help-17

[thin vale]

would you

[autumn coral]

ik

[thin vale]

and also

[autumn coral]

two solutions

[thin vale]

what about Ce^(sqrt(2)x)

does that work?

C being a constant

[autumn coral]

where did C come from

[thin vale]

or De^(-sqrt(2)x)

D being a constant

[thin vale]

maybe you should start going to class

imagine though that we restarted this whole entire process

and our trial solution was instead

y=Ce^(lambda x)

then when we get to the factoring step

we'd factor out Ce^(lambda x)

and the C would've also dissapeard

when we didvided

So really, any constant times e^(sqrt(2) x)

is also a solution

[autumn coral]

ye

[thin vale]

and same with any potentially different constant times e^(-sqrt(2) x)

[autumn coral]

ok i got that

[thin vale]

flock is this for a test

[autumn coral]

nah

assignment

[thin vale]

Okay good

All the other questions

are the exact same

except skip the e^(...)

just go straight to the

r^2-2=0

find r

then you have the solutions are e^(r...)

yknow

Oh and the last thing there is one more thing you need to find

[autumn coral]

Which is

[thin vale]

What is amanda?

[formal pond]

Funny

And you should also note

Idk if it’s even correct

But I BELEIVE the sum of the solutions is also a solution

[thin vale]

it is

$$y=Cy_1+Dy_2$$

[twin meteor]

Austin

[thin vale]

where y_1 and y_2 are the complentary solutions and C, D constants

that's the general solution

[autumn coral]

Ye

Well

Thanks

Im slow w this

[formal pond]

Do you know what to do now?

[autumn coral]

bc of work

i gotta go for a bit tho

[thin vale]

okay

[autumn coral]

Ill try to do the rest on my own

ill comeback if i need

[thin vale]

well the other problems are literally the same

[autumn coral]

yea

[thin vale]

so I hope to not see u back here

[autumn coral]

ye

[formal pond]

Yes come back (it’s 1 am I’ll be asleep)

[autumn coral]

Sorry :/

[formal pond]

All g

[thin vale]

flock

[autumn coral]

Thanks for the help

[thin vale]

do 1 thing for me though

[formal pond]

My sleep schedule fucked anyway

[autumn coral]

ye

[thin vale]

go to class

or read the book if ur gonna skip

[autumn coral]

I cant man

[thin vale]

or find a youtube video

XD

[autumn coral]

ill just get austin and mortta to help me

[formal pond]

Organic chem tutor really good

[thin vale]

you wont always get blessed with the 2 best helpers on the server

[formal pond]

LOL

I don’t even help much

[autumn coral]

Im also teaching u

guys

to have more patience

[formal pond]

I’m the one asking 90% of the time

[autumn coral]

feel me

[thin vale]

You're teaching us how shockingly little calculus knowledge some diffeq students have

[autumn coral]

well

last time i did calc

was over a year

[formal pond]

What major r you?

[autumn coral]

electrical engineering gango

[formal pond]

Hell naw

That’s my brothers

[autumn coral]

Ye

[formal pond]

He’s an idiot

[autumn coral]

Its disgusting

im switching to computer engineering

[formal pond]

Ok just and FYI

[thin vale]

If you want to be an engineer of any kind

[formal pond]

you’ll be using diff equations so much

[thin vale]

you'll need to know diffeqs

[formal pond]

Next year

[thin vale]

Yeah

me and mortta are on the same wavelength

[autumn coral]

ik man i hate it

so much

[formal pond]

They be like 2nd nature to you

[autumn coral]

I hate engineering man

[thin vale]

https://ocw.mit.edu/courses/18-03-differential-equations-spring-2010/

here's a full diffeqs course

with lecture videos

notes

hws

exams

if you need resources

you have them

[autumn coral]

can i hire someone to do my assignments

instead

[thin vale]

that's where I learned diffeqs actually

[formal pond]

Ok

[autumn coral]

LOL

[thin vale]

highly recommend

[formal pond]

and flock

This is a very common topic

[thin vale]

no because then you'll need to hire someone to do your job for you in the future too

[formal pond]

There are calculators for this

With solutions

I’m not telling you to instantly use them

[autumn coral]

ye i get it

Anyways tho

[formal pond]

But if your really stuck and no one can help go for it

[autumn coral]

Sure!

Thanks for the help

gtg now

[thin vale]

.close

[autumn coral]

Sorry for the mess

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[faint carbon]

am i understanding combinations right?

[spiral inlet]

Not exactly. A combination is when you choose a group of objects, without putting them in a particular order, from a single bigger set

choosing an ice cream and a topping isn't a combination, because you're choosing those things from two different sets

There are some ice cream flavors and there are some toppings, those are two different sets.

@faint carbon

Another example:

"How many ways can you draw a 5-card hand from a deck of cards?"
This is a combination.

"How many ways can you draw one card from a blue deck and one card from a red deck?"
This is not a combination.

[faint carbon]

ok

[faint carbon]

so b would be Yes, since I am a part of a set of people

[spiral inlet]

yes

your teacher would be choosing a group of students from a larger set: the whole class

(without putting them in a particular order)

[faint carbon]

c: No, letters and numbers are different sets

[spiral inlet]

that's right

and also, you'd have to put them in a particular order to make a license plate

(and also, you could choose the same letter multiple times, unlike in a combination)

[faint carbon]

so has replacement

[spiral inlet]

right

[faint carbon]

d: Yes, two books is a smaller set of all books

[spiral inlet]

yep

[faint carbon]

ok now permutation

[spiral inlet]

permutation is the same, except you put the objects you choose in a particular order

I'm not sure what OSEE is

[faint carbon]

a code?

word

[spiral inlet]

One Student Each E_____?

idk

[faint carbon]

hmmm

[faint carbon]

b: No, topping order doesn't matter

[spiral inlet]

right

[faint carbon]

c: Yes, each ID order makes it unique

[spiral inlet]

but there is replacement

[faint carbon]

ah

[spiral inlet]

So it's not the same as choosing a set from a bigger set and putting them in order

[faint carbon]

d: Yes, there's no replacement

[spiral inlet]

yea

[faint carbon]

idk osee

[spiral inlet]

I'm not sure, I'd guess it's like assigning a unique number to each student

[faint carbon]

ill ask

[spiral inlet]

👍

[faint carbon]

thank you

[spiral inlet]

np 👍

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[vague vine]

I need help on a couple questions on delta math 🙏

[vague vine]

@spice sun

[vast shale]

ask away, what is ur specific problem?

[vague vine]

I need help

[vast shale]

okay you need to elaborate on WHERE you are stuck and what you have tried so far

[vague vine]

I don’t understand parent functions

And I’ve tried almost everything photo math, symbolab, desmos

And I don’t understand if I flip it over the y or x axis etc

[vast shale]

Well do you want me to link you resources on the subject?

[vague vine]

Yes

[vast shale]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:exponential-growth-decay/x2f8bb11595b61c86:exponential-decay/v/graphing-basic-exponential-functions#:~:text=For exponential functions%2C the basic,also shifts to x%3Dk.

watch this

[vague vine]

Thank you

Close

.close

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[silver berry]

I understand u gotta use the alternating series test and for that u gotta check the 3 requirements

[silver berry]

which is basically

[sly sierra]

ok

[silver berry]

but at k = 1 bk = 0

but for tthe requirements bn > 0 for all k >= 1

why is that not a problem

[sly sierra]

that condition is overly strict

you only need it to be "eventually true"

for large enough k

[silver berry]

oh condition sohuld be something like bk >0 for large k

[sly sierra]

here's why:
you can always delete the first N terms (whatever N you choose) from the start of a series,

and that won't change whether it converges or diverges

(if it converges, it may change the value to which it converges)

[silver berry]

oh because since it is finiite it is a numberr

[sly sierra]

yeah, pretty much

[sly sierra]

yes correct

[silver berry]

are the other 2 conditions overly strict?

they arre fine right?

[sly sierra]

the first two only need to be true for large enough k

the last one is a statement about the limit so by definition it only involves large k

so basically yes, the behavior for any initial (finite) segment of the series has no effect on convergence

so if there are some small k's where the conditions aren't satisfied, you can ignore them

[silver berry]

the reason forr this is because of a theorem right i forgot the name but basically it's decreaasing and it has a lower bound which is >= 0 so it converges

[sly sierra]

yea, it's slightly more complicated than that, when you include the (-1)^k it actually oscillates back and forth between positive and negative, but what you said is part of the proof for sure

[silver berry]

hmm okay I think I understand, thank you for helping

[silver berry]

1 more quesiton the alternating series test is in this form(-1^n-1) so the first term is positive in my question the first term is negative it doesn't matter right? only thing that matters is that it is alternating right?

[sly sierra]

yes correct, you can use similar reasoning to what I said earlier
if you delete the n=1 term then the first term is positive and the other conditions still hold

or

you can factor out a -1 from your series

and you'll get $$(-1)\sum_{n=1}^\infty (-1)^n b_n$$

[twin meteor]

Bungo

[silver berry]

ah I see many ways to manipulate to the correct form

[sly sierra]

yep!

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[long flame]

what rule is being used for r cos (theta + pi/6) = r cos theta cos pi/6 - r sin theta sin pi/6 ?

forgive format im on my phone

[dawn kestrel]

angle addition formula. cos(a+b) = cos(a)cos(b)-sin(a)sin(b)

and the r just distributes

[long flame]

great, ok!

hmm ok so r cos(a+b) = r cos(a) cos(b) - r sin(a) sin(b)

i didn't know that it distributed as such

ik obv a (b+c) = ab + ac

[dawn kestrel]

yes its just rcos(a+b) = r[cos(a)cos(b)-sin(a)sin(b)] = rcos(a)cos(b) - rsin(a)sin(b)

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[tardy orchid]

hello

[tardy orchid]

can someone explain this to me real quick

how can a function's absolute value be greater than the absolute value of its own antiderivative

[glossy maple]

think of f(x) as a function that becomes positive and negative a few times

the absolute value of the integral is just the signed area, in absolute value

meaning you add area above the x-axis, subtract area below the x-axis

if you take the integral of the absolute value of f(x), you are ONLY adding areas

[tardy orchid]

how is that any different from say

|f(x^3)|

ur doing the exact same thing

are u not

[glossy maple]

what

[tardy orchid]

the function f = x^3, if u were to take it's absolute value then you'd only be left with positive numbers

[glossy maple]

what are the integral bounds?

this is about area

[tardy orchid]

does it matter? since im assuming said rule is true for all bounds

but for the sake of it

[glossy maple]

here is an example

[tardy orchid]

yea

[glossy maple]

im just to show u intuitively why this is true

[tardy orchid]

okay

[glossy maple]

Let's say these are the areas

So our integral is A - B + C, right?

and then we take the absolute value.

this is

$\abs{\int_a^b \pmap{f}{x} \dd x}$

[twin meteor]

RedstonePlayz09

[tardy orchid]

yeah

[glossy maple]

so now let's look at |f(x)|

which is the red graph here

so if we take its integral, we have

A + B + C

[tardy orchid]

ahhhhhhhhhhhhhhhhhh

yeah

i get u

[glossy maple]

so we are not subtracting the areas under the x-axis anymore

we are ADDING them

[tardy orchid]

yeah yeah yeah

i get it now

[glossy maple]

so obviously we get a bigger value then if we subtracted some of them

that's the intuition

[tardy orchid]

yeah or an equal one at the least

ty

[glossy maple]

np

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[agile meteor]

Kinda need somebody practical with statistics. I have a question regarding confidence intervals.

[agile meteor]

The majority of the exercise i have asks me to compute confidence intervals at a certain percentage, say 95%

so the first thing i do is usually set

1 - alpha = 95%
=> alpha = 0.05

and now im usually supposed to use either gaussian, chi square or t student

say i wanna compute a bilateral interval

for a gaussian distribution with known sigma and mu = x bar

does q need to be Phi(alpha/2)^-1 or Phi(1 - alpha/2)^-1 (^-1 as in inverse function)

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[wraith stirrup]

I need help

[sly sierra]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wraith stirrup]

1

[sly sierra]

do you have any thoughts about f([0,1])? do you know what the notation means?

[wraith stirrup]

I have this but they are wrong

[sly sierra]

can f(x) ever be 0?

[wraith stirrup]

no I guess

because x will be positive

[sly sierra]

x can be anything

that's what f : R -> R means

x is any real number

[wraith stirrup]

yeah but after |x| + 2

[sly sierra]

oh, that's a 2?

i read it as |x| + 1

[wraith stirrup]

yeah

[sly sierra]

ok

what is the smallest value that |x| + 2 could possibly have?

[wraith stirrup]

2

[sly sierra]

right good

and that happens when x is what?

[wraith stirrup]

0

[sly sierra]

right

how about when x is 1

what's the value then

[wraith stirrup]

3

[sly sierra]

yes

so f(0) is 2 and f(1) is 3

what about when x is between 0 and 1

what can you say about f(x)

[wraith stirrup]

Idk

f(x) will also be real number?

[sly sierra]

well a real number in what range

[wraith stirrup]

(0, +inf)

[sly sierra]

if x is between 0 and 1

how big can f(x) be

[wraith stirrup]

well (0, 3)

[sly sierra]

can it go below 2?

[wraith stirrup]

no

[sly sierra]

so you mean (2, 3)?

[wraith stirrup]

yeah

[sly sierra]

so what should the first answer be

given what we just discussed

[wraith stirrup]

?

[tranquil lichen]

f([0,1]) damn we can put an array to a single variable now?

[sly sierra]

no, that's not what the notation means

[sly sierra]

well (2,3) is an open interval

it does not include the endpoints

should they be included?

[wraith stirrup]

well from (2, +inf)

[sly sierra]

wait, how did we get above 3

[wraith stirrup]

Wait wait 0 and 1 are ranges for x right?

well then 2 and 3 should be included [2, 3]

[sly sierra]

yes

[sly sierra]

^ yes this is the correct answer

[wraith stirrup]

ok what about f^-1

[sly sierra]

do you know what f^-1({2}) means? can you say it in words?

[wraith stirrup]

inverse function

[sly sierra]

not exactly no

inverse image

what does that mean

[wraith stirrup]

Idk

[sly sierra]

is this notation defined somewhere in your notes/book/course?

[wraith stirrup]

I only know that f(x) = y

so if its inverse

[sly sierra]

i recommend looking it up, i find it hard to believe someone assigned you a problem without defining the notation

[wraith stirrup]

(-2, +inf)?

[sly sierra]

are you just guessing things now?

[wraith stirrup]

thats the inverse image of |x| + 2 right?

[sly sierra]

say in words what f^-1({2}) means
there's no way you're gonna get it right if you don't understand what is being asked for

[wraith stirrup]

|x| + 2 > 0; |x| > -2

[wraith stirrup]

@sly sierra can you confirm this?

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[blissful gale]

need help w physics,ik this is a maths channel

[sly sierra]

fire away

[blissful gale]

what formula

[sly sierra]

wait, that can't possibly be the entire question

they tell you nothing about the light bulb?

[blissful gale]

lol i was shocked too

[sly sierra]

show what is above this

[blissful gale]

nothing

💀

[sly sierra]

no way

[blissful gale]

i just circled the c option

[sly sierra]

i can even see the shadows of some characters, there is something there

[blissful gale]

😭

so u have no clue?

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[pearl axle]

hey guys, how do i solve this problem?

[sly sierra]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pearl axle]

I have begun but got stuck

I know that in order for the augmented matrix to have no solutions, the rightmost column of the last row should be a leading entry of 1

@sly sierra

<@&286206848099549185>

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[uncut blaze]

Can someone help me with integration?

[uncut blaze]

I am looking to find the area between the curve, and need to integrate e^x - xe^x

between the 0 and 1.

I keep getting lost through the integration and am not understanding how the integration calculators are deriving their answers

[cobalt ocean]

e^x - xe^x = (1-x)e^x so you can just do integration by parts to get rid of the 1-x

[uncut blaze]

Mind explaining why I would do that? Apologies, I'm new to this and don't quite see how that makes it easier

And thanks for the help

@cobalt ocean

[cobalt ocean]

wdym

[uncut blaze]

why would I rewrite it to (1-x)e^x?

[cobalt ocean]

basically you have then integral f(x)g(x)

where the derivative of f is just like -1

so when you do the IBP and differentiate f and integrate g it becomes simpler

and then it just works out

like whenever you have integral of two things multiplied together you should be looking to try IBP basically

differentiating whichever one disappears when you differentiate it

[uncut blaze]

Okay I'll give that a try. Really appreciate it

[vocal sleet]

@uncut blaze Has your question been resolved?

[uncut blaze]

@cobalt ocean Wow I just did it and it was so much easier that way! Any advice on how I can know how to write a function differently (i.e. factor something out or just find a way to isolate the x or 1-x in this case) so that IBP is easier?

Just want to understand how to look at it logically and figure that out on my own

[cobalt ocean]

i mean tbh it didn't really matter

you could integrate e^x and then do IBP on xe^x separately

and that would also work

maybe minutely easier if you factorise first but yeah

i just factorised because it was a little simpler

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[uncut blaze]

.reopen

[vocal sleet]

✅

[uncut blaze]

@cobalt ocean Can you tell me where I went wrong here? I did it correctly when it was written (1-x)e^x but when I try as its written I keep getting stuck

[cobalt ocean]

i can't see that file

[hushed pewter]

convert heic to png or jpeg please

[uncut blaze]

Ok will do

@cobalt ocean @hushed pewter is this visible?

[cobalt ocean]

yes

[hushed pewter]

yup

[cobalt ocean]

your working isn't super clear

$$\int xe^x dx = xe^x - e^x$$ is correct

[twin meteor]

Kaisheng21

[cobalt ocean]

i think it's just an algebra mistake

it's integral **-**xe^x

so it should be like e^x - (xe^x - e^x)?

not e^x + (xe^x - e^x)

[uncut blaze]

That makes so much sense...

Really appreciate your help

.close

[vocal sleet]

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[urban edge]

party time

[thin vale]

I'm having some trouble proving this and likely that's just because I'm not understanding the definitions/notation.

[patent nymph]

What is your progress so far

[thin vale]

really no progress

I've tried a few times using my definition

M(T) is the mxn matrix whose entries are given by

Tvk=A_(1,k)w1+...+A_(m,k)wm

but nothing from it

[flat whale]

Write down M(T+S)

[patent nymph]

Is w_i the basis?

[thin vale]

yes v_1,.... v_n a basis for V, and w_1, ... w_m a basis for W

[thin vale]

M(T+S) has entries given by
(T+S)vk=B_(1,k)w1+....+B_(m,k)wm

is that what you want?

[flat whale]

Sorta

What's A

[flat whale]

Shouldn't be the same A here

[thin vale]

I'm not sure what A is, I think it's just whatever we need it to be so the matrix works out

changed it to B

[flat whale]

Well you need to learn what A does

[thin vale]

any A_(i,j) is just a field element

A is like the top row of our matrix

well it's just

[thin vale]

A isn't anything I don't think, it's just the entries that define M(T)

[flat whale]

Yes okay

Actually it's easier to start by writing down M(S) + M(T) first

[thin vale]

okay so,
M(S) has entries
Sv_k=C_(1,k)w1+....+C_(m,k)wm

M(T) has entries
Tv_k=D_(1,k)w1+....+D_(m,k)wm

M(S)+M(T) has entries

Sv_k+Tv_k=C_(1,k)w1+....+C_(m,k)wm+D_(1,k)w1+....+D_(m,k)wm

=(C_(1,k)+D_(1,k))w1 + ...... (C_(m,k)+D_(m,k)wm

which I'm probably suppose to relate to M(S+T) somehow

but I don't see how because at this point I'm kind of lost in the sea of subscripts

<@&286206848099549185>

[flint frigate]

Poor guy 😭 , I don’t think more than 5 people in this discord can solve this problem

[wraith venture]

This is basic linalg, so there's hundreds of such people

Unfortunately, even though I am able, I need to sleep rn

[thin vale]

basic enough for everyone to comment on how basic it is

but not basic enough to help easily

:(

XD

[wraith venture]

(S+T)(vk) = sum of the Cik wi + sum of the Dik wi = sum of the (Cik + Dik) wi

[sly sierra]

do i espy an axler screenshot?

[thin vale]

I already wrote this out

am I done without realizing it

I'd buy that

[wraith venture]

The second is M(S)vk + M(T)vk
The third is M(S+T)vk by definition

[thin vale]

are you an axler fan

[sly sierra]

i'd say i'm 75% an axler fan

[thin vale]

the 25%?

[sly sierra]

the treatment of the characteristic polynomial is a shitshow

[thin vale]

haven't got there yet

[wraith venture]

Yeah that's usually linalg 2 material

[thin vale]

this is linalg 2

:(

we're just slow and I don't get this notation

should be an easy problem

[worthy citrus]

how to make people hate linear algebra 101

[thin vale]

characterstic polynomial?

[worthy citrus]

your question

[thin vale]

why

[worthy citrus]

are you having fun?

[thin vale]

no I'm just grinding notation

[sly sierra]

you mean besides making them do a bunch of row reductions and skimp on the theory like most linalg 1 classes?

[surreal basin]

at least if that happens its a very linear class

[worthy citrus]

i had to mark a bunch of row reductions from an alg exam this week, it was hell

[surreal basin]

even more linear than algebra with y = mx + b

[sly sierra]

ugh you have my sympathies, sounds bad

[sly sierra]

that's affine algebra

[thin vale]

I'm confused on if I've solved this or not

[wraith venture]

It's just linearity all the way

[flat whale]

.

You're done

[wraith venture]

Just notice the formula you have is indeed an isomorphism

[thin vale]

$$Sv_k=C_{1,k}w_1+\dots+C_{m,k}w_m$$
$$Tv_k=D_{1,k}w_1+\dots+C_{m,k}w_m$$
$$Sv_k+Tv_k=C_{1,k}w_1+\dots+C_{m,k}w_m+D_{1,k}w_1+\dots+C_{m,k}w_m$$
$$=(C_{1,k}+D_{1,k})w_1+\dots +(C_{m,k}+D_{m,k})w_m=(S+T)v_k$$

isn't it already obvious that Svk+Tvk=(S+T)vk

wtf am I even doing

[wraith venture]

You're literally hiding the proof between your equalities because it is actually that computationally obvious

[thin vale]

where did i prove anything

[worthy citrus]

you just need some more names for things, C be the name of the matrix for S, D for T and A (or whatever you want) for S+T

[wraith venture]

When you skipped saying Cik wi + Dik wi = (Cik + Dik) wi

[worthy citrus]

then you're just showing that A = C + D using the stuff you've been doing

[twin meteor]

Austin

[worthy citrus]

you need to write down Svk + Tvk with no reference to C or D first

[thin vale]

huh

how

[surreal basin]

certified proof moment

[sly sierra]

all this componentwise spam is obscuring what is fairly simple:

[worthy citrus]

just call the matrix that represents S + T, A

[thin vale]

I want to call it M(S+T)

can we do that

[worthy citrus]

thats going to look horrible

[thin vale]

okay whatever then it's A

[sly sierra]

the k'th column of M(S) is the vector S(vk), expressed in w coordinates
the k'th column of M(T) is the vector T(vk), expressed in w coordinates
S+T maps vk to S(vk) + T(vk)
the k'th column of M(S) + M(T) is the k'th column of M(S) plus the k'th column of M(T)
conclude

[wraith venture]

Letting A = M(S+T), notice that (S+T)vk = sum of the Aik wi
Therefore deduce that it must be that Cik + Dik = Aik
(Because it holds for all vk)

[worthy citrus]

and you know that (S+T)vk = Svk + Tvk

you have all the pieces now

its up to you

[thin vale]

conclusion
the k'th column of M(S)+M(T) is the vector S(vk)+T(vk) expressed in w coordinates which is mapped from (S+T)vk?

wait no

[sly sierra]

the k'th column of M(S)+M(T) is the vector S(vk)+T(vk) expressed in w coordinates which is the same as the vector (S+T)vk expresed in w coordinates
thus the matrix of S+T with respect to these bases is M(S) + M(T)

[thin vale]

the k'th column of M(S)+M(T) is S(vk)+T(vk)

which is (S+T)vk

was this ever confusing to anyone else or only me

[sly sierra]

it was always confusing to most people at first, just because the notation is messy and because when first encountering it, the idea of abstract linear maps as distinct from matrices is still a novel thing

[thin vale]

Okay I do think I understand this

Thank you Bungo et. al

I have office hours so I have to go now

.close

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[compact sluice]

I’m confused on how to find the coterminal for this problem

It says π−θ on the chart but I feel like that’s not the final solution since you need to find one negative and positive coterminal

[wraith python]

A coterminal angle has a difference of ±360 degrees, or ±2pi.

[compact sluice]

Ohhh ok

Tysm!!!!

.close

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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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[brisk dawn]

I am being asked to prove $$e^x = \sum_{k=0}^{n} \frac{x^k}{k!} + \frac{1}{n!} \int_{0}^{x} e^t(x-t)^n \text{d}t$$ for $n = 1$, then to prove it by induction for $n = n + 1$, assuming it works for the base case of $n$. I proved the equation for $n = 1$ as such:

$$e^x = \sum_{k=0}^{n} \frac{x^k}{k!} + \frac{1}{n!} \int_{0}^{x} e^t(x-t)^n \text{d}t$$
$$e^x = \frac{x^0}{0!} + \frac{x^1}{1!} + \int_{0}^{x} e^t(x-t) \text{d}t$$

Solving the integral by integration of parts, we get $e^x - x - 1$.

$$e^x = x + 1 + e^x - x - 1 = e^x$$

The identity holds for n = 1. However, when attempting to plug in $n + 1$ for n, I ran into difficulties trying to solve the integral. Could anyone help me with this?

[twin meteor]

PA

[brisk dawn]

so basically, figure out how to prove $$e^x = \sum_{k=0}^{n+1} \frac{x^k}{k!} + \frac{1}{(n+1)!} \int_{0}^{x} e^t(x-t)^{n+1} \text{d}t$$

[twin meteor]

PA

[hushed pewter]

Interesting that e^0 would not be the base case. It's certainly simpler.

[brisk dawn]

¯_(ツ)_/¯

i guess my professor wanted to challenge us

[hushed pewter]

My guess, if you're using induction, try integration by parts again

[brisk dawn]

i did

i got

[hushed pewter]

Perhaps you can transform the integral into one where your assumption comes into play'

[brisk dawn]

$$e^t((x-t)^{n+1}) - (n+1) \int e^t(x-t)^n \text{d}t$$

[twin meteor]

PA

[brisk dawn]

idk what to do with that integral

[hushed pewter]

Use your inductive assumption

[brisk dawn]

uhh

idk like

how to do that

my guess is that i dont really understand induction

dyk a useful video or smth, cuz i know that integral is the assumed case

[hushed pewter]

You are assuming $$e^x = \sum{k=0}^{n} \frac{x^k}{k!} + \frac{1}{n!} \int{0}^{x} e^t(x-t)^n \text{d}t$$
are you not?

[twin meteor]

SWR

[hushed pewter]

err

[brisk dawn]

yes

i am

im assuming it is proven for n

[hushed pewter]

$$e^x = \sum{k=0}^{n} \frac{x^k}{k!} + \frac{1}{n!} \int{0}^{x} e^t(x-t)^n \text{d}t$$

[twin meteor]

SWR

[brisk dawn]

im not sure exactly what to do with the assumed case

[hushed pewter]

Okay, so solve for $\int{0}^{x} e^t(x-t)^n \text{d}t$

[twin meteor]

SWR

[hushed pewter]

why do I keep getting 0 dammit?

$\int_{0}^{x} e^t(x-t)^n \text{d}t$

[twin meteor]

SWR

[brisk dawn]

use _

yea

ok wait lemme try integrating that

i think i tried before

and there was smth weird

[hushed pewter]

you don't need to

[brisk dawn]

y?

[hushed pewter]

You are assuming
$$e^x = \sum_{k=0}^{n} \frac{x^k}{k!} + \frac{1}{n!} \int_{0}^{x} e^t(x-t)^n \text{d}t$$
So you can use that to solve for $\int_{0}^{x} e^t(x-t)^n \text{d}t$

[twin meteor]

SWR

[brisk dawn]

wait

so

is it.

n! - summation of x^k/k!?

[hushed pewter]

don't forget the e^x

[brisk dawn]

oh

waiot

$$n! \cdot e^x - \sum_{k=0}^{n} \frac{x^k}{k!}$$

[twin meteor]

PA

[brisk dawn]

is that the integral?

[hushed pewter]

Almost. Your algebra needs a little correcting, but you seem to have the idea

[brisk dawn]

yea

ok wait

idk how to get rid of the summation term

$$n! \cdot (e^x \cdot n!) \cdot (n! \cdot \sum_{k=0}^{n} \frac{-x^k}{k!}$$

[twin meteor]

PA

[brisk dawn]

hold on

this is def wrong

$$e^x = \sum_{k=0}^{n} \frac{x^k}{k!} + \frac{1}{n!} \int_{0}^{x} e^t(x-t)^n \text{d}t$$

[twin meteor]

PA

[hushed pewter]

You're not trying to get rid of it

[brisk dawn]

why?

[hushed pewter]

Or at least, I never asked you too

[brisk dawn]

but i need to prove the base case to be true

where e^x = e^x

so i need to get rid of it

[hushed pewter]

I believe a communication error is happening between us

One moment while I try to clarify where we are both at right now

[brisk dawn]

i want rhs to be e^x right?

so i would need to make the integral equal something which makes rhs e^x

[hushed pewter]

To save on some writing, I'm gonna make a few shorthands. Let's define $$S(n)=\sum_{k=0}^n \frac{x^k}{k!}$$ and $$I(n)=\int_0^x e^t(x-t)^n dt$$
That means we are trying to prove $e^x=S(n)+\frac{1}{n!}I(n)$ for every $n$. Yeah?

[brisk dawn]

ohh

yes i understand that now

[twin meteor]

SWR

[hushed pewter]

Made a minor correction

[brisk dawn]

yes i understand

[hushed pewter]

So, you have already proven $e^x=S(1)+\frac{1}{1!}I(1)$, correct?

[twin meteor]

SWR

[brisk dawn]

yes

[hushed pewter]

Technically, you should've first proved $e^x=S(1)+\frac{1}{0!}I(0)$ first, but whatever. For fun, you can still do it if you want.

[twin meteor]

SWR

[hushed pewter]

Okay. Here's the next task for induction

You are going to ASSUME that $e^x=S(n)+\frac{1}{n!}I(n)$ is true for some particular $n\ge 1$. You are not trying to prove that this is true. It is just an assumption and we get use this is assumption in our following work.

[twin meteor]

SWR

[hushed pewter]

With this assumption, you must now PROVE that $e^x=S(n+1)+\frac{1}{(n+1)!}I(n+1)$ will also be true.

[twin meteor]

SWR

[brisk dawn]

oh ok i get it

so like

[hushed pewter]

Here are some hints on how you can do it. Recall that, for any sum, $$\sum_{k=a}^b u_k=\left[\sum_{k=a}^c u_k\right]+\left[\sum_{k=c+1}^b u_k\right]$$

[brisk dawn]

you need to have a base case to prove, then u have to prove the generic case plus 1 (n+1) in this scenario with the assumption of the generic case being true

[twin meteor]

SWR

[hushed pewter]

correct

[hushed pewter]

For any integer c

[brisk dawn]

can i keep this channel open in case i have questions?

[hushed pewter]

So, you can express $S(n+1)$ in terms of $S(n)$ and some other stuff.

[brisk dawn]

im going to try doing this rn

[twin meteor]

SWR

[brisk dawn]

ill close within 10

[hushed pewter]

And, using integration by parts, you can express $I(n+1)$ in terms of $I(n)$

[twin meteor]

SWR

[hushed pewter]

Keep it open as long as you need

[brisk dawn]

should i try to rewrite this as S(n) + 1/n! I(n) @hushed pewter

[hushed pewter]

If you can end up doing that, that would be sufficient, yes

[brisk dawn]

S(n+1) = S(n) + S(1)?

[hushed pewter]

Not exactly. It would be S(n+1)=S(n)+<something>, but not S(1).

[brisk dawn]

wait

ok

[hushed pewter]

Use this identity

[brisk dawn]

yup

wait is it

S(n) + S(1-1)? 😭

wait let me just

test that

oh

i see it

$$S(n + 1) = S(n) + \frac{x^{n+1}}{(n+1)!}$$

[twin meteor]

PA

[hushed pewter]

Brilliant

Exactly correct

Next job: I(n+1)

[brisk dawn]

ook

ok so $$\int fg' = fg - \int f'g$$

[twin meteor]

PA

[brisk dawn]

$$g' = e^t, g = e^t, f = (x-t)^{n+1}, f' = (n+1)(x-t)^n$$

[twin meteor]

PA

[brisk dawn]

ok so from this i get

$$e^t(x-t)^{n + 1} - \int (n+1)(x-t)^n e^t \text{d}t$$

[twin meteor]

PA

[brisk dawn]

n+1 is a constant

i can move it outside right?

[hushed pewter]

wait

you need to evaluate the definite integral

[brisk dawn]

theres no definite integral here tho

[hushed pewter]

But there is in your problem

[brisk dawn]

right

so i can just plug into fundamental thm of calculus

at the end

no?

[hushed pewter]

yes

so long as we don't forget

[brisk dawn]

yep

$$e^t(x-t)^{n + 1} + (n + 1) - \int e^t(x-t)^n \text{d}t$$

[twin meteor]

PA

[brisk dawn]

uh the issue is that

that integral isnt I(n)

cuz its not definite

[hushed pewter]

wdym? I don't follow

[brisk dawn]

$$\int e^t(x-t)^n \text{d}t$$ is not equal to $$\int_{0}^{x} e^t(x-t)^n \text{d}t$$

[twin meteor]

PA

[brisk dawn]

cuz one has bounds, one doesnt

[hushed pewter]

What does that change for you though?

[brisk dawn]

i cant express as I(n) then

unless u want me to use the fundamental thm of calculus now

[hushed pewter]

You can do it whenever you want

But I(n+1) uses a definite integral also, you eventually have to end up with a definite integral

[brisk dawn]

should i use the fundamental thm now?

[brisk dawn]

idk how to do it cuz i still have this indefinite integral

[hushed pewter]

yes

[hushed pewter]

If you're unsure, you should ask yourself how you ended up with an indefinite integral in the first place.

[brisk dawn]

i used integration by parts

[hushed pewter]

Yes but of what?

[brisk dawn]

oh

i started

with an indefinite integral

[hushed pewter]

When you should not have. Correct

[brisk dawn]

how would i integrate by parts with the bound then

every way of solving it ive seen uses an indefinite integral

[hushed pewter]

Are you familiar with the notation $\int_a^b f(x)dx=\left[F(x)\right]_{x=a}^b$?

[twin meteor]

SWR

[brisk dawn]

yep

idk how to apply that here

[brisk dawn]

= F(b) - F(a) right

[hushed pewter]

$$\int_0^x e^t(x-t)^{n+1}dt=\left[e^t(x-t)^{n + 1}\right]_{t=0}^x + (n + 1) \int_0^x e^t(x-t)^n \text{d}t$$

[twin meteor]

SWR

[hushed pewter]

F(b)-F(a)

[brisk dawn]

right

[brisk dawn]

so then i evaluate F(x) - F(0)

[hushed pewter]

yup

[brisk dawn]

$$F(x) - F(0) = (e^t(x-x)^{n+1}) - (e^t(x-0)^{n+1}) = e^t(1 - x^{n+1})$$

[twin meteor]

PA

[hushed pewter]

e^t should not remain

[brisk dawn]

oh right i forgot to plug it in 😭

$$F(x) - F(0) = (e^x(x-x)^{n+1}) - (e^0(x-0)^{n+1}) = e^x - x^{n+1}$$

[twin meteor]

PA

[hushed pewter]

Perhaps take another look at x-x

[brisk dawn]

x-x = 0

so

e^x(0)^n+1

= e^x

wait

💀

= 0

[hushed pewter]

yea

[brisk dawn]

$$F(x) - F(0) = (e^x(x-x)^{n+1}) - (e^0(x-0)^{n+1}) = - x^{n+1}$$

[twin meteor]

PA

[hushed pewter]

okay

So very close

[brisk dawn]

ok

so

$$S(n + 1) = S(n) + \frac{x^{n+1}}{(n+1)!}$$
$$I(n + 1) = -I(n) - x^{n+1} + n + 1$$

[twin meteor]

PA

[brisk dawn]

u need to multiply I(n+1) by 1/n!

wait

[hushed pewter]

Made a few mistakes in I(n+1)

[brisk dawn]

no way 😭

$$-x^{n+1} + (n + 1) - \int_{0}^{x} e^t(x-t)^n \text{d}t$$

[twin meteor]

PA

[brisk dawn]

isnt this just

$$-x^{n+1} + n + 1 - I(n)$$

[twin meteor]

PA

[hushed pewter]

Where did n+1 come from?

[brisk dawn]

when i moved it out of the integral

[brisk dawn]

.

[hushed pewter]

why is it an additive term?

[brisk dawn]

oh

$$-x^{n+1} + I(n)(n + 1)$$

[twin meteor]

PA

[hushed pewter]

huzzah

I(n+1)=...

[brisk dawn]

$$I(n + 1) = I(n)(n+1) -x^{n+1}$$

[twin meteor]

PA

[brisk dawn]

then

[hushed pewter]

👍

[brisk dawn]

oh i think

i see it

wait do i need to multiply 1/(n+1)! by this term

yes right

$$\frac{1}{(n+1)!}(I(n)(n+1) - x^{n+1}) = \frac{I(n)(n+1)}{(n+1)!} - \frac{x^{n+1}}{(n+1)!}$$

[twin meteor]

PA

[brisk dawn]

OH

i see it

u cancel (n+1) and (n+1)!

to get

1/n! * I(n)

then subtract. that term from S(n+1) to get S(n)

then u get the assumed case

[hushed pewter]

good work

Interesting problem

[brisk dawn]

tysm 😄

[hushed pewter]

Welcome to the server'

[brisk dawn]

.close

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[waxen hawk]

f(-x)=-f(x)

[waxen hawk]

that is the definition of odd functions

How to describe such definition in terms of geometry

[frail jungle]

graphs?

[waxen hawk]

yes

[inner osprey]

rotating the function 180 degrees about the origin gives the same graph

[frail jungle]

do you want an example or a definition

the best example is sin x and cos x

both of them show two different functional outputs for the same input

understanding these two would be enough to understand the concept

[waxen hawk]

rotate it clockwise or counter-clockwise

[woeful cobalt]

or x^3 is another example

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@waxen hawk Has your question been resolved?

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[cerulean brook]

whats next?

[pallid zenith]

do you understand what the problem is with just shrinking down the larger map?

why cant we just scale down a 9x15 map and put it on a 4x6 card?

[cerulean brook]

idk

do you think you can help me find the largest possible complete map?

[pallid zenith]

yea

im trying to think of a way to explain it

you have a problem with the length vs width

do you see why like

if you started with a cat picture thats a rectangle

and we wanted to fit this on a 2x2 card

we arent going to be able to this without squishing it right?

OR, its not going to cover the entire card

with me?

[cerulean brook]

yeah

[pallid zenith]

so the problem is we have to shrink the image down enough that the biggest dimension is small enough to fit on the card

[cerulean brook]

yea

[pallid zenith]

you can do this with ratios like, the ratio between the height and the length will always be 9/15

so, lets say you make the height 4 inches

how wide will it be?

you can do this by solving $$\frac{H}{W} = \frac{9}{15} = \frac 4x$$

[twin meteor]

jan Niku

[pallid zenith]

for x

[cerulean brook]

why is the 6 turned into x

[pallid zenith]

because if you divide the height by the width, it has to be 9/15

this is a requirement in order to avoid squishing or stretching the image

[cerulean brook]

i understand but im a bit confused because

[pallid zenith]

we use x instead of 6 for the same reason that the cat picture i showed can't fit on a square

[cerulean brook]

the ratio of the original was 9/15 and the smaller one wants to be put on 4/6

[pallid zenith]

right

so, like

say we shrunk the map

each side by half

so now it will be $\frac{4.5}{7.5}$

[twin meteor]

jan Niku

[cerulean brook]

yeah

[pallid zenith]

do you see how we dont get to decide both numbers?

i mean, we can change the height

but if we do that, it determines how we have to change the length

we cant make it $\frac{4}{15}$

[twin meteor]

jan Niku

[pallid zenith]

(from 9/15)

because then it will be squished

[cerulean brook]

x is approx 6.666

[pallid zenith]

so, we can choose to make the height 4, but we need to know what to make the length so that the bap isnt squished or stretched out

[cerulean brook]

do we round it to a whole number, 6

[pallid zenith]

,w solve 9/15==4/x

[twin meteor]

Results provided by WolframAlpha

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Requested by janniku

[pallid zenith]

,calc 20/3

[twin meteor]

Result:

6.6666666666667

[pallid zenith]

close enough

[pallid zenith]

no

[cerulean brook]

uhhh

[pallid zenith]

the answer is telling you this

if you make the map 4 inches high

it will be over 6 inches wide

so will it fit on the card?

👀

[cerulean brook]

no

[pallid zenith]

yea

so maybe we can make the width 6 inches instead

and try to figure out how tall it would be

can you tell what the equation would be for this?

[cerulean brook]

no

[pallid zenith]

so we want the ratio to be 9/15

but now we are flipped

$\frac{9}{15} = \frac x6$

[twin meteor]

jan Niku

[cerulean brook]

3.6

[pallid zenith]

yea

so it fits on a 4x6 card now?

[cerulean brook]

no

yes

[pallid zenith]

idk what to say to that haha

[cerulean brook]

it says 4:6

does the number need to be 4

or it can be under

[pallid zenith]

well, it has to be under right

like a rectangular cat picture cant cover an entire square

[cerulean brook]

why cant it be just 4

[pallid zenith]

without us squishing thi image

[cerulean brook]

4 should fit

[pallid zenith]

well, we solved for that actually

if the map is 4 inches high

it would have to be over 6 inches wide

otherwise itd be squished

so, we have to make it less than 4 inches high

otherwise, its too wide

[cerulean brook]

too confusing

very lost right now

[pallid zenith]

lemme see if i can find a good image

man i wish i we still had streaming channels

can you hang out for a second? I will make an image

i just have to load up photoshop

[cerulean brook]

kk

[cerulean brook]

i can just close ticket and dm

[pallid zenith]

i mean a picture will work as well

[cerulean brook]

kk

[pallid zenith]

okay, so here is a map thats 9/15

if i just shrink this map down to 4/6, ill get (one sec

can you tell how it is squished?

everything is thinner and taller

its less dramatic than i was hoping

lemme see if someone with a better answer than me is around

[cerulean brook]

yeah

i opened the originali image

i see how its squished

[pallid zenith]

thats because we didnt maintain the ratio

[cerulean brook]

yes

[pallid zenith]

so thats the problem

thats why we cant make the height 4

what i mean is, if we make the height 4, and the width 6

[cerulean brook]

okay so how do we always maintain an aspect raito

[pallid zenith]

it will squish the image

[pallid zenith]

you pick either the height or the width

and you solve for the other

you use $\frac{H}{W} = \frac{9}{15}$

[twin meteor]

jan Niku

[cerulean brook]

and on how the fractions wiould look

if it asked you to resize from bigger to smaller

you would do 4/6 = 9/x right

[pallid zenith]

this would have you maintaining a ratio of 4/6

to be sure

not 9/15

[cerulean brook]

yea

[cerulean brook]

just a random question i made

[pallid zenith]

so yea, if youre given an image thats 4/6 and you want to rescale it from that size to a bigger canvas sure