help-36

[shrewd yew]

its funny this always happens

in the middle of me asking for help i suddenly figure out what i did wrong

[dapper hull]

lowk better than waiting for the help imo

[final saddle]

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[warm scaffold]

I need help with confidence intervals. The confidence interval calculated here is this answer: (95% confident between 0.44 and 0.64). Is this saying that 100% certainty that there is a 95% chance of the mean being within that interval from the sample mean? How can it 100% be the case that: (there is a 95% confident between 0.44 and 0.64) if the sample mean is whats used to calculate that interval (which is just and estimate of the mean) and not the actual mean (because it is uknown)

[warm scaffold]

the image is from a kahn academy video

[final saddle]

@warm scaffold Has your question been resolved?

[dense badge]

maybe a rewording will help. if you fix p, then repeating this setup with different new data generated this way like 100 times, you should expect 95 of the 100 interval produced to contain p. That’s what’s guaranteed here. Picture:

[warm scaffold]

that makes sense

but just so that I understand this clearly

the answer (there is a 95% confident between 0.44 and 0.64) is just an unbiased estimates, it is not 100% the case that (there is a 95% confident between 0.44 and 0.64), it would only be 100% true that: (there is a 95% confident between 0.44 and 0.64) if the confidence interval was calulated using p and not p hat. BUT the answer (there is a 95% confident between 0.44 and 0.64) is an unbiased estimate

[cyan kayak]

Well, technically not guaranteed. Randomness is a bit of a jerk that way.

[dense badge]

emphasis on expect

[warm scaffold]

would it be the case using the law of large numbers?

[cyan kayak]

It would be almost certain with a large enough sample size to get arbitrarily close to 95%

But yes the confidence interval is typically calculated using an unbiased estimator.

[warm scaffold]

i dont understand though, wouldnt that only be true if you used the actual standard deviation to calculate the interval and not the sample sd??

[cyan kayak]

Sample sd is unbiased as well

(provided you use the correct sample sd formula)

[warm scaffold]

so because the sample sd is ubiased (although not perfect unlike the actual sd) if you were to repeat the calculation with a new sample infinitely many times, 95% of those infinite calculations would contain the mean in its interval

[cyan kayak]

I guess you can think of this as a Bernoulli random variable. If you have multiple confidence intervals for the same underlying distribution, whether or not the confidence interval contains the mean is a Bernoulli random variable with mean p where p is the confidence interval width (0.95 in our example).

If you draw an infinite number of times from a Bernoulli random variable, you would expect to have about 0.95 of them be successes.

[warm scaffold]

why do you say about

because of dependence bias?

[cyan kayak]

But, of course, there are always pathological cases.

[warm scaffold]

aside from dependence bias, if we drawed infinity times it should be exactly 95%

[cyan kayak]

Like, it is technically possible to always have the the confidence interval not contain the mean simply by random chance

The chance that this happens, to be clear, is 0, but something happening with 0 probability is not always impossible.

So what we say is that it is almost certain that as we draw more and more samples the probability that we are within some epsilon of 0.95 approaches 1.

[warm scaffold]

your saying its not impossible to have 0 expected intervals contain the mean after infinity draws?

[cyan kayak]

Exactly, it is not impossible, just unlikely.

[warm scaffold]

bruh

[ripe jewel]

it's like how if you throw a dart at a target, if you model the needle as 1d, the chance it hits the center is 0

[warm scaffold]

hmm ok

[cyan kayak]

In fact, the chance that the dart hits any specific point is 0

[warm scaffold]

is there like a version of limits we can use for this?

[ripe jewel]

probabilities

[cyan kayak]

But of course it has to hit some point, doesn't it?

[ripe jewel]

it's not really based on limits

it's based on a concept called measures

[warm scaffold]

like it does converge to something

[cyan kayak]

Measure theory

[warm scaffold]

like i agree with your logic, but it must objectivel converge to 95% is what im saying

[cyan kayak]

Formalizing probability is something you're going to want to do after taking a course in real analysis.

[warm scaffold]

I went from being confused about the answer from the video to defending it, so I guess Im satisfied

[cyan kayak]

My quibble really is a very technical one, if you're not interested in the details it will probably be easier and almost as good to believe this.

[warm scaffold]

in some sense it must be true

[cyan kayak]

Sure

[warm scaffold]

even if i dont know how to state it formally

[cyan kayak]

It's just not a "guarantee" because there is a possible sequence of events that can cause what you stated to be untrue.

However, that sequence of events, although possible, is unlikely.

[warm scaffold]

I think i understand

[cyan kayak]

So let's say we do this experiment, and draw 95% confidence intervals and test them to see if they contain the mean. Maybe we pull the first 20

TTTTTTTTTTFTTTTTTTTF

We can see that our percentage of F actually happens to be 90%. But that's ok, that's within what we would expect by random chance.

Let's say we continue to draw from this distribution, and what you expect happens, i.e. as we draw more and more we approach 95% more and more closely.

We do this for an infinite number of samples, and find we arrive at exactly 95%, exactly as we drew up.

We have a particular sequence of draws now, what is the probability that we get this exact sequence when doing this draw?

(it's 0)

(which is the same as drawing all F, and also the same as drawing few enough Ts, such as one for every prime number, such that the ratio converges to 0 instead of 0.95)

Hopefully this explains why it's almost certain and not certain.

[final saddle]

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[warm scaffold]

thank you guys

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[wind narwhal]

abcd is a parallelogram, e and f are two points on a line parallel to ab. AE, BF and DE,CF intersect at p and q respectively. prove that pq is parallel to ad

[wind narwhal]

i drew the diagrams for this question asw

[severe verge]

show the diagram

[wind narwhal]

im planning to approach this question using similarity

[severe verge]

nice diagrams

[wind narwhal]

but im not sure about it, nor can i identify what triangles to start with, im just thinking if i should go for random triangles until i just get one

[wind narwhal]

thanks

[severe verge]

Now let's do a bit of analysis

Start with parallelogram ABCD

what properties do you get

[wind narwhal]

the opposite sides being equal, the opposite angles

[severe verge]

are the opposite sides parallel?

[wind narwhal]

yh

[severe verge]

ok

next, EF//AB

[wind narwhal]

yea i got that

[severe verge]

so obv there are 3 parallel lines

[wind narwhal]

i honestly planned to use similarity on the triangle pef, and the smaller triangle above it

[wind narwhal]

yeah

[severe verge]

now let's introduce P

so we got ∆PAB and EF//AB

what do you get here?

[wind narwhal]

can you tell me which diagram you're using

[severe verge]

left one for ease

[wind narwhal]

oh pe/ae = pf/bf

[severe verge]

not only that

you have learnt similar triangles right?

[wind narwhal]

= ef/ab

[wind narwhal]

yeah

[severe verge]

no

PE/AE is not equal to EF/AB

[wind narwhal]

ef is parallel to ab right

[severe verge]

yes

[wind narwhal]

oh yh alr got it

[severe verge]

Now answer my question again

[wind narwhal]

i get the whole thing asw pe/pa = pf/pb

[severe verge]

no

[wind narwhal]

the angles pab and pef being equal asw

[severe verge]

you're missing sth

it's crucial to solving this

[wind narwhal]

smth about similar triangles ?

[severe verge]

this is equal to one more ratio

[wind narwhal]

we're considering the pef triangle and pab triangle

[severe verge]

yes

I agree that they're similar

but you're missing one ratio that is crucial to solving this

PE/PA = PF/PB = ?

[wind narwhal]

do you mean the heights of the triangles being in the same ratio

[severe verge]

no

[wind narwhal]

oh the whole thing i think should be equal to ef/ab

[wind narwhal]

this should be equal to ef/ab ?

[severe verge]

correct

PE/PA = PF/PB = EF/AB

now do the same thing with ∆QCD

[wind narwhal]

oh yeah didnt i initially say that asw ?

oh wait no i considered that equal to pe/ae myf

[wind narwhal]

alr got it

QF/QC = QE/QD = EF/CD

[severe verge]

nice

now notice one crucial thing about EF/AB and EF/CD

[wind narwhal]

ab and cd being equal

[severe verge]

yes

[wind narwhal]

i can get a lot of equalities from that

[severe verge]

true

All 6 ratios are equal

[wind narwhal]

yeah

[severe verge]

now let's get to the important part

how do we prove PQ//AD?

one clear problem we can see is that we don't know anything about the angular properties of PQ

[wind narwhal]

thats what i couldnt figure out, i was planning to somehow arrive at a few equalities involving ratios for which i can use the converse of thales

[severe verge]

nor do we know its length properties

[wind narwhal]

fair

yh true

[severe verge]

so what do we need to prove here?

[wind narwhal]

uhhhh something involving the quadrilateral peqf ?

[severe verge]

nope

do you know the converse of the intercept theorem

or whatever it is called

[wind narwhal]

like

if the ratios are equal, then the lines are parallel ?

[severe verge]

correct

[wind narwhal]

yeah thats the converse of thales thats what im trying to use, but i cant figure out where

[severe verge]

now I should mention

[wind narwhal]

maybe construct a line to get the required triangles with those lines ?

[severe verge]

a common misconception is proving PQ/AD = ED/EQ leads to PQ//AD

[wind narwhal]

yeah no thats not gonna work

oh i see it now

[severe verge]

the problem is there exists a segment not parallel to AD yet still satisfies the equal ratio property

[wind narwhal]

so i gotta somehow prove the ratio of the sides of triangles aed and peq to be equal

[severe verge]

yep

[wind narwhal]

okay

[severe verge]

what ratio should you prove to be equal here?

[wind narwhal]

i think i can arrive at ed/eq = ae/pe from those previous equalities we got

[severe verge]

ayyyy you got it

[wind narwhal]

if we consider all of them equal and tinker with them a bit

[severe verge]

so now the problem is

[wind narwhal]

the last remaining side

[severe verge]

how do we get from PE/PA = QE/QD to PE/AE = QE/DE?

try to tinker with the ratios a little

you're almost there

[wind narwhal]

i think you can just recirprocal it and

subtract 1

that should work

[severe verge]

nice you got it

and that finishes the problem

[wind narwhal]

cold so we get similar triangles

and so we can say

the last remaining sides are parallel asw

[severe verge]

[wind narwhal]

thanks man, i couldnt find the appropriate triangles

[severe verge]

no problem

!done

[final saddle]

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

[wind narwhal]

i was initially thinking of the right triangle to pick but i couldnt figure out the aed and peq triangles

.close

[final saddle]

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[tight glade]

can you help pls🙏🏻🙏🏻

[tired walrus]

!msgdel

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[terse crypt]

bruh

[wind narwhal]

dont you just have to multiply

do you have the answer ? is it 972 ?

[final saddle]

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[tight glade]

it is 1458

[wind narwhal]

that's weird

[terse crypt]

how did u get that

[wind narwhal]

oh i made a small mistake

wait ill get back to you

[wind narwhal]

yeah the answer's 1458

do you want an explanation

[tight glade]

yes plss

[terse crypt]

Bruh

[wind narwhal]

okay so for f(2) if you put 2 instead of x you get ab^2/n = 6

[severe verge]

[wind narwhal]

and for f(5) if you put 5 instead of x you get ab^5/n

to find f(7) you gotta find possible values of a and b^7/n

lets first work on the b term

if you divide f(5) by f(2) you get ab^5/n over ab^2/n

if you solve this you get b^3/n = 162/6

which is 27

so b^3/n = 27

raise it to the 1/3rd exponent, which will get you b^1/n = 3

and b = 3^n

[tight glade]

oh I think I got it, thanks a lot 🙏🏻🙏🏻

[wind narwhal]

yeah js put 1 instead of n to find a possible value of b, and then find the value of a according to that, and then substitute all those for f(7)

which will give you 2 times 3^6

i think

[wind narwhal]

wut

[severe verge]

nothin

[wind narwhal]

[serene panther]

No?

B is 3

[wind narwhal]

yh ik it isnt the answer i made a stupid mistake kinda

[wind narwhal]

yeah i got it later

[serene panther]

Oh ok

[wind narwhal]

oh i should close the ticket

!done

[final saddle]

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

[wind narwhal]

@tight glade

[tight glade]

.close

[final saddle]

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[final saddle]

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[north frigate]

my answers seems too wrong idk, am i correct

[north frigate]

this is the graph clearer

[terse crypt]

Cuz they are

[north frigate]

oh

[terse crypt]

Well not all of them

[shrewd meadow]

you got 4 of 7 incorrect

[tired oak]

hi guys

im from portugal and i have a questions someone help?

[dusty quarry]

!occupied

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[shrewd meadow]

please read #❓how-to-get-help, @tired oak

[dusty quarry]

please use another channel to get help!

but welcome!

[shrewd meadow]

welcome to mathcord, @dusty quarry! would ya like to help our current helpee?

[tired walrus]

unless i missed something?

[dusty quarry]

wot

[tired oak]

thx

[shrewd meadow]

i would like to amend my count to "4 of 7 correct"
i miscounted and agree with @tired walrus's check

[north frigate]

ohh

[shrewd meadow]

there's a helpee here lol
i was memeing, but you may help him

[north frigate]

wait so my number 3 is not dne?

how so>

[dusty quarry]

yes i know there is, hence why i told the other person to use another channel

[north frigate]

theres no shaded lines in the y axis of -3

[dusty quarry]

f(-3) has no white circle

its value is defined there

[shrewd meadow]

i am talking about the helpee in this channel
thx for helping him

[north frigate]

isn't f(-3) this one?

[dusty quarry]

yeah

[shrewd meadow]

that's f(-3), yea

[dusty quarry]

eh

are you perhaps looking at y = -3

that is f(-2)

[north frigate]

nono, x=-3

[dusty quarry]

x = -3 is a piece of the graph

it exists

[shrewd meadow]

do you see it now?

[north frigate]

OHHHH

so even without

the circles

it counts?

[dusty quarry]

ok wait

[shrewd meadow]

it always counts 😭

[dusty quarry]

do you know what those circles mean?

[north frigate]

not really

[dusty quarry]

those circles indicate that the value at that point is not the value at that circle

aka an exclusion

[north frigate]

ohh, all i know is shaded circles are included

[dusty quarry]

but if there's just a normal line there

[shrewd meadow]

here, they are hollow/empty circles

they are NOT shaded circles

[dusty quarry]

then of course the lines are the values

like a normal graph

[shrewd meadow]

a line going through is equal to a shaded circle

in fact, those "circles" are points

[shrewd meadow]

you're all clear!

[north frigate]

thankss

i get it now

!close

.close

[final saddle]

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[terse crypt]

Oh my gyat

If they have common roots

Then the linear factors of the first equation

Must be in the second equation

I wonder what the linear factors of the first equation expand into

[untold sail]

Ye like you would try to eliminate x ^2 here and get a linear but we have a cubic now

[terse crypt]

.

Hint

[untold sail]

Can you elaborate

[terse crypt]

Say the product form of the third equation is

[ (x - \alpha)(x - \beta)(x - \gamma) = 0]

[soft zealot]

k

[terse crypt]

There is no coefficient in front of the product

Cuz the coefficient in front of x is 1

Now

[untold sail]

Ye

[terse crypt]

We know that ${(x-\alpha)(x-\beta) = 0}$ is the solution to the first equation as well

[soft zealot]

k

[terse crypt]

Therefore

The expansion of this should be the same as ${x^2 - 5x + 5}$, no?

[soft zealot]

k

[untold sail]

But aren't you assuming another root common here

[terse crypt]

This.

[terse crypt]

How so?

[untold sail]

Alpha and beta

[terse crypt]

Oh Shi

I read it as ‘common roots’ sorry

[untold sail]

It's alr

[terse crypt]

I think i have an idea

Using quadratic formula, we can find that

[ x = \frac{5 \pm \sqrt{5}}{2}]

Find rational numbers are closed under addition

[soft zealot]

k

[terse crypt]

$\frac{5 \pm \sqrt{5}}{2} + \beta + \gamma = -a\in \mathbb{Q}$ and $\left(\frac{5 \pm \sqrt{5}}{2}\right)\beta\gamma = -5$

Also

[ \left(\frac{5 \pm \sqrt{5}}{2}\alpha\right) + \frac{5 \pm \sqrt{5}}{2}\gamma\right) + \alpha\beta = b \in \mathbb{Q}]

[soft zealot]

k
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[terse crypt]

Sarin, help me plz

[tribal jackal]

Since one common root is irrational, and irrational roots occur in conjugate pairs, both roots will be common

[terse crypt]

But what’s gamma

That’s beta

Oh Shi

U can js use this

[terse crypt]

Thx

[untold sail]

Ye in this case it should work as the a,b are rational

It worked for cubic too tho?

[tribal jackal]

We can use this along with $\alpha \beta \gamma = -d/a$ to find gamma

[soft zealot]

Sarin

k

[tribal jackal]

Then use the other 2 relations to get a and b

[untold sail]

Yo @tribal jackal this works for cubic too?

You know that irrational roots occurs in conjugate pairs

[tribal jackal]

It should...

[untold sail]

Ic

[tribal jackal]

Since you can factor a cubic as a linear factor and a quadratic

[terse crypt]

Ye it works

[untold sail]

Alr

[tribal jackal]

As long as a and b turn out to be rational, this should work. If they are irrational, we have a problem

[untold sail]

Ye i never knew that could be extended

[terse crypt]

Both are rational

[tribal jackal]

Oh its given in the question

[terse crypt]

It’s not 😔

It’s ‘root’ not ‘roots’

But u proved its actually true anyway. Thank u for u

(At least in this case - i was running through possible ways to make the thing rational)

[tribal jackal]

But they didn't specify if it was exactly one root. They just said one root was common

[terse crypt]

Why not ‘at least one root’ ;-;

[tribal jackal]

they want to trick you

by being vague

[terse crypt]

Bruh

[final saddle]

@untold sail Has your question been resolved?

[final saddle]

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[tranquil lake]

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

Is my strategy valid? Like would it work in other cases?

I know I got the right answer, but my method was not the same

They wanted me to see that it was conservative

First I need to prove the del phi is conservative

$\del \phi = (2x,-2y)$

[soft zealot]

smeagol

[tranquil lake]

we can call this vector field f

since we know that it equals the gradient of phi

phi can be the g scalar field

then we can plug in the start and end points

which would be 0-0

So I get what the answer key says

but did my strategy also work or is it just coincidence?

[tranquil lake]

my method ^^

did I make any mistakes in my method

[final saddle]

@tranquil lake Has your question been resolved?

[tranquil lake]

<@&286206848099549185> is this a correct method?

[limpid crown]

hmm

how did you get these amswers

[tranquil lake]

I broke the curve into 4 parts

each with t 0 to 1

then I found the gradient

and plugged each little c in

and took the dot product with the gradient of each little c

$C_{1}=\int_{0}^{1}\left(0,-2t\right)\cdot\left(0,1\right)dt$

[soft zealot]

smeagol

[tranquil lake]

for example

and added each of the big Cs

[limpid crown]

and how did you do that

[tranquil lake]

gradient is $( \frac{\partial}{\partial x}(x^2-y^2) , \frac{\partial}{\partial y} (x^2 - y^2) ) = (2x, -2y)$

[soft zealot]

smeagol

[limpid crown]

ah ok nvm i see it now

yup

yess yess

everything makes perfect sense to me

[tranquil lake]

so my method isn't wrong

its just too many steps?

[limpid crown]

yess its correvt

[limpid crown]

i wouldnt say so

[tranquil lake]

cause I should've realized its conservative

[limpid crown]

its ok either way

good job

[tranquil lake]

thank you

.close

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[blazing linden]

got some math test revision uh its about patterns and stuff im stuck with the part where it asks about rules please help

[blazing linden]

js ping me if u decide to help me !

[faint fern]

Given a number of posts, how many rails will you need?

that's what it's asking for

[blazing linden]

which sheet r u reffering to?

[faint fern]

oh sorry, the black one

[blazing linden]

it goes up in threes

so uh

every 2 posts there are you will need three rails

I MEAN 1 POST

im so cooked

🤑

[faint fern]

and how do you account for the case of 1 post?

1 post - 0 rails

[blazing linden]

yes

and 2 posts equals 3 rails

then 3 posts 6 and so on

[faint fern]

so for every post you add, you gain 3 rails (starting with 1 post)
if P is the number of posts, you must have added (P - 1) posts to your initial post

so from that, how many rails does P posts give us?

[blazing linden]

im sorry what

[faint fern]

You start with 1 post, 0 rails
To have say 5 posts in total, you must add 4 posts right?
For each new post, you need 3 rails

so for 5 posts, you need 4 x 3 = 12 rails

[blazing linden]

ohh i get it

cause every post needs a second one to go with it

which would mean it needs to be a even number of posts

am i right?

[faint fern]

no

you can have 3 posts

you add 2 new posts to the initial one, that requires 2 x 3 = 6 rails

[blazing linden]

okay..

and then

[faint fern]

do you see the calculation I'm doing to find the number of rails?

[blazing linden]

ur using timestables

so what im getting is
you start with 3 posts
you add another two posts
making it 5 posts

and then to connect them all together

i didnt understand from there

brb cause i need to get ready

[faint fern]

Idk how to make it much clearer. I'm just taking (P - 1) which is the number of extra posts we need to add
1 post + (P - 1) posts = P posts
and multiplying it by 3

3(P-1)

[final saddle]

@blazing linden Has your question been resolved?

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[strong flax]

help

[final saddle]

Please don't occupy multiple help channels.

[strong flax]

can someone explain why

its a negative

-98/16

i thought a negative number raised to the power of 2 will result in a positive number

negative x negative = positive no?

[dapper hull]

+2(49/16) - 2(49/16)

[atomic moon]

This

https://discord.com/channels/268882317391429632/908076667224854568

Close your previous channel before opening a new one

[strong flax]

.close

thjx

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[golden hollow]

help

[golden hollow]

i have no idea how to solve this using quotient rule

i get to here (if its even correct) and have no clue where to go next

[bleak granite]

it appears you applied quotient rule correctly, beyond that it's just some algebra to simplify...

depending on your instructor they might or might not have you not leave it in that form but it varies

[golden hollow]

🥀 no instructor but the book simplifies it to and i have no clue how

[whole halo]

dont forget parentheses

you have $\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}$

[soft zealot]

mtt

[whole halo]

since $(\sqrt x)'=\frac1{2\sqrt x}$ and $(5x+8)'=5$,

[soft zealot]

mtt

[whole halo]

,,\left(\frac{\sqrt x}{5x+8}\right)'=\frac{\frac1{2\sqrt x}(5x+8)-\sqrt x\cdot 5}{(5x+8)^2}

[soft zealot]

mtt

[whole halo]

so other than forgetting the parentheses, this is about as correct as you can get

now you need to simplify it

[whole halo]

you can see here in the answer theres a 2 sqrt(x) in the denominator
so you can begin with multiplying the fraction by 2 sqrt(x) / (2 sqrt(x)) and see where that goes

[golden hollow]

i fear i dont quite understand

i dont get how id apply that

[whole halo]

,,\frac{\frac1{2\sqrt x}(5x+8)-\sqrt x\cdot 5}{(5x+8)^2}\cdot\frac{2\sqrt x}{2\sqrt x}

[soft zealot]

mtt

[whole halo]

Im multiplying by something over itself, which would simplify to 1 and so would not affect the answer

now go multiply this across and see if you can get closer to the book answer that way

[golden hollow]

i see i get to this

so the bottom part of the fraction is like the book answer

but i dont understand how the top answer turns into that

[final saddle]

@golden hollow Has your question been resolved?

[scenic notch]

You forgot to make the whole thing negative

You differentiated -g(x) instead of g(x)

[golden hollow]

youre right

i still dont understand how it simplifies into that even if it were negative

[scenic notch]

10x-(5x+8)

[golden hollow]

OH

i forgot to multiply the 5 sqrtx by 2sqrtx

🫠

i see it now

🙏 thank you all for helping

[final saddle]

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[waxen bear]

did i mess up somewhere? the answer isnt 0...

[atomic moon]

Suspicious factorisation

[waxen bear]

mh?

with the 1/64?

[atomic moon]

Yes

[waxen bear]

damn... alway up there...

[atomic moon]

25 Should have been divided by 64 when doing this

[waxen bear]

wait why?

ohhhh

nvm

i got it

[atomic moon]

The rest is coherant wrt to the mistake

[waxen bear]

?

[atomic moon]

Nothing wrong except the final result due to that mistake

In the reasonning i mean

[waxen bear]

ohhh gotcha

how does that change the arctan?
would it be arctan (x/5/8) || arctan (8x/5)

[delicate sorrel]

Can you send the integral

[waxen bear]

$\lim_{a\to\infty} \frac{1}{64} tan^{-1}(\frac{8x}{5})|_{5/8}^{e^b}$

that didnt work as i wanted

how do you make that line?

after an improper integral

nvm got it

[soft zealot]

YEET-MA-KNEES

[delicate sorrel]

We didnt study that in high school

[waxen bear]

im in college

[delicate sorrel]

Ik

[waxen bear]

?

[delicate sorrel]

I know

[waxen bear]

like i said

"?"

[delicate sorrel]

I tought we study same thing as college

In math atleast

[waxen bear]

no????

[delicate sorrel]

Nvm

[waxen bear]

there wouldnt be a reason for maths in college then

[delicate sorrel]

Yeah true

[waxen bear]

.

[delicate sorrel]

I dont have any idea

[waxen bear]

well this, this is what it should be fixed to right?

[waxen bear]

i wasnt really asking you.. ToT

[waxen bear]

.

[delicate sorrel]

I know im still not in college

Wait until i go to college and answer u

[waxen bear]

...

[delicate sorrel]

Dw i got u

It will be fast just couple of years

[waxen bear]

literally how

i genuinely dont know how it was 160

[fair ginkgo]

what is the original question?

[waxen bear]

$\int_{ln\frac{5}{8}}^{\infty} \frac{e^x}{64e^{2x}+25}dx$

[soft zealot]

YEET-MA-KNEES

[waxen bear]

original quesiton

[fair ginkgo]

in the second line when you group 64, you leave out the 64 from 25

[waxen bear]

thats already taken care of

[fair ginkgo]

should be e^2x + (5/8)^2

[waxen bear]

.

[waxen bear]

.

hello?

not you

[delicate sorrel]

Did u try to set y(x)= e^x8 ????

8 not in the exponent

[fair ginkgo]

I'll try to see if I can find where you messed up

[waxen bear]

i didnt pull out the 5 (25)

could that be why?

[fair ginkgo]

yeah you have to factor the term without x to get to a tan form, yes

you factored the 64, but you should've really factored the 25

[waxen bear]

why is that?

wasnt it $64e^{2x}+25$

[soft zealot]

YEET-MA-KNEES

[fair ginkgo]

exactly, factoring the 64 does nothing for you

you need to get to someting like 1/(alpha^2 + 1)

so to get +1 in the denominator you'd factor the 25 and not the 64

[waxen bear]

so then, $u = 8e^x$?

[soft zealot]

YEET-MA-KNEES

[fair ginkgo]

that doesn't really matter until later, I would reccomend to do u=e^x as the first substitution and later sub again with another variable

but the best sub would be 8/5 e^x

[waxen bear]

ah damn a nested sub

[waxen bear]

huh?

righttt you took out the 25

[fair ginkgo]

dont worry about it, just sub twice and you'll get there

if you can u sub once you can easily do it twice lol

[waxen bear]

true lol

[delicate sorrel]

What about this?

Oshi forgot an 8

1/40

[fair ginkgo]

yeah and also e^x should be in the numerator

[delicate sorrel]

Why?

U sure?

[fair ginkgo]

yes

[delicate sorrel]

In the final result ? Or somewhere else

[fair ginkgo]

you factor in the deonimator also the e^x, leaving you with 1+(5/(8e^x))

[delicate sorrel]

Thats what i exactly wrote wdym

[fair ginkgo]

but when you create its derivative you put 5/(8e^x)

which is not (5/(8e^x))'

so you really shouldn't also factor e^x or you could also multiply by the derivative correctly

[delicate sorrel]

I think its correct

The derivative of 5/(8ex) is - 5/(8ex) isnt it?

I multiplied by 5 and divide by it and then the - too

[fair ginkgo]

no e^x is in the denominator

[delicate sorrel]

And i already have the 8

[fair ginkgo]

derivative(1/e^x)

is not e^x

[delicate sorrel]

-1/ex

Bro hold up

Do u agree?

Same shi as this

[final saddle]

@waxen bear Has your question been resolved?

[waxen bear]

how would i solve the arctan thats left?

since arctan x is different than that apparently

[fair ginkgo]

yeah you're right, your approach also works

[delicate sorrel]

Man good news it turn out i didnt need to go to college to help u

U dont need to wait

🙂

:)))

[waxen bear]

you're really helping my inferiority complex : D

[delicate sorrel]

What does suppsed to mean

That*

Oh

Sory man didnt meant to do that

But i think ur jk

[waxen bear]

.

the question still stands

[delicate sorrel]

Does x go to infinity?

Just replace the arctan expression with x with pi/2 and youre done

[waxen bear]

so this isnt true then????

[delicate sorrel]

X and the expression inside the arctan behave both the same way in infinity so yeah its not true

[waxen bear]

THEN WHY IS 1 SO DIFFERENT

[delicate sorrel]

Huh?

[waxen bear]

arctan (1) is fucking pi/4

[delicate sorrel]

Because the 1 isnt infinity ig?

[waxen bear]

so isnt 73/24

[delicate sorrel]

No

I never talked about that one

Bro

U aplly the limite only in the x expression

Leave the other arctan alone

[waxen bear]

i wish to say some socially acceptable words

[delicate sorrel]

By the way i didnt had the same result as you :))

[waxen bear]

that either means 3 things

[fair ginkgo]

you factored 3 when you really needed to factor 9

also I don't know what kind of multiplication got you 14/24 on line 4

[waxen bear]

indont remember typing a 14 anywhere

[fair ginkgo]

is that a 19?

17

[waxen bear]

can you circle which you are talking about?

[waxen bear]

yea no i dont see it

wait you mean the tiny horn at the top of the 17??

[delicate sorrel]

Yo thinder why he didnt had the same result as me

[waxen bear]

yea those 17 brodie

[fair ginkgo]

so the first thing is grouping 3 instead of 9

and how did you get a prime number on the top

you have 1/3 * 3/8 = 1/8

but since you grouped 3 and not 9, you also should have 1/3 * 1/8 = 1/24

[delicate sorrel]

Can i ask you why 1/3 isnt 25=5^2

[fair ginkgo]

this is a different problem @delicate sorrel

[delicate sorrel]

Oooooh thats why

[fair ginkgo]

look at the very top

[waxen bear]

?

[fair ginkgo]

are you trying to solve the integral with 64e^(2x) + 25 or 64e^(2x) + 9 in the denominator?

[waxen bear]

i, somconfused

[delicate sorrel]

Bro rilax pls

Take life easy

[waxen bear]

where did the 25 come prom

[fair ginkgo]

is this exercise number 4?

[waxen bear]

you mesn this?.?
since ingot the question wrong it gabe me another question
it jus so happen to have 64 again

[fair ginkgo]

ok, so first thing first group 9, from the denominator

do you follow?

[delicate sorrel]

Yes sir!!

[waxen bear]

where did I comefrom

oh thats a 1 nvm

[fair ginkgo]

yes that's a 1/9

we group 1/9 to get (something) + 1 in the denominator

we set t=8/3 e^x

t is the same as u sub, but with t

[waxen bear]

back to this now

cant solve the second arctan

[fair ginkgo]

bro it should not be 3/8

but 1/24

and the second arctan should be of (1) = pi/4

[waxen bear]

fuckkkkkkkkkkkk i forgot the 1/9

[waxen bear]

but how would the second one turn into 1?

[fair ginkgo]

x=ln(3/8) gets you 1/24 arctan(1) = pi/4 * 1/24

[waxen bear]

but shouldnt you change the limits since we did usub?

[fair ginkgo]

yes but first correct this

and send your work again

[waxen bear]

well i just changed the last line since it only matters then

[fair ginkgo]

ok that's fine but you also have to correct your arctan

how did you get 13 inside?

or did you sub already?

[waxen bear]

there are no 13

[fair ginkgo]

what did you write inside the arctan

in the last line

[waxen bear]

73/24

[waxen bear]

this one right?

[fair ginkgo]

how did you get 73

yeah

[waxen bear]

8/3 * 3/8

[fair ginkgo]

that's exactly 1

24/24

[waxen bear]

i seemed to have mathed wrong

OH FUCK

CROSS MULTIPLY IS FOR ADDITION

IM A FUCKING IDIOT

[fair ginkgo]

that arctan at +inf is pi/2

[waxen bear]

finalllyyyyy

this fuckass question took me,with my fuckass brain, 5.75 hours to finish.

.close

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[digital turtle]

Hey does a non filled in circle mean it isn't a spot on the graph

[digital turtle]

Like for example f(4) doesn't equal 3

[dense coral]

indeed, f(4) is undefined

[digital turtle]

Alright thanks

this is actually summer AP calc homework but I forgot everything

😥

[robust mulch]

i like to think of it as placing a piece of spaghetti o the line and if the spaghetti can go through the hole, its undefined

[craggy plank]

Hi, yes

[digital turtle]

okayyy thanks ❤️

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[opaque ember]

https://i.gyazo.com/9e238b1f3daa9b7d581dda2bf8ff6d4b.png

[opaque ember]

@wary current

a typical requirement on solutions is regularity aka "nice" behavior

in this case it means no blow ups at r=0

$\lim_{r\to0}u(r,t)$ exists and is finite

[soft zealot]

ロケット・ジャンプ

[final saddle]

@opaque ember Has your question been resolved?

[opaque ember]

no

[wary current]

intuition tells me v(0,t) = 0

[opaque ember]

i gave a limit for u, how does it become a limit for v?

[wary current]

I'm sorry I'm just really lost :') Here is what I thought: If limit exists, u(0,t) = 0. v(0,t) = 0*u(0,t) = 0

[opaque ember]

there should be limits not plugging r=0

[wary current]

I genuinely have no clue what you are even speaking of :(

I will return to my textbook and do some more reading

Thank you through I'll revisit if I have questions

[opaque ember]

write $\lim_{r\to0}$ instead of $u(0,t)$

[soft zealot]

ロケット・ジャンプ

[opaque ember]

i started by saying $\lim_{r\to0}u(r,t)$ exists and is finite

[soft zealot]

ロケット・ジャンプ

[opaque ember]

$\lim_{r\to0}v(r,t)=~?$

[soft zealot]

ロケット・ジャンプ

[final saddle]

@opaque ember Has your question been resolved?

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[round cedar]

hey, is this a good way to show that the limit for XCOS(1/X) exists?

[round cedar]

when x=1/t

[tired walrus]

lim[t->infty] cos(t) does not exist

so no this is not good

the substitution by itself is OK but what you do with it afterwards is a no

[round cedar]

but cost is between -1 to 1

[lofty sinew]

Hmmm... Breaking the limit into both the top and bottom of the fraction might not be a good idea, yea

[tired walrus]

ok then apply squeeze theorem.

don't refer to nonexistent things!

[lofty sinew]

Talking about how cost is bounded by 1 and dividing by an ever increasing value goes to 0 is solid though

You might wanna also include t -> -infinity, since going to positive infinity is the same as x going to 0 from the positive side

[round cedar]

btw i mean showing theres a limit for x-->0 but sry idont mean that x=0 is continuous for this function

so i just need to show that the limit exists

for x--->0

[lofty sinew]

You don't need to show that the limit is the same regardless of the direction?

[tired walrus]

i think i have made myself clear but i can repeat myself if anyone wants to

[round cedar]

yep

[round cedar]

thats a good idea

so starting with cosx between -1 and 1 is good?

[lofty sinew]

Yeah

[tired walrus]

yes. that's how you put your intuition of cos being bounded into something rigorous

[final saddle]

@round cedar Has your question been resolved?

[round cedar]

@tired walrus @lofty sinew do i need to show it for both x--->+-0 ?

[pliant parcel]

!noping

[final saddle]

Please do not ping individual helpers unprompted.

[round cedar]

?

[tired walrus]

n/a here

[tired walrus]

yes you do

[round cedar]

ok and then if for both t--->+-∞ its ok then theres a limit for x--->0?

[tired walrus]

yes

[round cedar]

ty !

.close

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[round cedar]

hey i need to prove limit for x--->0 so i did x=1/t with the squeeze theorem but im having hard time cuz t-->+∞ for +-1/t is +-0..

[tired walrus]

±0 is 0...

[dapper hull]

plus minus 0 finally

[round cedar]

k thx

.close

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[round cedar]

i need to show that g is continuous for every x. i proved for x=0 now i need for ≠ 0 . i know that theres a limit for x-->a for 1/x and that polynominals are continuous. is that ok?

[marsh temple]

i know that theres a limit for x-->a for 1/x
I'm not super thrilled with this wording

but 0 is the only spot you have to concern yourself with

[round cedar]

0 i did

earlier

with squeeze theorem

for x--->0 ^

[final saddle]

@round cedar Has your question been resolved?

[terse crypt]

https://www.dpmms.cam.ac.uk/~wtg10/easyanalysis4.html

[lofty sinew]

https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Yet_Another_Calculus_Text__A_Short_Introduction_with_Infinitesimals_(Sloughter)/01%3A_Derivatives/1.05%3A_Properties_of_Continuous_Functions

[round cedar]

k

.close

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[jade fable]

<x1>=1,2,3,4
<x2>=1/2, 1, 3/2, 2.

[jade fable]

Let's say the standard deviation of <x1> is d

[fiery bluff]

These are vectors?

[jade fable]

what is the standard deviation of <x2>

[fiery bluff]

Hmm... these are probability distributions?

[jade fable]

just some data set

[severe verge]

wonder why they gave d?

[fiery bluff]

And x2 is from scaling down x1, right?

[jade fable]

because it is not important to calculate the exact value of d?

[jade fable]

yes

[fiery bluff]

Do you remember the expression for standard deviation

[jade fable]

I have to reopen it

.close

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[fiery bluff]

From there you can figure how it changes when data changes

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[golden hollow]

i need to find the tangent? (not sure if correct translation) where it goes through the X-axis in a place thats not (0,0) with the function f(x)=x^2+2x-3 and ive got no clue how to go about it

[terse crypt]

do u know

how to find the tangent line

https://tenor.com/view/cat-gif-12450451651611168773

[golden hollow]

i know when given the value of X but without it im lost

[terse crypt]

do u know how to find derivative

[golden hollow]

f'(x) ? yeah

f'(x) = 2x-2 in this case

[terse crypt]

2x+2

[golden hollow]

oh yeah oops

[terse crypt]

ok

let our point of interest be (a,f(a))

so the tangent eq would be

[ y = (2a + 2)(x-a) + f(a)]

right?

[soft zealot]

k

[golden hollow]

yeah

[terse crypt]

good

for this thing

to pass through the x-axis thats not (0,0)

whats the condition

[golden hollow]

(2a+2)(x-a)+f(a)=0 while x=/=0

[terse crypt]

in other words (x,y) = (0,0) mustnt be a solution to this

since y is linear, it will always cross the x-axis anyway

so

$0 = (2a + 2)(-a) + a^2 + 2a - 3$ should have no solution

[soft zealot]

k

[terse crypt]

@golden hollow

[golden hollow]

huh now im lost again

[terse crypt]

(0,0) mustnt be a solution to this

so if we plug x=0, y=0 in

we shouldnt get an answer, right?

else

if we get an answer, the thing crosses (0,0)

[golden hollow]

oh okay

[terse crypt]

can u find when this has no solution

[golden hollow]

in almost all cases it has no solution no?

[terse crypt]

hmm lets see

this simplifies to

$0 = -a^2 - 3$. So the discriminant is ${0^2 - 4(-1)(-3) = -12}$

which is negative...

[soft zealot]

k

[terse crypt]

lemme take a step back

and think for a bit

[golden hollow]

👍

[terse crypt]

oh

it never crosses (0,0)

which agrees with our result

i thought the result was wrong lol

@golden hollow it can never cross (0,0)

as our working has suggested lol

deductive reasoning is quite smth

[golden hollow]

wait holdon this made me realize i looked at the wrong value for this one

in said point X=1

[terse crypt]

oh

[golden hollow]

wait holdon let me read over this again im sorry

[terse crypt]

so (1,0)?

[terse crypt]

alrighty

[golden hollow]

😔 trying to translate while not understanding the maths made me mess up

alright so earlier i had to find line K in point A (1,0) which was Y=4x-4.
the next question goes like this "the graph of F, not only cuts through the X-axis in point A (1,0) but also point C(?,0) . The line M touches f(x)=x^2+2x-3 in point C(?,0) . make the function for line M

[terse crypt]

im having trouble understanding that

could u put ur question through google translate or smth ;-;

[golden hollow]

ill do that

The graph of f intersects the x-axis not only at point a but also at point c. Line m touches the graph of f at point c. Write down the equation of m.

point a in this case being (1,0)

[terse crypt]

any more information abt f?

[golden hollow]

f(x)=x^2 + 2x - 3

so f'(x) = 2x + 2

[terse crypt]

so f = x^2 + 2x -3

?

[golden hollow]

yep

[terse crypt]

alrighty

and the tangent line of this mustnt pass (1,0)?

[golden hollow]

correct

[terse crypt]

ok

[ y = (2a + 2)(x-a) + f(a)]

[soft zealot]

k

[terse crypt]

plug (x,y) = (1,0) in

and make it not have real solutions

[golden hollow]

im ending up on 0=-30

[terse crypt]

,w simplify (2a+2)(1-a) + a^2 + 2a - 3

[soft zealot]

Results provided by WolframAlpha

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Requested by nuclearinteraction

[terse crypt]

,w 4 - (4)(-1)(-1)

hm

wait

[ -(a-1)^2 \implies a = 1]

[soft zealot]

k

[terse crypt]

a = 1 is the only soln

which agrees with our result

so other than a = 1, the tangent line will not cross the x-aixs at (1,0)

[golden hollow]

hmm okay

[terse crypt]

So uhh

Anything else?

[golden hollow]

nope thats it

oh

thanks!

.close

[final saddle]

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[jagged flare]

for all positive integer $n$, $f(n)=\sum_{d\in\bZ, d>0, d\equiv 1 \mod 2, d\mid n}(-1)^{\frac{d-1}{2}}$. Prove that:

[soft zealot]

skissue.in.a.teacup

[jagged flare]

for all positive integer $a$ then $f(2^a)=1$

[soft zealot]

skissue.in.a.teacup

[jagged flare]

if $p>2$ is prime and $b$ is a positive integer then $$f(p^a)=\begin{cases}
1+a & \mbox{if}; p\equiv 1\mod 4 \
1 & \mbox{if}; p\equiv 3\mod 4, a\equiv 0\mod 2 \
0 & \mbox{if}; p\equiv 3\mod 4, a\equiv 1\mod 2 \
\end{cases}$$

[soft zealot]

skissue.in.a.teacup

[jagged flare]

if $\gcd(M,N)=1$ then $f(MN)=f(M)f(N)$

[soft zealot]

skissue.in.a.teacup

[jagged flare]

im stuck on the last one

[lime crest]

is the question all of these 3

also

gcd = 0???

[jagged flare]

part 1,2,3

[lime crest]

do you mean = 1 vro

[jagged flare]

wdyt

[spring haven]

seems like a super convoluted way to write the number of odd factors of a number

[jagged flare]

odd factor 1 mod 4-odd factor 3 mod 4

[spring haven]

ah wait I misread the question

mb

[lime crest]

hmmm

$d = d_Md_N$ with $d_M,d_N$ odd and show that$$\ (-1)^{\frac{d-1}{2}} = (-1)^{\frac{d_M-1}{2}}(-1)^{\frac{d_N-1}{2}}$$

[soft zealot]

Copter

[lime crest]

or something

(by checking exponent mod 2)

[final saddle]

@jagged flare Has your question been resolved?

[jagged flare]

what is dM and dN supposed to even represent

[lime crest]

from this thingy

oh my gosh my keyboard is glitching

cya man 😭

[final saddle]

@jagged flare Has your question been resolved?

[thin remnant]

!xy

[final saddle]

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.

[thin remnant]

if it isn't resolved yet, may I help

[dusty quarry]

the og problem is here

[thin remnant]

for all positive integer $n$, $f(n)=\sum_{d\in\bZ, d>0, d\equiv 1 \mod 2, d\mid n}(-1)^{\frac{d-1}{2}}$. Prove that: for all positive integer $a$ then $f(2^a)=1$

[soft zealot]

Ren🍓

[jagged flare]

thats part a

[jagged flare]

this is b

[jagged flare]

this is c

im stuck on c

[thin remnant]

oh so you need help proving if gcd(M,N) = 1 then f(MN) = f(M)f(N) ?

[jagged flare]

yes