3A_Riemann_Sums

[pearl pond]

@torpid basin Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[toxic lichen]

what's $2^{\cdot}$ ?

[crimson haven]

Idk which to take out

[jolly parrot]

Ann

[crimson haven]

Omg

[toxic lichen]

what's 2 to the power of "dot"

[crimson haven]

It’s Ann

Yay

[crimson haven]

Oh that was a mistake

[toxic lichen]

ok then what should go in there instead

[crimson haven]

One

[toxic lichen]

right

anyway, 2^(-3x) is best kept as is

$2^{1} \cdot 2^{2x} \cdot 2^{-2}$ should be combined into a single power

[jolly parrot]

Ann

[crimson haven]

Oh

[toxic lichen]

can you erase the line starting with "(2^x)^-3 ...", do what i said, and stop there

[crimson haven]

Yes pls

[toxic lichen]

... then go and do it lol

and show what you get

[crimson haven]

yes ms

Can I send the 2 to the other side

[toxic lichen]

i clearly said to do something and stop there

so my answer to that is: not yet.

[crimson haven]

oki

[toxic lichen]

girl don't overthink it

[crimson haven]

I still don’t know how to factories this

[toxic lichen]

ok, so you did not follow the instruction of "do what i said and stop there". you did not, in fact, stop there.

you jumped ahead.

...though your steps are correct

you got it down to 5x = 1 and divided by 5, so..

what's the issue exactly?

[crimson haven]

Understanding how to solve this

And I’m sorry for jumping

I rlly rhgouht ie as doing what u said

Well except the stop sort

Part

[toxic lichen]

girl you got it down to literally 1 step away from the solution

even less than that

bc you basically executed the last step

[crimson haven]

I did it unconsciously though

[toxic lichen]

if you were just told to solve the equation 5x=1 and that was the entire question, would you know what to do?

[crimson haven]

Something it feel like I’m solving but my Brian Isnt there

[crimson haven]

Yes

Wait no

[toxic lichen]

ok then do it

[crimson haven]

yes

[toxic lichen]

.

[crimson haven]

No no I mean with lower

You don’t like upper case

[toxic lichen]

????

[crimson haven]

You said to stop tlajing w capital

[toxic lichen]

where and when did i say that?

either you misunderstood me somewhere, or you are straight up imagining things i said which i never did

[crimson haven]

I think you did

But it’s fine

Asking as your happy

[celest rivet]

yo

[toxic lichen]

if you were just told to solve the equation 5x=1 and that was the entire question, would you know what to do?

answer me this now. yes or no. please don't flip flop.

[crimson haven]

Hello

[celest rivet]

how r u

[crimson haven]

Yes

[crimson haven]

Ik good wby

[toxic lichen]

hello and welcome to the server. #discussion is our place for chat

[celest rivet]

good i want to lock in for gcse

[toxic lichen]

ok then do it

[celest rivet]

oh my bad

[crimson haven]

Ok

Uh you divide Bith side by 5

So you got 1/5 =c

X

[toxic lichen]

yes

x = 1/5 is correct

[crimson haven]

Yay

Does this seem correct pls

[glossy blade]

you can just put the value in the question to recheck

[crimson haven]

Well no I don’t honk it’s right

Or Im notninouting well

[pearl pond]

@crimson haven Has your question been resolved?

[pearl pond]

@crimson haven Has your question been resolved?

[pearl pond]

Channel closed

Closed by @crimson haven

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vestal thistle]

Hello, could someone check if this proof looks good please?

\begin{Definition}[Natural numbers]
The set of the \emph{natural numbers} is the set $\bN = \{1,2,3,4,5,\dots\}$.
\end{Definition}

\begin{Theorem}
Suppose $n \in \bN$, then $5 \mid (n^5 - n)$.
\end{Theorem}

\begin{proof}
Let $n \in \bN$. We proceed by induction.\\

\underline{Base Case.} The base case is when $n = 1$, and $5 \mid (1^5 - 1)$ as desired.\\

\underline{Inductive Hypothesis.} Let $k \in \bN$, and assume $5 \mid (k^5 - k)$.\\

\underline{Induction step.} We aim to prove that the result holds for $k + 1$.
That is, we wish to show that $5 \mid ((k + 1)^5 - (k + 1))$.
To do this, we begin with the expression $(k + 1)^5 - (k + 1)$:
\begin{align*}
(k + 1)^5 - (k + 1) &= k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 - k - 1 \\
&= k^5 - k + 5k^4 + 10k^3 + 10k^2 + 5k
\end{align*}

By the inductive hypothesis, $5 \mid (k^5 - k)$, and so, by definition of divisibility, 
$k^5 - k = 5l$ for some integer $l$.
Then,
\begin{align*}
k^5 - k + 5k^4 + 10k^3 + 10k^2 + 5k &= 5l + 5(k^4 + 2k^3 + 2k^2 + k) \\
&= 5(l + k^4 + 2k^3 + 2k^2 + k)
\end{align*}

Since $k$ and $l$ are integers, so is $l + k^4 + 2k^3 + 2k^2 + k$.
Then, $(k + 1)^5 - (k + 1) = 5m$ where $m = l + k^4 + 2k^3 + 2k^2 + k$.
Thus, by the definition of divisibility, $5 \mid ((k + 1)^5 - (k + 1))$.
Therefore, by induction, $5 \mid (k^5 - k)$ for all $n \in \bN$.
\end{proof}

[jolly parrot]

Mor Bras

[autumn fossil]

seems fine to me

[toxic lichen]

seems ok if your goal was to prove it by induction

[haughty mural]

you worked the goal to the hypothesis and you could consider working the hypothesis to the goal

it's a style difference

equally valid

[autumn fossil]

This is actually quite interesting, I think it could be used to prove fermat's little theorem by binomial coefficients + induction

the statement p | n^p - n is valid for every prime btw

[vestal thistle]

For sure

[haughty mural]

To work the hypothesis to the goal, you'd start with 5 | k^5 - k and have 5 | k^5 - k + 5 m for all m. Then specialize for m = 5k^4 + ... and prove 5 | (k + 1)^5 - (k + 1) by rewrites.

[vestal thistle]

I feel that it's more straight forward than showing that 0 is not in the set of values modulo p and that every element in that set doesn't appear more than once

Thank you for the comments!

.close

[pearl pond]

Channel closed

Closed by @vestal thistle

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[boreal laurel]

Let $\theta_1, \theta_2, \theta_3, \dots, \theta_n$ be $n$ angles. Let $s_r$ denote the sum of the tangents of these angles taken $r$ at a time (e.g., $s_1 = \sum \tan \theta_1$, $s_2 = \sum \tan \theta_1 \tan \theta_2$, etc.). Prove that:$$\tan(\theta_1 + \theta_2 + \dots + \theta_n) = \frac{s_1 - s_3 + s_5 - \dots}{1 - s_2 + s_4 - \dots}$$

[jolly parrot]

Swastik

[pearl pond]

@boreal laurel Has your question been resolved?

[boreal laurel]

there is no one that helps me

😭😭😭😭😭

[warped violet]

It's been 20 minutes. Be patient 🙂

Also, you can mention @ Helpers after 15 minutes

[boreal laurel]

<@&286206848099549185>

[dusk swift]

Hi?

[boreal laurel]

solve it

[dusk swift]

What do u need help with?

[boreal laurel]

its a question from sl loney

[dusk swift]

What genre this is?

Calcus or somethin

Kinda lazy

[boreal laurel]

trignometry and theory of equations and something

[dusk swift]

I have 3+ on that sry🙏😭😭😭

[boreal laurel]

ok np just be safe while you can

[dusk swift]

🤩

[boreal laurel]

let me put some dirt in your eyes

[boreal raptor]

idk maybe try induction

[pearl pond]

@boreal laurel Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[jolly kernel]

dipole
q = 5 uC
d = 3cm
therefore p = 1.5E-7 C m j

Field
k = 9E9
q = -10 uC
r = 1.323 cm -- from sqrt(2^2 - 1.5^2)
E = -5.1419E8 i

j x -i = k

but we just need magnitude so
|p*E| = 77.14 N m

but its saying wrong

any idea what I missed?

[ionic merlin]

The electric field due to the dipole at the perpendicular bisector is :

Kp/r^3

[jolly kernel]

It says only find the torque from the point charge right?

[jolly kernel]

but it says find the torque exerted by the -10uC charge?

[ionic merlin]

Ohk

Then use the formula I told for Electric field

And u have to find a dipole by each charge on 10uC

[pearl pond]

@jolly kernel Has your question been resolved?

[jolly kernel]

I gotta get to sleep Ill try again tomorrow

thanks for stopping by

.close

[pearl pond]

Channel closed

Closed by @jolly kernel

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wild fable]

i am a little confused, is this saying rho=0, phi=0,pi and theta=0,2pi is allowed or is not allowed here?

[lilac jackal]

technically a change of variables requires an invertible transformation with invertible derivative but the text says it still works in this case by taking limits

[wild fable]

so my i guess my question is when we integrate we are allowed to use these bounds e.g phi=0,pi and theta = 0,2pi

[lilac jackal]

yes its valid and the text tells us we can use limits to prove that its valid

[wild fable]

okay thanks

[lilac jackal]

np

[wild fable]

.solved

[pearl pond]

Channel closed

Closed by @wild fable

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fading ledge]

the cases are not clear to me

amp(-z) how did this come?

[toxic lichen]

... and what does amp(z) mean at all

[fading ledge]

i guess it is principal argument

[toxic lichen]

you GUESS?

the book never tells?

[fading ledge]

yes it tells

lol

[toxic lichen]

show

[toxic lichen]

!noai

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[toxic lichen]

ok so in fact amp(z) means arg(z) only they decided to use the rarer or older word for whatever reason.

[fading ledge]

i see

[toxic lichen]

anyway do you know what z and -z look like on an argand diagram

[fading ledge]

z=x+iy

-z=-x-yi

[toxic lichen]

that's not "looks"

geometrically what does it mean to go from point z to -z

[pearl pond]

@fading ledge Has your question been resolved?

[lilac jackal]

@fading ledge pick a random complex number and call it z. plot z and -z. whats the visual difference?

[fading ledge]

Rotation π clock or anticlockwise

@lilac jackal

[toxic lichen]

great

[fading ledge]

Am I wrong?

[toxic lichen]

that is exactly what you get by adding or subtracting pi to the argument

[lilac jackal]

in other words the difference in their args is pi

[fading ledge]

Sometimes Ann's compliment feels fear

[lilac jackal]

thats the visual intuition for this claim

[toxic lichen]

you see me say "great", not slap an ❌ react, and think you might be wrong?

[fading ledge]

So what are these cases?

[toxic lichen]

in order to remain in the principal argument range you need to not cross the negative x axis in your rotation

[fading ledge]

Here

[toxic lichen]

so these cases represent when the necessary rotation is half turn CW vs. half turn CCW

[fading ledge]

Thanks

.close

[pearl pond]

Channel closed

Closed by @fading ledge

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fading ledge]

e^itheta1=e^-itheta2

??

[dusty jungle]

whats the q

[fading ledge]

i did not understand the solution

[dusty jungle]

what part of it

[fading ledge]

three lines

le me visualise it

that is correct butt

[toxic lichen]

conjugate of re^(iθ) is re^(-iθ)

[fading ledge]

ohh true

thanks ann

.close

[pearl pond]

Channel closed

Closed by @fading ledge

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vestal pelican]

im solving an integral but im stuck on this

[heavy onyx]

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[vestal pelican]

solve the integral

[worldly jewel]

Uhm, A = B = 1/2?

It's just a system of linear equations

A+B=1 and 4A-4B=0

[vestal pelican]

why is A+B=1 and 4A-4B=0

[worldly jewel]

because you compare the coefficients on both sides?

you're doing partial fractions right

[vestal pelican]

yeah

[boreal raptor]

equate coefficients :p

[vestal pelican]

im still lost

[boreal raptor]

on the left it says x^2

on the right we have (A+B)x^2 + (4A-4B)

we can look at the lhs and see that we need an x^2 on the rhs

hence, i need to set A+B = 1 so that on the rhs i have an x^2

similarly, i can do the same thing with the 4A-4B

since there's no constant term, I can say that 4A-4B = 0

[vestal pelican]

im sorry but why

[boreal raptor]

wdym

lhs has x^2
so we equate and say we need an x^2 on the rhs

[vestal pelican]

oh we want x^2 = x^2

and lhs is x^2 + 0 which would mean that 4A - 4B=0

and A+B=1 becasue 1*x^2 is x^2

[worldly jewel]

Write the LHS as 1x^2 + 0x + 0

you want the coefficients to match in order for the equation to hold for all x, right?

this is equating coefficients you need to remember this from middle school

[vestal pelican]

i was getting confused for no reason ngl

[worldly jewel]

No worries we get confused by strange things sometimes

[novel vale]

Who helps here?

[worldly jewel]

#❓how-to-get-help

[vestal pelican]

.close

[pearl pond]

Channel closed

Closed by @vestal pelican

Use .reopen if this was a mistake.

[fierce sky]

You don't need to perform partial fractions

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fierce sky]

@vestal pelican, write it as $\frac{1}{2}\int_{ }^{ }\frac{x^{2}-4+x^{2}+4}{\left(x^{2}-4\right)\left(x^{2}+4\right)}dx$

[jolly parrot]

Roy

[fierce sky]

.close

[pearl pond]

Channel closed

Closed by @fierce sky

Use .reopen if this was a mistake.

[vestal pelican]

Where'd the 1/2 come from

@fierce sky

[fierce sky]

We forced it out so we can do that to the numerator

Or else we can't get two copies of x^2

[vestal pelican]

Can you show me a detailed way how u got there pls

[fierce sky]

it isn't so hard to arrive at that

\begin{align*}
&\frac{x^{2}}{\left(x^{2}-4\right)\left(x^{2}+4\right)} \
&=\frac{1}{2}\frac{2x^{2}}{\left(x^{2}-4\right)\left(x^{2}+4\right)} \
&=\frac{1}{2}\frac{x^{2}-4+x^{2}+4}{\left(x^{2}-4\right)\left(x^{2}+4\right)} \
&= \frac{1}{2}\left(\frac{1}{x^{2}-4}+\frac{1}{x^{2}+4}\right)
\end{align*}

[jolly parrot]

Roy

[fierce sky]

@vestal pelican

[vestal pelican]

Still confused with that 1/2

[fierce sky]

What are you exactly confused about

$\frac{1}{2}\cdot2=1$

[jolly parrot]

Roy

[vestal pelican]

Aha got it

[vestal pelican]

Now confused with the last part how did it split and how did the numerator become 1

[fierce sky]

$\frac{\left(x^{2}-4\right)+\left(x^{2}+4\right)}{\left(x^{2}-4\right)\left(x^{2}+4\right)}=\frac{x^{2}-4}{\left(x^{2}-4\right)\left(x^{2}+4\right)}+\frac{x^{2}+4}{\left(x^{2}-4\right)\left(x^{2}+4\right)}$

[jolly parrot]

Roy

[fierce sky]

Then cancel out common factors

[vestal pelican]

Ohhhh

Well this is amazing

Why would people choose that annoying ass way

[fierce sky]

as someone who hates partial fractions I will take any opportunity to avoid them

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vestal pelican]

I hate them too

[vestal pelican]

Gawd dayum

I mean they're fine when it's not a very weird integral but this one was a pain

[fierce sky]

you can do something similar to $\frac{1}{\left(x+a\right)\left(x+b\right)}$

[jolly parrot]

Roy

[fierce sky]

in this case you divide and multiply by b - a or a - b

[autumn fossil]

You can do this trick everytime you'd do partial fractions, you just need to guess how to rewrite the numerator

[fierce sky]

$$\frac{1}{b-a}\frac{b-a}{\left(x+a\right)\left(x+b\right)}$$
$$=\frac{1}{b-a}\frac{\left(x+b\right)-\left(x+a\right)}{\left(x+a\right)\left(x+b\right)}$$
$$=\frac{1}{b-a}\left(\frac{1}{x+a}-\frac{1}{x-b}\right)$$

[jolly parrot]

Roy

[fierce sky]

if you practice it a lot you'll get some intuition for what to divide

@vestal pelican if you've got no more questions, you can close this helpdesk with .close

[vestal pelican]

Ok thanks roy

.close

[pearl pond]

Channel closed

Closed by @vestal pelican

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[warm patio]

So I had an exam where I simplified the problem to the following recurrence and I dont know how I am supposed to solve this recurrence:
$ a p_{j} = b p_{j+1} + c p_{j-1}$ where $b + c = a$, but im also interested in the general case.

[worldly jewel]

,tex $ ap_j = bp_{j+1} + cp_{j-1} $?

[jolly parrot]

k

[warm patio]

yes

sorry

[worldly jewel]

have you tried to solve the characteristic equation

[warm patio]

$p_{m1} = 1$ and $p_{-m2} = 0$

idk what characteristic equation is

[jolly parrot]

k

[worldly jewel]

,tex
If you have a recurrence relation
[ x_{n} + a_1 x_{n-1} + a_2 x_{n-2} + \cdots + a_k x_{n-k} = 0 ]
Then the characteristic polynomial is
[ x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_k ]

[warm patio]

oh okay

and roots of the polynomial are like roots of the recurrence or soemthing?

[worldly jewel]

If q_1, q_2, ..., q_j are distinct roots to the characteristic polynomial P then { q_1^n, q_2^n, ..., q_j^n } is a linearly independent set of solutions to x_n

In general, for roots with multiplicities say q_1 with multiplicity m then q_1^n, nq_1^n, ..., n^(m-1) q_1^n are linearly independent solutions

[warm patio]

alright

is the proof for this hard?

imma read it first

then

[worldly jewel]

This gives you a basis for the solution space and by plugging in initial conditions x_1, x_2, ..., x_k you can determine the general form for x_n

[worldly jewel]

You can look up "linear homogeneous recurrence relations".

I think only some linear algebra is required to understand the proof

[warm patio]

okay then I can manage

[worldly jewel]

Sorry, the n's in the polynomial should be changed to k's

[warm patio]

oh yeah no worries I got the idea

[pearl pond]

@warm patio Has your question been resolved?

[warm patio]

thank you

.close

[pearl pond]

Channel closed

Closed by @warm patio

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[scenic raven]

i need help

[olive fractal]

?

[scenic raven]

how does this work

can someone tell me where i can find some good course explaining polar coordinates and its differentiation on youtube

i searched a lot

couldnt find

what i wanted

[eager jewel]

sorry what does delta represent here

p is dipole right

[scenic raven]

yes

basically

they are trying to find

force of inteaction

between the dipoles

[eager jewel]

ohh alr alr

[scenic raven]

i just always differentiate the field and multiply with the dipole

but the thing is

i wann properly

understand

the del operator

and how it works on polar coordinates

[jolly kernel]

sorry my discord lagged and showed it as open

[scenic raven]

thats alright

[eager jewel]

hmm alr alr sorry then i thought ill be able to help but havent learnt this wrt polar coordinates

[scenic raven]

mm

.close

[pearl pond]

Channel closed

Closed by @scenic raven

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[green aurora]

help to understand this site pls

[green aurora]

I don't understand what the graphs on the right represent

for example now I clicked on draw card and that green square with 6.25 written on it came out, but I don't know what it means

the cards are from 1 to 10

[toxic lichen]

you drew a 3 so the value of (x-mu)^2 is (3 - 5.5)^2 = 6.25
and since that's your only drawn card so far, this squared difference is the only one being averaged

eventually if you draw more cards you will see the green square going up and down a bit as it reflects the running average value of (x-5.5)^2

[crude pollen]

i think the intent is just to show that the two values converge eventually

but the axis has nothing to do with it

i suppose

the "value" axis is kinda weird

[green aurora]

What does it mean to do E[X^2]?

i know what E[X] is

it would be the average over many attempts

if X="number of people taking their hat" , how do you calculate the variance?

[toxic lichen]

"number of people taking their hat" is a bit unclear as a description

[green aurora]

lemme fix

[toxic lichen]

ah.

so the hats are shuffled and redistributed, and we want the variance of the number of people who get their own hat back.

there are two approaches here that i can see:

1. find the pmf of X and then the mean and then finally the variance formulaically
2. mess around with a sum of indicator variables for "the k'th person receives the k'th hat" and somehow get the variance from that (but keeping in mind these indicators won't be independent so there can be some difficulty)

(these are NOT steps, these are an either/or choice!!!)

[green aurora]

thanks

.close

[pearl pond]

Channel closed

Closed by @green aurora

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[primal bloom]

<@&286206848099549185>

[manic oriole]

Hello there

[primal bloom]

Hello\

[prisma kernel]

!15m please

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[timid lodge]

!15m

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[timid lodge]

whoops sent it twice

[prisma kernel]

it was me

[manic oriole]

1 only probably

Like a person can want to not answer the question, because of bais

Maybe

[autumn fossil]

I'd say none

[manic oriole]

Rest are maybe not related to bias

[alpine cipher]

I think none 2 only

[manic oriole]

There's probably 1 or 2 or both

[autumn fossil]

The margin of error is likely not even related to systematic biases

[manic oriole]

Refusing to answer is different though?

Not related to the system

Related to the person

Who has bias

Possibly

[timid lodge]

Is there any more context for this problem

[twin portal]

Conducted by phone or telephone?

[primal bloom]

no thats all

[primal bloom]

a phone isnt a telephone which means unrepvoive bais

[twin portal]

Most people have phones, not telephones.

[primal bloom]

but the people had telpeohnes couldnt be contacted

[timid lodge]

I think telephone aand phone are referring to the same thing here

[primal bloom]

no they arent

[manic oriole]

Probably

[timid lodge]

These all seem like biases that would need a larger margin of error to account for

[twin portal]

What kind of question is this?

[timid lodge]

oh then II perhaps

waait no

[manic oriole]

Yeah

But maybe it is low

Hm

[autumn fossil]

Does it even make sense to assign a number to correct such sources of bias? Margin of error usually only corrects sampling error

[timid lodge]

I haven't taken statistics or whatever this is yet so take what I say w/ a grain of salt

[primal bloom]

nope

[primal bloom]

I gave you the formula

[manic oriole]

Alright

[autumn fossil]

Yeah, that's why I think hte answer is c, but it's been a long time since i did stats

[timid lodge]

Yeah no I'm unknowledgable here then

[primal bloom]

it was none of them

apperntly

[autumn fossil]

If we look at this, none of the parameters are related to I, II or III

It should make sense, like do you just go "Oh, people without telephones could not be in the sample, let's just randomly throw in 10% margin of error to account for that

that'd be quite random

[pearl pond]

@primal bloom Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[jovial ember]

can some1 share with me all kinds of trigonometric formulas and advanced trigonometry like quadrants of which ratios are positive negative and maths related to it for practice as I'll also need to understand the concept of it first

[plush bramble]

This isn't a math question

Ask in #geometry-and-trigonometry or just Google any trig book

#book-recommendations probably has some

[pearl pond]

@jovial ember Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tulip cradle]

Is this right

[dense jasper]

yes

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[surreal pumice]

sec

[pearl pond]

@tulip cradle Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[graceful garden]

My professor didn't teach us how to prove this specific type of question (if...then)

I proved based on what I thought was logical

Is this correct?

[vestal surge]

it is

[graceful garden]

Wait

Really

[hasty field]

Ah, set theory. Unfortunately, I dont know much literature related to set theory.

[vestal surge]

i feel so

[hasty field]

But I will take a look at what you have

[graceful garden]

Maybe I actually am a genius

[hasty field]

Looks good to me. Though the handwriting could use some improvement

[graceful garden]

I would be lying if I said I'm working on it

[hasty field]

Set theory is really fun math.

Its the first math I actually started enjoying

[hasty field]

Lol thats okay me neither. Heres a pic

[graceful garden]

Man

Is English your first language

[hasty field]

Ive tried making it more legible but I just dont have the ability.

[hasty field]

It is!

[graceful garden]

I feel sorry for you

[graceful garden]

Same

Though I can't say anything cuz my handwriting is even worse in my main language

We should close this channel

[hasty field]

Yeah me too. Unfortunately I just dont have the manual dexterity to write well.

[graceful garden]

Thank you!

[hasty field]

Good idea!

Youre welcome.

[graceful garden]

.close

[pearl pond]

Channel closed

Closed by @graceful garden

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fathom oxide]

would these not both be correct

the only number for real roots it could have are 5 3 and 1

which are all odd

[dense jasper]

it says it contains the term 8x^5

not that the leading term is 8x^5

[fathom oxide]

ah

right

ok thank you

[dense jasper]

[fathom oxide]

.close

[pearl pond]

Channel closed

Closed by @fathom oxide

Use .reopen if this was a mistake.

[ivory basin]

This channel is currently available.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[green meteor]

How can i calculate the exact value of log using a calculator that isnt log_10

[green meteor]

Like log_3 (100)

[ivory basin]

,tex .log rules

[jolly parrot]

Xavier 🌺

[ivory basin]

The last one

[west sapphire]

oh nice

[green meteor]

Wait

The book hasnt taught me this yet

[ivory basin]

#1071226369071136979 for your perusal

[green meteor]

Huh

[ivory basin]

!original pls then

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[green meteor]

So log_3 100 is just log 100 / log 3

[ivory basin]

Yes

[green meteor]

Okok

Thanks

.close

[pearl pond]

Channel closed

Closed by @green meteor

Use .reopen if this was a mistake.

[proper nova]

that's the base change proof

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fading ledge]

How do I solve it?

Hints

[toxic lichen]

does the next page have options (3) and (4) for this question? both of the ones you gave seem impossible to conclude.

also in the book

[pearl pond]

@fading ledge Has your question been resolved?

[fading ledge]

Yeah

[toxic lichen]

ok can you show options 3 and 4

[fading ledge]

Given answer is B

2

@toxic lichen

Gone?

[toxic lichen]

hm, so they're saying that the laurent series cuts off at some negative-power term?

[toxic lichen]

"OH YES LET ME ASK IF ANN IS GONE LIKE 3 FUCKING SECONDS AFTER I PING HER AND THEN SEE HER TYPING"

that was rude and upsetting mate

[fading ledge]

I didn't see u typing actually

Sorry ma'am

I'll not ping u again

[toxic lichen]

the issue is not the ping

the issue is that you pinged me and then gave me no more than like 3 fucking seconds to even see the ping and respond to it

and then already concluded that i was gone

[fading ledge]

Ohh it is just ny type of saying

Gone means nothing much

Don't make it wrong

Okay let's back to the question

[fading ledge]

Yes

-infinity to -1

And 0 to infinity

[toxic lichen]

so "gone" does not mean "gone"

am i to understand you thusly

ok anyway

they're saying it's holomorphic in 0 < |z| < ε (with poorly typeset epsilon)

... i guess this means that there's no essential singularity at 0

[fading ledge]

Hmm yes

Essential

i see

So??

[toxic lichen]

if option 2 were incorrect then we would have nonzero coeffs on negative powers all the way down and then maybe see something resembling e^(1/z) and then i... wanna say it would not be holomorphic???

but i am honestly not 100% sure on this at all

[pearl pond]

@fading ledge Has your question been resolved?

Channel closed

Closed by @fading ledge

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[opaque void]

let A a non-empty set, how can I i prove that there's no bijection between A and pertation of A??

[burnt helm]

use a diagonal arguement

[opaque void]

what are they

[still hamlet]

partition?

[opaque void]

yes

[toxic lichen]

wait what do you mean by "pertation" or "partition" or whatever

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[opaque void]

i meant p(A)

for example: A={1,2} so p(A)={ø,{1},{2},{1.2}}

[toxic lichen]

oh ok you mean the power set.

the set of all subsets of A.

[opaque void]

exactly

[toxic lichen]

for that: have you heard of cantor's diagonal argument

[opaque void]

can you do more details, im really looking forward

[feral sedge]

A can be potentially infinite?

This seems like a pretty difficult problem to be assigned without any hints

[toxic lichen]

im not even sure how to hint at it without giving the key idea away

[feral sedge]

The trick is to assume the existence of such a bijection from A to P(A), and then use it to try to construct an element not in the image

[pearl pond]

@opaque void Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[opaque void]

@feral sedge @toxic lichen @burnt helm im so sorry for the mention, but do you got any idea?

[burnt helm]

which

where

cantor's theorem ig

you could use

[pearl pond]

@opaque void Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[green meteor]

How do i do this wth😭

[mortal seal]

Do you know how to get the number of digits using log_10 ?

[green meteor]

No

[mortal seal]

Let’s say $\log_{10} x = y$
Then $10^y = x$

[jolly parrot]

Starlord

[ivory basin]

Look at log_10 of 10, 100, 1000, etc and see the pattern

[mortal seal]

How many digits would x have depending on y’s value ?

[green meteor]

Uh

X has one more digit than y if u use powers of 10?

Wait so

Log 4^100 =100 log 4

So about 60

And then u plus 1

so is it 61?

[thorn agate]

yes

[green meteor]

Uh

How to do b

[thorn agate]

i forgot but

for a similar type of question we had used this
https://math.stackexchange.com/a/2006181

[pearl pond]

@green meteor Has your question been resolved?

[thorn agate]

ok so i searched up yh that is how it is done

[green meteor]

Wait

Is this correct

4^100=10^a+b where a is an integer and b is a fraction from 0-1

So log 4^100 =60 + 0.206….

So 4^100=10^60 + 10^0.206

And u have a and b

The questions says 10^b is less than 2 but it is also more than 1 cause b>0

[thorn agate]

yh

[green meteor]

So 1<10^b<2

Make it 4^100

By times 10^60

So 10^60<4^100<2 x 10^60

And so 4^100 between 1….. and 1.999x …

so it starts with 1?

[thorn agate]

yh

[green meteor]

Am i supposed to know the turning into 10^a+b

Cause

Ive never been taught

It

[thorn agate]

lmao

yh it seems given the values in question, you are supposed to do this way

[green meteor]

How are u supposed to learn this b4 log

Wth

It went from log inequalities to these questions

[thorn agate]

avg professor moment

you should know log if you are doing these type of questions

[green meteor]

Yea i do

But not using it in this way

[thorn agate]

ic

[green meteor]

Damn pure 2 might be the end for me😭

.close

[pearl pond]

Channel closed

Closed by @green meteor

Use .reopen if this was a mistake.

[green meteor]

Thanks

[thorn agate]

pure is wierd icl

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[robust lynx]

[white estuary]

um?

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wary tusk]

How do I do h because its not giving the same answer

As the answer book

[west sail]

if you have an attempt, please share it!

[west sail]

well, that 3x somehow became a 3 after factoring a 3 out.

also, you can divide both sides by 3 straight away to get rid of it.

[wary tusk]

Oh okk

Im still getting decimals

[west sail]

what have you corrected your work to?

[fathom sequoia]

one second

[wary tusk]

I took out the three

[fathom sequoia]

show me the revised work please 😄

[west sail]

can you show your corrected work?

[west sail]

you did not fix the first issue I pointed out.

[west sail]

this issue is still present.

[wary tusk]

Ohhhhhhhhhhhhhhhh

[frank dawn]

Yes

[fathom sequoia]

should be x instead of 3

dont know how that happened

[west sail]

I'll let the other helpers handle this then.

[frank dawn]

Did you know the slept the middle term

??

[fathom sequoia]

what

[frank dawn]

Haa

It is the Indian method to slove the quadratic equation

[wary tusk]

Its still a fraction

[fathom sequoia]

its just completing the square

[fathom sequoia]

show revised work pls

[frank dawn]

Yaa

[west sail]

please show your corrected work.

[wary tusk]

Oh wait

I divided by 3 wrong

[frank dawn]

Wrong

[wary tusk]

Yes

[fathom sequoia]

where'd 2/3 come from?

[frank dawn]

2/3 is wroung

[fathom sequoia]

imo, just disregard the 3 right after you factor it out to avoid confusion

so from 3x^2 + 3x - 6 you get x^2 + x - 2

then work your way from there

[fathom sequoia]

hm

i reccomend this

one moment please

[wary tusk]

Oki

[fathom sequoia]

in completing the square, i reccomend transposing the constant term to the other side as to avoid confusion with all other variables.

[frank dawn]

This is the real

[fathom sequoia]

that's one way to do it

but i dont think he knows this method

the colloquial way of doing it is transposition of the constant term to the right side then proceeding

[frank dawn]

Yaa this is Indian way

[wary tusk]

The question said to complete the square

[frank dawn]

Yaa i know

[fathom sequoia]

he employed factoring w/ regrouping

hold on let me actually do the problem myself

[frank dawn]

There are 4 way to slove

[fathom sequoia]

that method is inefficient imo

[crystal dew]

!nosols

[pearl pond]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[fathom sequoia]

@wary tusk

[wary tusk]

Yes

[fathom sequoia]

i see your problem now

[frank dawn]

I am right ??

[fathom sequoia]

where'd this come from?

[wary tusk]

Thats from the completing square method

[fathom sequoia]

well, it's wrong

you know how to do CTS yes?

[wary tusk]

It says to (b/2)^2

[fathom sequoia]

what's b in this equation

[wary tusk]

It became 1

[fathom sequoia]

ah no

[frank dawn]

Can I help ?

[fathom sequoia]

b = 1

so therefore

(1/2)^2

=1/4

[frank dawn]

Can I help ?

[fathom sequoia]

thats why you got 1

did you get it @wary tusk ?

[wary tusk]

Let me correct it

[fathom sequoia]

👍

show me the revised solution when you're ready

[crystal dew]

you don't need to ask for permission to help so long as you aren't tossing the full solution directly, unprompted.

but also, keep in mind that others are already helping, so try not to derail the current helper