help-0

[remote heron]

i think so, yea

[late parcel]

Can you elaborate as to how that works

[remote heron]

the fundamental theorem of calculus?

in this case you have a function that describes the total accumulation of f'(x)

[late parcel]

Hmm I’ll look that up, thanks

[remote heron]

namely f(x)

[late parcel]

Oh ok

[remote heron]

you could just take f(6), the upper bound

but you have a problem that you may have started with some accumulation

so you subtract what was there already when you started counting

f(6)-f(0)

in this case, you started at -22, and ended at 2

so we must have accumulated 24

[late parcel]

Interesting

[remote heron]

the proof takes some real analysis

you can find decent explanations that dont require all of that though

[late parcel]

Gotcha, thanks

[remote heron]

https://www.youtube.com/watch?v=rfG8ce4nNh0

[lone heart]

@late parcel Has your question been resolved?

[lone heart]

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[alpine sable]

the question and working

this is the answer

how does 1/2sqrtx occur???????????

[glass lichen]

x/sqrt(x)=sqrt(x)

and d/dx sqrt(x)=1/(2sqrt(x))

[alpine sable]

after breaking the bracket you would get x^3 +1

[glass lichen]

You would not

as I just said

x/sqrt(x) = sqrt(x)

not 1

review your exponent laws

[alpine sable]

x^3 is right tho for the first part?

so what would you put down for the second bit?

wait.

nvm, i coped the q wrong, i’ll retry that again, sorry

ah i see

right okay thanks again

.close

[lone heart]

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[rich basin]

q5

[glass lichen]

what about it?

[rich basin]

I am confused on it

I am stucvk on one part of it

@glass lichen what do you think i shjould do?

[glass lichen]

You never posted a question

so idk what you expect me to say.

[rich basin]

I said question 5

on the image

[rich basin]

here

[glass lichen]

Not a question.

[glass lichen]

That's just the finite geometric series formula

What is the question?

[rich basin]

It says at thevery top of the page "Prove by mathematical induction"

[glass lichen]

That's the question.

but anyway, where are you stuck?

[rich basin]

From the image i sent you, I'm stuck at trying to simplify it

[glass lichen]

$\frac{a(1-r^n)}{1-r}+ar^{n}$

[ocean seal]

Mosh

[glass lichen]

by the IH

[rich basin]

@glass lichen How does (1-r^n+1) divide 1-r into that?

[glass lichen]

what?

also: It's easier to assume k-1 and prove k

just from a writing stand point

[rich basin]

why is that?

[glass lichen]

cause the LHS of the RTP statement ends at n-1

assume that $\sum_{i=1}^{k-1}ar^i=\frac{a(1-r^k)}{1-r}$

[ocean seal]

Mosh

[glass lichen]

Your assumption is also written wrong

[rich basin]

what did i do wrong?

[glass lichen]

the general term in the summation is clearly not what you have written

[rich basin]

what general term

the counter?

[glass lichen]

wrong

[rich basin]

isn't it right?

[glass lichen]

It isn't

[rich basin]

I have a j above the r

[glass lichen]

as I just said

the general term of the LHS is ar^j

[rich basin]

oh right

shouldn't it be ar^j-1

[glass lichen]

yes, same difference though

[rich basin]

Thanks manged t oget the anwer

and also fort (6)

what is a good way to do it?

when you have numbers that can already be factored by 2304 can I just get rid of them?

[lone heart]

@rich basin Has your question been resolved?

[lone heart]

@rich basin Has your question been resolved?

[rich basin]

I am raelly confused on what to do next

<@&286206848099549185>

@crisp iron

.

.

[lone heart]

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[rich basin]

.reopen

[lone heart]

✅

[rich basin]

<@&286206848099549185>

@crisp iron

[lone heart]

@rich basin Has your question been resolved?

[alpine sable]

Where's the question

[rich basin]

@alpine sable this is the queston

it is at the very top of the page

it is to prove this by induction

[urban pine]

the first picture you sent is cut off, it starts at question 2
that's why no one knows what you want (there is no question in that picture...)

[rich basin]

@urban pine the previous part is not part of the latest image i've sent

the latest image is at the top but it is prove by mathematical induction

[urban pine]

oh sorry i see now
but is this not finished? what do you mean to do next?

[rich basin]

@urban pine the image above is my working out

and I'm confused how can i get it to the RHS

[urban pine]

well first off: you're coming from the rhs and trying to show that equals the lhs (otherwise you have assumed what you're trying to prove)

second: the bracketed expression you have substituted with your inductive hypothesis isn't valid (notice the terms aren't exactly the same; e.g. the first term is x^{k-1} c but there is no c in your IH), and even still you'd have to distribute that (x - c) term in the next step.

is there any reason you don't want to use the proof from below? does it not make sense or are you trying to do it independently?

[lone heart]

@rich basin Has your question been resolved?

[rich basin]

@urban pine I don't get the proof from below

[urban pine]

which step is confusing?

[urban pine]

i gotta go now, but i do think the lower proof is quite clear
i'd recommend writing it out by hand and following what they do
try to justify each line
i'm sure others here can help you too 🙂

[rich basin]

how do we do (b)

<@&286206848099549185>

@crisp iron

[vale wigeon]

15 min rule...

but also this is literally the inductive step

[rich basin]

@vale wigeon Im' really confused what to do next?

it doesn't give you what sn+1 equals to ?

[vale wigeon]

k, not n. let's not mangle the book's notation unnecessarily.

also you should know that $s_{k+1} = s_k + u_{k+1}$ literally by the definition of ``sum of the first $k$ terms of the series''

[ocean seal]

Ann

[rich basin]

okay that makes sense

@vale wigeon so do i substitute that identity

and then reason it as "given"

[vale wigeon]

reason it as "given"
are you forced to make a two-column proof for this or something??

[rich basin]

Shouldn't we give a reason why we can subtitute that?

and for (c) can isay from (b)

substitute n

and we get that n/2n+1 which is equal to the expression they have given us?

[vale wigeon]

the sum of the first k+1 terms is the sum of the first k terms plus the (k+1)'th term

i don't know how else to say it to you

if you refuse to use this then by all means feel free to struggle along

[rich basin]

It should be fine, pretty sure the reason system should be lenient

what about for (c)

[vale wigeon]

induction lmao

[rich basin]

@vale wigeon how can i use inductino?

[vale wigeon]

in part (b) you literally proved the inductive step of part c

[rich basin]

@vale wigeon i'm confused on how to solve q6 \

[vale wigeon]

do i understand correctly that you have decided to abandon the previous question?

[rich basin]

i ithkn i solved (c)

(b) I managed to get it down to the rhs

[vale wigeon]

so you've decided to continue on your own but chose not to tell me

not even "i think i got it from here, thanks"

kinda rude

[rich basin]

Sorry, I thought I got it

[rich basin]

<@&286206848099549185>

[shell widget]

@rich basin Try using sin(A+B) = sin(A)cos(B) + cos(A)sin(B)

[rich basin]

@shell widget Where?

[shell widget]

q6

[rich basin]

I used it at the beginning

[shell widget]

Yes so you have sin(x)cos(180n)

are you required to do induction?

[rich basin]

yeah

[shell widget]

ok so you just need to show that cos(180n) = (-1)^n

which can be done easily

[rich basin]

I think i got it

Thanks and also do you know how to find the limit of ii)

[shell widget]

yeah its correct

have u found the sum?

[rich basin]

yeah

I proved the sum in the previous question

but not too sure about the limit

[shell widget]

what is it

[rich basin]

q6 is that

and it is equal to ther second image above

[shell widget]

so take the result for the sum from the previous part which im guessing is (1/5)(n)(n+1)(n+2)(n+3)(n+4)

divide this by n^5 and then take the limt as n goes to infinity

[rich basin]

yeah

[shell widget]

you can write n^5 as n n n n n

and so you put each n in each bracket

[rich basin]

yeah

[shell widget]

so you have (1/5)(n/n)[(n+1)/n][(n+2)/n][(n+3)/n][(n+4)/n]

then do some dividing

[rich basin]

I'm confused by your n+1/n part

[shell widget]

?

[rich basin]

so you divided by n?

wouldn't you have to divide the sum as well?

[shell widget]

i divided the sum of the series given which is (1/5)(n)(n+1)(n+2)(n+3)(n+4) by n^5

[rich basin]

yeah

Okay so what do we do next?

[shell widget]

so you divide

[rich basin]

@shell widget I confused what to do nmext?

[shell widget]

.

[rich basin]

so it's just 1/5?

[shell widget]

yes

[rich basin]

q9 should be an easy one

just I can't seem to factor it down

[rich basin]

<@&286206848099549185>

[rich basin]

@crisp iron

[rich basin]

<@&286206848099549185>

[lone heart]

@rich basin Has your question been resolved?

[small flint]

help

@crisp iron

[wispy wing]

i feel very stupid for not knowing these

index notation

7^2 + 4^3 is wrong

[robust geode]

Where did your + come from ?

[wispy wing]

idk, is it times?

[robust geode]

Answer my question

You had a reason

I'm trying to understand it

[wispy wing]

it was more or less a guess because i knew it wouldnt be timsed

[noble sinew]

#❓how-to-get-help

Don’t steal a channel

[robust geode]

Yeah create your own

[noble sinew]

You want to show 6(1^2+…+(k+1)^2)=(k+1)(k+2)(2(k+1)+1)

Not what you wrote down

[robust geode]

You didn't write the proper thing you had to prove

[lone heart]

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[stuck geyser]

I'm trying to graph a sphere on a 3D point cloud but I'm a bit confused on x^2 + y^2 + z^2, if I looped through each point and ran it through that equation, the result would be... ?

[lethal tendon]

how do you plan on looping through each point?

[bleak ridge]

Well that itself isnt even an equation

[stuck geyser]

Just using python

isnt the format a = x^2 + y^2 + z^2?

[bleak ridge]

r^2 = x^2 + y^2 + z^2

Its just 3d pythagorean theorem

[lethal tendon]

looping through each and every point would theoretically give you that sphere but you'll be working in discrete intervals, so maybe you'll have to introduce a tolerance to the r value

[bleak ridge]

Because you want an equal distance to a certain point

The definition of a sphere

[stuck geyser]

Ohh yeah I see

So for context I'm doing Stand up Math's Christmas tree thing, and I want to change the rbg values of a given LED (point) if it lies on the edge of the sphere, I suppose I would have to write something to turn on the LED if it's near the given point since it's discrete @lethal tendon

[lethal tendon]

yup that's the right idea

when deciding on the size of that tolerance value, you can maybe consider the average distance between LED pixels, just an idea

[stuck geyser]

Yeah I figured, I already have some code that finds nearest neighbor but its O(N*M) so it's pretty slow, I think just messing around with a straight distance value would be a lot faster with a decent result.
So to be clear, r^2 = z^2 + y^2 + z^2 gives the distance from the center of the sphere. I should generate a template sphere, and then run all my coords through the same equation and if r is similar enough then change the color

[lethal tendon]

that only gives the distance from the centre of the sphere if the sphere is centred at the origin

also, the computation speed shouldn't matter too too much right? I heard stand up math's doing pre-rendered displays this year

[stuck geyser]

Okay I think I got a good start, yeah correct but my code is getting pretty slow and I would rather have it optimized a bit for testing the csv a bunch

thanks for the help 👍

[lethal tendon]

np, I look forward to seeing your display

[stuck geyser]

I'll ping you when I get it working

[lethal tendon]

noice

[stuck geyser]

!close

[echo socket]

It's .close

[stuck geyser]

.close

[lone heart]

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[dull onyx]

yes i know that the limit is the derivative definition

but idk how to go on abt writing it

[urban pine]

that's not the limit definition of the derivative

i think you'll use that info tho

[dull onyx]

oh really?

[urban pine]

it's similar looking

[dull onyx]

i got confused

[tawdry saffron]

do you have f(x)?

[dull onyx]

no

[urban pine]

i don't think you need it

https://andymath.com/wp-content/uploads/2018/12/v65.jpg

[dull onyx]

hhh

i dont get it

[urban pine]

the definitions?

[dull onyx]

no wjat im supposed to do

[urban pine]

i think if we find f'(2) using that definition and that hint above we have shown the limit exists and so it's differentiable there, and also have found the value

[dull onyx]

what limit

first of all i made a typo

[prisma copper]

im pretty sure you have to use the rules of lospital

[urban pine]

we haven't shown f is differentiable yet

[echo socket]

For that you need to prove that f(x) is differentiable

[prisma copper]

oh right, sorry

[urban pine]

i'd start with the lower definition i posted earlier (deriv at c) and see if you can evaluate it

[dull onyx]

yea well idk how to use that

without a function

[urban pine]

just write f(x); that's your function

[echo socket]

Wait I think I got it

The limit at 0 of f(2 + h)/h would exist iff f(2 + h) approaches 0, no?

Otherwisely it'd approach ±infinity

[dull onyx]

yes

[echo socket]

So we can say that limit at 2 of f(x) is 0

But f(x) is continuous so we can plug in x=2 there

[dull onyx]

and bc its continuous

[echo socket]

So f(2) = 0

[dull onyx]

yep

[echo socket]

This f(2 + h) - 0 = f(2 + h) - f(2)

[dull onyx]

that’s what i was thinking but idk how to write it correctly

[echo socket]

Thus*

[dull onyx]

yep so it is the derivative limit

[echo socket]

I think the rest is trivial

[dull onyx]

so i say

that the limit of f(x) as x approaches 2 needs to be 0?

[echo socket]

Yes

Because if it approaches some nonzero value

Then f(2 + h)/h would go to infinity

[dull onyx]

ohh yes

[echo socket]

So by contradiction, it approaches 0 instead

[dull onyx]

so f(2) = 0

[echo socket]

Yup

[dull onyx]

then the limit definition of derivative is right there

so its differentiable and the derivative is 3 at x=2

[urban pine]

isn't that circular reasoning?
we don't know that the limit exists, that's what we need to show it's differentiable

wait i see that now

[echo socket]

f is continutous at 2, so the limit of f at 2 exists and is equal to f(2)

[lone heart]

@dull onyx Has your question been resolved?

[urban pine]

\begin{align*}
f'(2)
&= \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} \\
&= \lim_{h \to 0} \frac{f(h + 2) - f(2)}{h} \\
&= \lim_{h \to 0} \frac{f(h + 2)}{h} - \lim_{h \to 0} \frac{f(2)}{h} \\
&= 3 - 0
\end{align*}

[ocean seal]

citrusmunch

[prisma copper]

I know its not my question, but why is

$$\lim_{h \to 0}\frac{f(2)}{h} = 0$$

??
I really liked the questions and answers you have all given, so it just picked my interest 🐵

[ocean seal]

Arne

[lone heart]

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[silk steeple]

help

[silk steeple]

Question 16

[glass lichen]

what have you tried/thought?

@silk steeple

[lone heart]

@silk steeple Has your question been resolved?

[lone heart]

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[silk steeple]

.reopen

[lone heart]

✅

[silk steeple]

sorry I went to eat

my bad

so I used the discriminat

then I got it to

4-4k^2+2k=0

the hard thing is

In the book for some reason there’s no answer for this question

strangely

.close

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[ruby summit]

Hello, I have a problem

[ruby summit]

How do I solve x^2/2^x = a

If I know 'a'

$\frac{x^2}{2^x} = a$

[ocean seal]

Bman

[ruby summit]

And is it possible to extend the solution to $\frac{x^b}{b^x} = a$

[ocean seal]

Bman

[ruby summit]

Where I know 'a' and 'b'

[gilded olive]

What is the value of a

[ruby summit]

We can imagine that we know

So we can find x in relation to a

[marsh rapids]

I.e. a in R+

[ruby summit]

Or we could try an example of 1

[ruby summit]

Yes

[marsh rapids]

Do you know of the Lambert W function ?

[ruby summit]

No

[raw shard]

oh no not the lambert w function

[ruby summit]

I can google it but that may take a while, do you have a summary?

[raw shard]

those are impossible to set up

[marsh rapids]

It's quite simple here: 2 = e^ln 2, put the denominator as a negative exponent and take the square root of the whole thing. The rest is easy

[gilded olive]

You will simply multiply by the "cross" method, you will level the foundations and work with the exponents

[marsh rapids]

The W function is defined as the inverse of x e^x, so u e^u = k <=> u=W(k)

[ruby summit]

So $x^2 = a2^x$ Then, $x^2 = ae^{Ln(2)}x$

[ocean seal]

Bman

[marsh rapids]

So it's all about putting the equation in the right form, taking W of it, then solving for x

x^2 e^(-x ln(2)) = a, then you can take the sqrt to get rid of the square on the outer x

[ruby summit]

Ohhh

[marsh rapids]

Then multiply by -ln2 /2, apply W to isolate -x ln2 /2 and you get x from that

[ruby summit]

But what happens to $e^{-xLn(2)}$ when I take the square root

[raw shard]

sorry i’m not understanding this and i really wanna see the solution

[ocean seal]

Bman

[raw shard]

could you type it out

[marsh rapids]

(a^b)^1/2 = a^(b/2)

[ruby summit]

Sorry, about me not making that connection

[marsh rapids]

Haven't written it out but I think it's -2/ln(2) * W(-ln(2)/2 * sqrt(a))

[ruby summit]

And thank you,

Is there a more, I guess elegant, solution to W(x)?

Rather than saying W(x)

[marsh rapids]

There are some dirty formulas, either series or integrals

[ruby summit]

Okay

Thank you for your time, patience, and help!

[marsh rapids]

You can find them on Wikipedia

[prisma copper]

this sounds like the rule about cheating in assignments on this server...

[raw shard]

thanks now i know not to help you

@ruby summit can you do .close

if you’re done

[prisma copper]

not to help johan?

Think about the people who actually solve this by themselves. Also if it isn't "on points" you shouldnt bother and wait for an answer.
It's an "olympiade" that should say something about fair play.. no?

But there is a timelimit on your math skills? Also your math skills force you to beg in EVERY other help channel? Ngl. this is just getting even more bad mannered at this point

[glass lichen]

Just stick to 1 channel and wait.

No one's entitled to help you as well

Cause you're asking a contest question.

Good idea.

@prisma copper drop it.

[prisma copper]

didnt see the answer, my bad

[marsh rapids]

Also for a question like this, you'd probably have a better chance in the math olympiads server

[lone heart]

@ruby summit Has your question been resolved?

[lone heart]

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[pine bronze]

Anyone here to help?

[pine bronze]

Oh hello

Can you help me with this geometry question

<@&286206848099549185>

<@&286206848099549185> can you please help me with this question

<@&286206848099549185>

@crisp iron

Please help

[swift jolt]

@pine bronze use the process of elimintation

do you know what supplementary angles are?

a is incorrect because L1 and L3 are parallel not supplementary

and you can use this process to eliminate the rest

[lone heart]

@pine bronze Has your question been resolved?

[lone heart]

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[hollow pelican]

hi guys for this question when i do u sub, why i do not need to change the limits of u?

[gray isle]

you should be changing your bounds when doing sub

[hollow pelican]

but if i change my bounds the ans is wrong tho

[gray isle]

how so?

the work written on the page is invalid

[hollow pelican]

the online calculator didnt change the bound too

[gray isle]

the online calc is doing indefinite integrals (where there are no bounds)

as opposed to a definite integral which you have which does have bounds

[hollow pelican]

opps i forgot to screenshot the final working

[gray isle]

technically you don't "need" to explicitly change you bounds
but if you use that approach you would need to explicitly write
**x=**1 for your lower bound
and **x=**3 for the upper bound

[hollow pelican]

but when i change the bounds i keep getting wrong answer

[gray isle]

you're conflating the two approaches

when updating the bounds, the bounds are now related to u
and you don't sub back x anymore

[hollow pelican]

but what do i sub back with?

[gray isle]

technically once you reach
your upper bound is u=4.0986
and your lower bound is u=3

so sub those values in directly

[hollow pelican]

i subbed in 4.0986 and 3 tho

but its wrong

[gray isle]

you unnecessarily subbed x=3 + ln|x| back in

where if you updated your bounds earlier such that you have

your upper bound is u=4.0986
and your lower bound is u=3
that is no longer required

[hollow pelican]

ohh so i just sub 1 and 3 as the value again?

[gray isle]

wdym

[hollow pelican]

in this part

[gray isle]

if you choose the route where you sub back u=3+ln|x| then you'd use
x=3 and x=1 for your bounds

if you use that route, you'd need to explicitly write x=3 and x=1 for your bounds while your integral and expressions are in u

[hollow pelican]

okay

[hollow pelican]

i see , thanks

[lone heart]

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[wind hawk]

What's the difference between:
$\sum^{\infty}{i=1} ...$
and
$\lim{n \rightarrow \infty} \sum^{n}_{i=1} ...$

[ocean seal]

THEBIGTHREE

[glass lichen]

nothing

[severe sluice]

when people say $\sum^{\infty}{i=1} ...$ they usually mean $\lim{n \rightarrow \infty} \sum^{n}_{i=1} ...$

[ocean seal]

just bob

[glass lichen]

prior is defined as latter

[wind hawk]

ah alright so its basically just a shorthand

[glass lichen]

yes

[severe sluice]

yeah

[wind hawk]

sweet, thanks

.close

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[severe sluice]

(wow, sync)

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[royal patrol]

I know how to work out everything but i just dont understand how to work out the shape of the curve?

[lone heart]

@royal patrol Has your question been resolved?

[royal patrol]

<@&286206848099549185>

[lone heart]

@royal patrol Has your question been resolved?

[thorny patio]

So for a question like this you can follow a general strategy

Factor completely

Find zeros

Find multiplicity of the zeros (odd or even)

Find y intercept

You know the end behavior from the degree

It either vaguely resembles a parabola or a cubic (even degree vs odd degree)

Check for lead coefficient if it's negative you have to flip upside down (reflect over x axis)

In this case

Degree is 4

It's like a parabola sort of

Draw from the top left. It reaches the zero on that side:

(-5,0)

At this point you check multiplicity.

Factored form tells you multiplicity is 1 (exponent on that factor)

Odd multiplicity means you cross through the x axis at that zero

You know that it must touch y intercept down there so you go all the way down to that point

Then you know it must come back up to the next zero:

(4,0)

Multiplicity is 1 again

Odd multiplicity means you pass through x-axis

Pass through and finish from there @royal patrol

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[pseudo mesa]

Show that for any real numbers a, b, c, the following inequality is true:
a^2 + b^2 + c^2 >= ab + ac + bc

[pseudo mesa]

Moving them on one side:
a^2 - ab - ac + b^2 - bc + c^2

a(a-b) -c(a-b) + b^2 + c^2

(a-c)(a-b) + b^2 + c^2

No idea what to do now

(a-c)(a-b) + (b+c)^2- 2bc, mby

[vale wigeon]

there is a geometric solution to this that involves reducing your inequality to Cauchy-Schwarz for three-dimensional vectors

[pseudo mesa]

i'll leran vectors later

it's time to practice algebra

[vale wigeon]

consider: $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$

[ocean seal]

Ann

[vale wigeon]

consider also that instead of $a^2 + b^2 + c^2 \overset?\geq ab+bc+ca$ you can prove $2(a^2+b^2+c^2) \overset?\geq 2(ab+bc+ca)$

[ocean seal]

Ann

[vale wigeon]

er wait no nevermind

that won't work as i intended it to

i thought i had it but no

[pseudo mesa]

ann

i looked up the tip to this problem

[pseudo mesa]

and it's precisely this

so that's probably what they wnat me to do

[gray isle]

consider
LHS = 1/2(a^2 + b^2 + a^2 + c^2 + b^2 + c^2)

[vale wigeon]

you can't do shit with a^2 + b^2 + c^2 - 2ab - 2bc - 2ca though can you?

it's not the square of a three-term sum

that's what i was (erroneously) trying to hint at

[vague coral]

you can't

[vale wigeon]

oh but yeah ramonov has it

yes i see it now

[gray isle]

the same problem has come up multiple times here

[pseudo mesa]

(a-b)^2 + (b-c)^2 + (a-c)^2
a^2 - 2ab + b^2

• b^2 -2bc + c^2
• a^2 - 2ac + c^2
  =
  2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac | /2
  a^2 + b^2 + c^2 - ab - bc - ac
  so
  1/2((a-b)^2 + (b-c)^2 + (a-c)^2) >= 0, cause every term is >= 0

this?

@gray isle @vale wigeon

well, thanks

.close

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[noble jacinth]

I have a question related slightly to college applications. The school I’m applying for wants to see that I’m learning matrices, linear systems of equations and determinants. This is the current curriculum I’m learning/presenting to them. Is this satisfactory to the requirements? What else should I know and brush up on? For context, I’m applying to a data science course.

[lone heart]

@noble jacinth Has your question been resolved?

[supple tundra]

You watch the linear algebra course from MIT by gilbert strange

and you should find a good thicc textbook

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[dull onyx]

prove f(0) = -2a 😐

[echo socket]

Plug in y = 0

The equation follows immediately

[dull onyx]

i feel so stuoid thank u😭

when i differentiate something like this do i treat x as a constant

[echo socket]

Depends on with respect to what variable are you differentiating

[dull onyx]

i gotta prove f’(x) = 2x + f’(0)

yep i differentiated with respect to y

and plugged 0 in and got it

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[small portal]

how would you choose bin sizes / the amount of bins for a time series? There seem to be several ways and some academic papers even say that they just chose bins based on visualisation. My time series looks like this:

timestamp value
0 2016-01-01 00:00:00+00:00 168.409707
1 2016-01-01 01:00:00+00:00 178.316160
2 2016-01-01 02:00:00+00:00 231.811009
3 2016-01-01 03:00:00+00:00 218.932619
4 2016-01-01 04:00:00+00:00 159.493899
... ...
8758 2016-12-30 22:00:00+00:00 291.019066
8759 2016-12-30 23:00:00+00:00 275.181294

[8760 rows x 2 columns]

standard deviation: 86.979259
min value : 34.635974
max value: 582.499458
mean: 228.99427

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[silk steeple]

help

[silk steeple]

question 19

I multiplied the top by the bottom to get rid of the denominator

And I’ve gotten

P=6x^3 -3x^2+6x-3

[gray isle]

can you show all work leading up to that

[silk steeple]

okay

1 second

I multiplied (x^2+1) by 3

[gray isle]

that's no longer p

[silk steeple]

wait

Oh…

Do I need multiply p by that too?

[gray isle]

thats p*(2x-1)^2

the simplest approach would be to sub in your p to the lhs and see what happens

[silk steeple]

I see

let me do that 1 second

hm

[glass lichen]

now do what Ramonov suggested

[silk steeple]

I don’t know how to do that

sorry

[gray isle]

don't overthink it

apply basic substitution

like how would show if x=3,
then x-1>0

[silk steeple]

Am I going in the right direction

[glass lichen]

No

You're ignoring what Ram is saying

[silk steeple]

I don’t know how to substitute

[glass lichen]

.

Answer this then....

[silk steeple]

The discriminant?

[glass lichen]

??????

since when was x-1 a quadratic??

[silk steeple]

OH

you would sub it in

[glass lichen]

Yes

x-1
3-1
2>0

[silk steeple]

I need to do it here

I’m stuck

[gray isle]

don't overthink it

the basic principle of sub is to replace something with something of equivalent value

so your first step could be to first replace/substitute the ps in the lhs with whatever you're told p is

[silk steeple]

wait

this?

[gray isle]

wdym?

[silk steeple]

I actually don’t know what to do

I see nothing to substitute except for the question

[gray isle]

in case it wasn't clear,
replace the ps in
p^2 - 3(p+3)
with whatever you're told p is.

i.e with 3(x^2+1)/(2x-1)

[silk steeple]

Wouldn’t p = -3

I think

[gray isle]

why?

[silk steeple]

(P+3)

[gray isle]

what?

[gray isle]

you're overthinking and/or not reading what I'm saying

[silk steeple]

I’m trying to understand

[gray isle]

like how would show
if x=3, then x-1>0

[silk steeple]

P = 3(x^2+1)/(2x-1)

wait

3-1>0

we sub the 3 for x?

I’m not sure

[gray isle]

yeh...

[silk steeple]

hm

[lone heart]

@silk steeple Has your question been resolved?

[lone heart]

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[tame light]

i should be able to do this by myself but somehow stuck lol

[vague coral]

did you try squaring a/b + b/a = 5 ?

[tame light]

and then..?

[lethal tendon]

maybe try rewrite both given equations in terms of (a + b) and ab

[lone heart]

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[warped spruce]

why does it true? can someone please give me an example with numbers?

[tight locust]

3/2 = 3/2

So 3=3

[echo socket]

Didn't see that both denominators are AB

[warped spruce]

yes but it could be 6/3, 9/3 and this is not the same number, or is it?

[vale wigeon]

6/3 ≠ 9/3. the top equality doesn't hold.

[tight locust]

This is what we call a biconditional statement

It goes both ways

The top is true only if the bottom is true and the bottom is true only if the top is true

Haha

[warped spruce]

hahahahh

but I didn't understand why the top is true

[vale wigeon]

well without context we can't tell you either

it might be that you ripped this out of a geometry problem

but then you would have to show us a diagram with the points A, B, E, F, G, H labeled

[tight locust]

If x = y then x/z = y/z. We take this as an axiom of the reals

[vale wigeon]

this is not an axiom

do not spread misinfo

but also is there anything else given about this diagram

[tame light]

can we have the whole question lol?

[warped spruce]

yes but I was able to prove this

[warped spruce]

.

so I just need to understand why if the button is equal so the top is equal

[vale wigeon]

???

[tame light]

lmao what

[warped spruce]

idk why this is true

[vale wigeon]

so first you say you couldn't understand why EG/AB = HF/AB, but now you say you were able to prove it?

[warped spruce]

nonon hahahah I didn't prove it, you can say it.. it is a rule. but I dont undersatnd why this rule is true

[vale wigeon]

......

so what was "yes but i was able to prove this" about?

[warped spruce]

I was able to prove that EG/AB = HF/AB

it wasn't given in the question

[tight locust]

It's true by reflexion haha. If x=y then x/3 = y/3 because x/3 = x/3. You can't get much simpler than that

[warped spruce]

I am a little bit dumb to be honest hahahah

[vale wigeon]

what's "reflexion"

are you trying to spread misinfo / terminology misuse again

[warped spruce]

hahahahaha

[vale wigeon]

@warped spruce do you understand that if x = y then 2x = 2y

yes or no

[warped spruce]

yes

[vale wigeon]

ok

do you understand that if x = y then cx = cy (for any number c)

yes or no

[warped spruce]

yes

[vale wigeon]

okay

so

EF/AB = HG/AB

[tight locust]

The reflexive property

[vale wigeon]

"If x=y then x/3 = y/3" has nothing to do w/ equality being reflexive

anyway

[warped spruce]

6/3 != 9/3

[vale wigeon]

EF/AB = HG/AB
multiply both sides by AB
(EF/AB) * AB = (HG/AB) * AB
simplify
EF = HG

[warped spruce]

so you can't say that if the button is equal so the top for my opinion

[tight locust]

"Because x/3 = x/3"

[vale wigeon]

@warped spruce i think you might want to remind yourself of what "IF ___ THEN ___" means

[warped spruce]

now I understand because of the math you did

but idk if you put numbers like EF =6 and HG = 3, you will get a wrong answer no?

6 / 3 != 3/3

[vale wigeon]

okay, and?

that doesn't mean the argument is invalid

[warped spruce]

this means that the 2 above numbers aren't for sure equal

[vale wigeon]

it's like when someone says "if it rains then my daughter stays inside" and you respond with "but it's sunny and she's playing outside, so you're lying!"

we're taking EF/AB = HG/AB as a premise

we assume it is true and everything we do afterwards follows from it

[warped spruce]

so without numbers?

[vale wigeon]

we do not give values to any of the three lengths involved in the equation, no.

[warped spruce]

yes you can prove it by multiplaying both sides with AB sure

but if you put numbers it wont worj

work

[vale wigeon]

if you feed bullshit in you'll get bullshit out

i fail to see how this is in any way surprising

[severe sluice]

generally we can perform any valid action on both sides of an equation to get another equivalent equation

[warped spruce]

alright I think I got it, tyyyy!

@severe sluice @severe sluice 🙂

[severe sluice]

that's both me??

[warped spruce]

ohhhh XD

@vale wigeon TYYY

I am so out today

hahahah

anyway Ty guys!

.close

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[warped spruce]

wait @vale wigeon what happens when you multiplay AB/HG * AB

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[warped spruce]

how does the equestion will see before you simplify

[vale wigeon]

we never had any AB/HG? we had HG/AB...

[warped spruce]

my problem is that idk how to simplify stuff

so ye

it doesn't matter

just for the example

[vale wigeon]

you do not know how to simplify (HG/AB)*AB into just HG?

[warped spruce]

I forgot hahahah

Ik you can do it because I mermorized it

and I want to understand the math behind it

[vale wigeon]

there is nothing to understand

division and multiplication by defn undo each other

[warped spruce]

but you can do it only if there is not + or - in the exercise

no?

[gray isle]

wdym

[warped spruce]

I don't understand why you can reduce the a

like the math behind it

you divide by a each side?

[gray isle]

assuming a isn't 0,

divide each side by a

[warped spruce]

sheshhhhhhh

tyyyy bro!

🙂

so like a/ sinB * a = 2a/ a right @gray isle ?

and then you just reduce the a

[gray isle]

it feels like you're missing parentheses

[warped spruce]

wdum?

[gray isle]

because it looks like you're multiplying the left side by a
but dividing the right side by a

[warped spruce]

so is it a/ sinB/a?

[gray isle]

no

(a/(sin(B))/a
or (a/ (a * sin(B))

parentheses are required when typing stuff

\verb|a/ sinB * a| is interpreted as $\frac{a}{\sin(B)} \cdot a$

[ocean seal]

ℝamonov

[gray isle]

even though you may have intended
$$\frac{a}{\sin(B)\cdot a}$$
typing stuff in a line of text does not have the benefit from horizontal fraction lines

[ocean seal]

ℝamonov

[gray isle]

so parentheses are required to communicate clearly

[warped spruce]

if you want to divide AB/CD by a, you do AB/(CD * a)

[gray isle]

when communicating in plain text, yeh

when writing it out on paper, (or using math type)
$$\frac{AB}{CD \cdot a}$$
isn't an issue

[ocean seal]

ℝamonov

[warped spruce]

alright and multiplaying AB/CD * a in a paper: AB * a / CD right?

[tired pagoda]

Bro this is wrong

[warped spruce]

why bro?

[tired pagoda]

AB/CD*1/a

[gray isle]

they're equivalent singh

[warped spruce]

yes because you can make in 2 fractions by multiplication

I think

[gray isle]

$\frac{AB}{CD} \cdot a$ and $\frac{a\cdot AB}{CD}$ are both fine

[ocean seal]

ℝamonov

[warped spruce]

and when you divide by a?

how does it look like?

[gray isle]

choose whichever: \
$\frac{\br{\frac{AB}{CD}}}{a}$ \ \
$\frac{AB}{a\cdot CD}$ \ \
$\frac{AB}{CD}\cdot \frac1a$

[tired pagoda]

How this bot works

[ocean seal]

ℝamonov

[gray isle]

basic guide can be found in #resources

[tired pagoda]

Thanks

[warped spruce]

Thanks bro

[gray isle]

basic understanding of functions, mathematical notation, parentheses usage helps a lot

[warped spruce]

@gray isle thanks a lot bro 🙂

[lone heart]

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[dusky osprey]

idk wgat to do

[echo socket]

Is this a test?

[dusky osprey]

no

drfrost

[echo socket]

Here depth of the bowl would be the minimum value of y

[dusky osprey]

is that bc itd yhe lowest point

[echo socket]

Yes

[dusky osprey]

Ight

what do i do next

[echo socket]

A parabola y = ax^2 + bx + c reaches its maximum or minimum (in this case minimum) at x = -b/2a

[dusky osprey]

-b/2a?

[echo socket]

Yes

[dusky osprey]

Why is it that

[echo socket]

Where a,b,c are coefficients as shown here

[echo socket]

Oh, you can prove that using calculus

Have you been introduced to derivatives yet?

[dusky osprey]

Im not even sure

it came up as mock gcse revison so i should know it

oh

r using the quadratic formula

[echo socket]

You can either substitute x = -b/2a into the equation to get the lowest value of y

Or define value M such that $\frac{x^2}{10} - 3x \ge M$ for all x

[ocean seal]

Touch Our Beans

[echo socket]

(Clearly the first way is simpler)

[dusky osprey]

r u using this formula to get it

-b/2a?

[echo socket]

Well if you view x = -b/2a as midpoint between the quadratic roots, then yeah

[coral moss]

I think differentiation is either in GCSE further maths or A-level (can't remember) so at gcse you have to complete the square to find the minimum/maximum point

[dusky osprey]

this is the mark scheme but they done complete sqaee

[dusky osprey]

yeah i think so cuz idk what that other stuff is

[coral moss]

the method using x = -b/2a is simpler to do but uses "more advanced concepts"

tbh if it's GCSE id say just complete the square cuz they might not give you the marks for differentiating even though it's technically better

[dusky osprey]

Ight

why is it -225 on the end

15^2

[coral moss]

yeah you need to get rid of the 15^2

like here -> x^2 + 14x = (x+7)^2 - 196

because if you do (x-15)^2 you get left over with + 225

[dusky osprey]

Oksy

[alpine sable]

Completing the square seems silly

[dusky osprey]

I havennt been taught anything else

[coral moss]

yeah it is but it's the way we had to do it at GCSE

[dusky osprey]

and idk if id get marks

i think i get it now

ty

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[quaint pine]

is 1/A+1/B=AB/A+B? NO it isn't but what if we flip the value of the left side of the equation? meaning if we plug in values for A and B in the left side of the equation and then we flip the fraction is the above identity equal or no?

[quaint pine]

I tried it for a few values and it worked but idk if it works with ALL NUMBERS

can someone confirm?

[median dirge]

Hi.

Do you mean to say...?

$\frac{1}{A}+\frac{1}{B}=\frac{A+B}{AB}$

[ocean seal]

Muhammad Hussaini

[quaint pine]

no the right side of the equation is flipped

AB/A+B

[median dirge]

Well.

[quaint pine]

it wouldn't be equal ik

[median dirge]

$\frac{1}{A}+\frac{1}{B}\ne\frac{AB}{A+B}$

[ocean seal]

Muhammad Hussaini

[quaint pine]

Yes

but my question is

if we plug in values for A and B in the left side of equation and we got 3/5 for example

and then we flip it so we get 5/3

would that ALWAYS be equal to the right side of the equation if we plug in the same values

[median dirge]

Lool

Look

When you flip a valu

Value

It is like this

Say.

I flip 2

What will i get

[quaint pine]

1/2

[median dirge]

Nice

And if i flip 1/2

[quaint pine]

2

what's your point

[median dirge]

So

The thing i

Is

If you flip

(AB)/(A+B)

In general form

So...

[median dirge]

The equation will be true only for this

[median dirge]

The way youre saying that i flip this value...

Means

You arrive at what i showed first

[quaint pine]

ok thanks محمد ❤️

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[cedar bridge]

can anyone help me with descriptive geometry?

[echo socket]

No need to ask that question in several channels at once

[cedar bridge]

I can't find anyone that knows about it

[lone heart]

@cedar bridge Has your question been resolved?

[glass lichen]

Don't ask to ask........

[cedar bridge]

im looking for help to answer question on the subject dumbass

why would i just want to know that information

[glass lichen]

How do you expect to get help on a question you haven't asked?

[cedar bridge]

you cant answer it if you dont know descriptive geometry

[glass lichen]

How do you know I am not be able to help?

Like... do you see the absurdity of what you're saying?

Drop the attitude and ask a question like the typical user.

[cedar bridge]

3.Represent, by its projections, a regular triangular prism, with the base [ABC] contained in a ramp plane .⎯hefdistant from the axis x 4 and 5 cm, respectively;⎯the edge [AB]of the base is contained in a straight line oblique whose frontal projection is 75º with the x(opened to the left) axis; the apex is 4.5 in height; the edges of the base of the prism measure 5 cm; the height measures 6 cm.

thats what i thought

you cant answer it

[late bridge]

whether they can answer or not is not the point, asking to ask is a waste of time, you could've posted the question right away to begin with

[cedar bridge]

people on discord are dumb as fuck

ill just uninstall this

[lone heart]

@cedar bridge Has your question been resolved?

[lone heart]

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[high raven]

i have a question about dice probability. I've tried to phrase it in a general mathematical way, but let me know if you want me to explain what I'm trying to do in terms of the game.

If i have x number of y-sided dice
how would I calculate the probability that any two dice from the dice rolled would add up to the number rolled by a 2y-sided die

[high raven]

(for example, if I rolled 5 4-sided dice, what is the probability that two of them add to 6 (the result of an 8-sided die)

[arctic gust]

you also threw the 8 sided die?

[high raven]

yes

When I say "any two dice" I mean the best choice.

[arctic gust]

k

[arctic gust]

i don't think i can answer this with the given information.

because you didn't mention exactly what number appears on the 2y sided die

it could even be a 1

[high raven]

Sorry, I should be more clear

I'm not looking for a solution, I'm looking for an equation, or a place to start building an equation that would let me see the answer for a number of variables

I'm mostly confused on how to put dice rolls into an equation at all.

[arctic gust]

i think there's no chance cos you cant get a 1 from the sum of 2 dice

[high raven]

I think I'm explaining it badly. I'll close this and come back when I have more time to show what I mean

[arctic gust]

ok

[high raven]

Thank you!! Sorry for the confusion

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[swift jolt]

I need help with this

I am not sure how to solve this

[urban pine]

i like to solve them individually and then plot them on a numberline

since it's OR you have to include the solutions from both

[swift jolt]

Ummm that's kinda confusing

[icy trail]

solve them individually first

[swift jolt]

ok

thx

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[alpine sable]

when 3 planes have a single intersection

[alpine sable]

is that a sheaf?

when there is a single point where they intersect

either thats a sheaf, or its a sheaf when they all intersect along a single line

[knotty sleet]

A sheaf is when they intersect in a line

[alpine sable]

i get confused between the two

ah so sheaf is line

[knotty sleet]

Yeah that

[alpine sable]

what is a single point?

does it have a name?

[knotty sleet]

Idk if there's a name

If there is I don't know it

[elfin snow]

"point of intersection"

ezpz

[alpine sable]

what is a prismatic intersection

is that when its a triangle?

[knotty sleet]

Basically yh

[alpine sable]

can it be a prismatic intersection if the 3 planes form a matrix which is singular?

[knotty sleet]

So the planes bound a triangular prism

[alpine sable]

i thought it couldn't

but im looking at this question rn

which is singular but apparently its prismatic intersection

[knotty sleet]

Yeah if it's prismatic then there's no solution

Because there's nowhere that lies on all three planes

[alpine sable]

hmm im slightly confused wont lie

[knotty sleet]

So the matrix is singular

[alpine sable]

so if matrix is singular

there is no solution

[knotty sleet]

No

[alpine sable]

but can it be a ohhhh wait i think i got it

wait no i didnt

o

[knotty sleet]

The other way around

[alpine sable]

no unique solution?

idk what that means

prismatic means no solution though?

[knotty sleet]

If there's no solution then it's singular

[alpine sable]

and its singular?

no solution = singular ok alright

[knotty sleet]

Not quite, singular means determinant 0

So it's not invertible

[alpine sable]

which means it can either be no solutions or infinitely many solutions?

[knotty sleet]

Yea

I just means no unique solution

*it

[alpine sable]

so its sheaf or prismatic

as long as they aren't parallel planes

[knotty sleet]

THey also can't be the same plane

[alpine sable]

otherwise if determinant is not zero, then it is a single intersection

[knotty sleet]

And can't be two parallel with the third cutting them

[alpine sable]

oh yes theres that shape too

2 parallel planes with 1 cutting them both

thats 2 lines of solution

so thats 5 different things they can be right:

[knotty sleet]

That's no solution