Mnaclaurin series question.

[plush dirge]

Hello everyone, it appears like I'll need some help. I have a question about the Maclaurin series that looks like this, and I'm not sure how to begin because I can't convert it into a product or anything. I'll need a few ideas here and there.

[real marsh]

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[vocal silo]

you took the n-th derivative yet?

usually when you evaluate it at zero there is some nice stuff that occurs

[plush dirge]

Here is what I did so far, I don't know if what I am doing in e^-X is correct or not

Since all their derivatives are 1 apparently

[vocal silo]

uhm

what?

no you take the n-th derivative evaluated at 0 of e^(-x)/(1-x)

try a base case n=0,1,2

then see if you notice a pattern

[plush dirge]

I need them in product form before I proceed with the maclaurin series, so I just separated them so they are now e^-X × (1-x)^-1 and I began to take their maclaurin series separately

[vocal silo]

both series are known

hmm

[plush dirge]

Oh

[plush dirge]

Could you explain this part for me then please

[vocal silo]

$M(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$

[minor plaza]

dark matter

[vocal silo]

the definition of maclaurin

[plush dirge]

Ummm

[vocal silo]

[plush dirge]

Isn't it supposed to be like

F(x)= f(0)+f'(0)/1!+....

[vocal silo]

it looks like we tryna prove $\frac{e^{-x}}{1-x}=\sum{n=0}^{\infty}\frac{x^n}{n}$ right

[minor plaza]

dark matter

[vocal silo]

what

nonono

[plush dirge]

Hm

Yea

[plush dirge]

Yea , by the nth derivative, you mean I should be differentiating the equations couple of times or?

[vocal silo]

the whole point of maclaurin is to make an approximate polynomial

how are constants gonna help with that

@plush dirge you there?

[dusk whale]

Hm

[vocal silo]

eh i’ll do the first derivative

[dusk whale]

We need an nth derivative of the function

Or we can just find a pattern and try to prove it via induction, whatever

[vocal silo]

$\frac{-e^{-x}(1-x)-(-1)(e^{-x})}{(1-x)^2}$

[dusk whale]

1st derivative is
xe^{-x}/(1-x)^2

[minor plaza]

dark matter

[dusk whale]

Nice cancellation occurs

[vocal silo]

yeah

wait

[dusk whale]

x = 0 we get wait ωτƒ

Ok what the hell

[vocal silo]

,texsp ||$\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}\cdot\sum_{n=0}^{\infty}x^n$||

[minor plaza]

dark matter

[dusk whale]

e^{-x} = (-x)^n/n!

[vocal silo]

i have a sneaky feeling this is a sum abuse notation

[dusk whale]

Yeah

[vocal silo]

but this wouldn’t make sense

[dusk whale]

Maybe try multiplying finite polynomials?

[vocal silo]

uhh

[dusk whale]

Nopw

[vocal silo]

,texsp ||$\sum_{n=1}^{\infty}\frac{x^n}{n}=\ln(1-x)$||

[dusk whale]

The maclaurin series is saying that the function is literally equal to ln(1-x)

It’s a minus

[minor plaza]

dark matter

[vocal silo]

what the heck

[dusk whale]

Yeah-

[vocal silo]

one second

[dusk whale]

So W(e^{-x}) = ln(1-x) ???

[vocal silo]

what

[dusk whale]

e^{-x} = (1-x)Ln(1-x)

Ohhhhh

[vocal silo]

wait a minute

differentiate both sides

[dusk whale]

Lmaoooo

[vocal silo]

that’s so off

wait

[dusk whale]

Wait how tf do you get factorials-

[vocal silo]

$\sum_{n=1}^{\infty}x^n=\frac{xe^{-x}}{(1-x)^2}$

[minor plaza]

dark matter

[vocal silo]

that’s what they’re saying

[dusk whale]

Something looks- wait divide both sides by x

[vocal silo]

oh yeahhhh

[dusk whale]

We got this as well

[vocal silo]

$\sum_{n=0}^{\infty}x^n=\frac{e^{-x}}{(1-x)^2}$

[minor plaza]

dark matter

[vocal silo]

oh my god w

[dusk whale]

Ermmmm

There would also be a 1/x term in there

[vocal silo]

where

[dusk whale]

LHS

[vocal silo]

what

are you okay

[dusk whale]

the infinite Σ was originally from 0 to ∞

[vocal silo]

i changed the lower bound

it was from 1 to infinity

[dusk whale]

What

[vocal silo]

ye

[dusk whale]

Wait

[vocal silo]

look at og expression he posted

[dusk whale]

There’s no x here

[vocal silo]

differentiate that

[dusk whale]

Oh

Bruhhhh

Ok

[vocal silo]

lol

[neat mist]

Cauchy product

[vocal silo]

so $1=\frac{e^{-x}}{1-x}$

[minor plaza]

dark matter

[dusk whale]

What 😭

[vocal silo]

nah

[dusk whale]

What the

[vocal silo]

that’s not real

thad literally what it says

[plush dirge]

Hmm

[dusk whale]

Multiply both sides by 1-x maybe

[vocal silo]

this is 100% false

no way

unless for sufficiently small x

[dusk whale]

Wait

Ωτƒ???

Wait I think we did an illegal move

We divided by 0 😭😭😭

[plush dirge]

Well I basically did this

[dusk whale]

,w Cauchy Product

[minor plaza]

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[dusk whale]

And that sums to-

[neat mist]

if $(a_k)$ and $(b_k)$ such that their series are absolutely convergent then $\sum_{k=0}^{+\infty} a_k \sum_{k=0}^{+\infty} b_k=\sum_{n=0}^{+\infty} \sum_{k=0}^{n} a_k b_{n-k}$

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[dusk whale]

Bruhhh I remember seeing this in analysis

[vocal silo]

shit i shoulda used that

[plush dirge]

@neat mist is what I did is right? Taking maclaurin series for them separately and then multiplying the results with each other in the end

[vocal silo]

ik there was some identity

wait

doesn’t this only converge for x€(-1,1)

[dusk whale]

Think so, yes

Yep

Yes

[neat mist]

Yeah

[vocal silo]

ok

product regardless

[neat mist]

Take this expression witj x^k you have the product of two power series gives you another power series

[dusk whale]

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

[minor plaza]

Ravioli ≥ ^• ∧ •^ ≤

[vocal silo]

bruh

[dusk whale]

???

No

[neat mist]

Yep the x^n can be factored in font

*front

[vocal silo]

wait no

becuase there is factorial

[neat mist]

Basically you have $\sum_{n=0}^{+\infty} c_n x^n$

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[vocal silo]

there isn’t supposed to be factorial in og expression

[neat mist]

With $c_n=\sum_{k=0}^{n} \frac{(-1)^{k}}{k!}$

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[dusk whale]

And the other sequence is literally 1

[neat mist]

Yeah

[vocal silo]

loooool

[neat mist]

b_n-k is just 1

[dusk whale]

Wait we still have factorials

[plush dirge]

Umm what are you guys doing?

[dusk whale]

Trying to murder this problem

Ok seriously

[neat mist]

It’s basically done

[dusk whale]

,w taylor series (e^{-x})/(1-x)

[minor plaza]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by inverse_cupid

[dusk whale]

How 😭😭😭

[neat mist]

You have this

[dusk whale]

But how does that simplify into what we have above

[neat mist]

So $\frac{e^{-x}}{1-x}=\sum_{n=0}^{+\infty} c_n x^{n}$

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[neat mist]

with cn being this

[dusk whale]

But that’s just the taylor series for e^{-x}

[neat mist]

No

Not exactly

Cn is a partial sum

[dusk whale]

Oh-

Didn’t see that, sorry

[neat mist]

I won’t forgive you

[dusk whale]

But then we have to evaluate that sum, right?

[neat mist]

You can’t simplify further I don’t think

[plush dirge]

Umm yea I don't know what to do there since our instructor didn't even say anything about this

[vocal silo]

so we realized your problem is almost impossible

[plush dirge]

Huh?

[vocal silo]

to prove using standard methods

[dusk whale]

I’ll attempt it on paper

First with spamming derivatives

[vocal silo]

this leads to a black hole

[dusk whale]

We did a division by 0 error actually

[neat mist]

I mean there isn’t a closed form for $\sum_{k=0}^{n} \frac{(-1)^{k}}{k!}$

[dusk whale]

We can’t divide by x

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[dusk whale]

Yeah- 😭😭😭

I’ll attempt this

[neat mist]

We just know it’s limit is e^-1 that’s all

[plush dirge]

Rotor did you check my screenshot?

[vocal silo]

because of power series

[vocal silo]

wrong ideas

[vocal silo]

here

its kinda illegal to avoid quotient rule

[plush dirge]

Well, it needs to be in product form

Otherwise, it's extremely difficult to attempt a maclaurin series on the equation

[vocal silo]

wdym

[plush dirge]

Maclaurin series in certain equations can't be done or it's very difficult if it's in quotient form

So by default, I just turn it into product form

[neat mist]

I mean you can use Leibniz’s formula if you want to evaluate the coefficients rigorously

You will get the same thing

[plush dirge]

Leibniz formula?

Kinda need to do a research about it now

[neat mist]

$(fg)^{(n)}=\sum_{k=0}^{n} \binom{n}{k} f^{(k)} g^{(n-k)}$

[plush dirge]

But I mostly use the talyer series

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[plush dirge]

Yeah...I think I am fine with the talyer series now

[neat mist]

I mean, you know that $f(x)=\sum_{n=0}^{+\infty} \frac{f^{(n)}(0)}{n!} x^n$

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[plush dirge]

Yep

That's what i am using currently

[neat mist]

so you must evaluate $f^{(n)}(0)$ for $f(x)=\frac{e^{-x}}{1-x}$

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[neat mist]

set $g(x)=e^{-x}$ and $h(x)=\frac{1}{1-x}$ you have $f=gh$

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[plush dirge]

Yes that's what I did in their actually, but still I don't know why I got it wrong

[neat mist]

So now apply this formula

[median needle]

This feels like there's been a lot more work here than needed to happen?

Sorry to interrupt.

[plush dirge]

Techie!!

[neat mist]

Oh so you applied Leibniz’s formula ?

[median needle]

Also, for the record, MacLaurin series are just a special case of Taylor series.

[plush dirge]

I mostly applied Talyer in their, liebienz was too complicated for me

[median needle]

You forgot to use the chain rule.

[plush dirge]

Wait where?

[median needle]

d/dx(e^-x) = -e^-x

[plush dirge]

Ahhhhh

[neat mist]

This is basically a formula to calculate the nth derivative of a product, but if you haven’t seen it, there is no helping it

[median needle]

But again, working out the series from first principles feels like way too much work.

We know the series for e^x and 1/(1 - x) right?

[neat mist]

Yes I already proposed the method with Cauchy product of two power series

[median needle]

Ah.

[vocal silo]

then that became a mess

[neat mist]

Not really though we had a nice answer

[median needle]

I mean, I won't say the alternating series of reciprocal factorials is pretty, but it should be relatively simple to prove by induction that it equals what we want it to?

At least, at first glance.

[neat mist]

Yeah

[plush dirge]

The question needs me to show that 1+....... , will your method have the same result as the question requires?

[neat mist]

Yes

When you calculate the nth derivative at 0 you obtain the alternating sum of the reciprocals of factorials

When dividing by n!

If you apply the formula correctly its relatively simple

But if you don’t know it and haven’t seen it in class, then maybe thats not what you should do

Like Techie said maybe induction is the answer

[plush dirge]

,rotate

[plush dirge]

Like this?

Since I need to show only 3 derivatives I assume

[neat mist]

Ah you aren’t supposed to get the full expression ?

of the power series

[plush dirge]

It only needs me to show 3 factorials

[neat mist]

Ahh okay

Well then just calculate the first three derivatives

I apologise, I misunderstood the problem

[plush dirge]

And then?

[neat mist]

Well then you are done, $a_0=f(0)$ $a_1=f’(0)$ $a_2=f’’(0)/2$ $a_3=f’’’(0)/6$

[minor plaza]

😑 ɿototoЯ | Rototor 😑
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[neat mist]

With $f(x)=a_0+a_1x+a_2x^2+a_3x^3+…$

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[plush dirge]

Is there an easier and quicker method than doing the derivative of this equation three times?

Since it's not that easy

[dusk whale]

Wait you don’t have to prove the series?

[neat mist]

$e^{-x} \underset{x \to 0}{=}1-x+\frac{x^{2}}{2} -\frac{x^{3}}{6}+o(x^3)$ and $\frac{1}{1-x} \underset{x \to 0}{=} 1+x +x^2+x^3+o(x^{3})$

[minor plaza]

😑 ɿototoЯ | Rototor 😑

[plush dirge]

Read the question

[neat mist]

No it was only the first 3 terms

[dusk whale]

Jumps off the Empire State Building

[neat mist]

😢

[dusk whale]

Bruhhh

[vocal silo]

bro

WHAT

smh

[dusk whale]

Here I was about to do some crazy induction

[neat mist]

@plush dirge do you know the little « o » notation ?

[plush dirge]

Umm no

[neat mist]

If you did, it would’ve made for a faster way to calculate the derivatives

[neat mist]

You can also apply the formula I gave for n=3

Or just do the calculations by hand and torture yourself

Bruh, this feels so unnecessarily long

[vocal silo]

ikr

real torture

[random seal]

@plush dirge The problem I have with the previously proposed expansion approaches is that we can't be fully sure that they're valid everywhere, that is, for every x. In fact, the geometric series expansion will most likely fail for |x| larger than 1, so that's a pretty big concern.

To address this, we have to find a method that will make your expansion valid for all x, or at least, as many as we can. For that, you can leverage the fact that the expansion of exp(-x) is actually valid for ALL x. That is, the series for exp(x) converges for all x

[neat mist]

The question was only to evaluate the first three terms actually

[random seal]

So if we name S(x) that RHS thing

[neat mist]

Not the whole expansion

[random seal]

it would suffice to check that (1-x)S(x) is equal to the expansion of exp(-x) for all x

[random seal]

Really? huh

[neat mist]

And yeah we only know radius of convergence of the series is greater than or equal to 1 not the radius

[neat mist]

We tortured ourselves for nothing

[random seal]

I misunderstood the statement of computing the three terms, I thought it was just checking for the three first terms but having to prove it for all of them

My apologies

[neat mist]

I thought so as well 😭 then I started talking about Cauchy product

[plush dirge]

Thank you all

+close

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