Series

[sand topaz]

Need help don’t understand

[faint wharf]

(a) r! = (r-1)!r, now use common denominators
(b) rewrite the (r-1)/r! using part a, you'll notice stuff cancels

[sand topaz]

I don’t understand is there anyway you could go through it?

I understand that r! = (r-1)!r but confused on what I’m doing with that

[sand topaz]

.

[faint wharf]

$\frac{1}{(r-1)!}=\frac{r}{r(r-1)!}=\frac{r}{r!}$, now just use common denominators

[night berry]

Civil Service Pigeon

[sand topaz]

Ok thank you understand now

@faint wharf would you be able to help with another series question?

[faint wharf]

sure

[sand topaz]

It’s kind of a longer one

I was told by someone else to factor out stuff

[faint wharf]

yeah assuming that's right, a bunch of stuff cancels (I'm assuming cuz idk what formulae you're given)

the ||(-1)^(r+1), x^(2r), and you can rewrite (2r+1)! like earlier||

you there?

ig I'll take that as a no?

well ping me if you come back and still have questions ig 🤷‍♂️

[sand topaz]

Oh sorry

Give me a sec

@faint wharf so does everything cancel to get to that?

Then you were saying rewrite the (2r+1)! which would be (2r+1)(2r)!

?

[sand topaz]

@faint wharf

Sad times

<@&727457814523674674>

[vague fox]

What class is this

Calculus 2?

This seems hard

[sand topaz]

Idk from the uk we don’t do school like that

In the uk you go college and choose 3 a levels and this is further maths which is basically uni work

[faint wharf]

ok idk what was up with me yesterday but I was kinda braindead

the logic with this series is that you're supposed to consider the full expansions

but a bunch of the terms will equal to 0 when x=0

So, the numerator can be expressed as $\frac{x^2}{3}-\frac{x^4}{30}+\frac{x^6}{840}-\cdots$

[night berry]

Civil Service Pigeon

[faint wharf]

Denominator can be expressed as $\frac{x^2}{2}-\frac{x^4}{24}+\frac{x^6}{720}-\cdots$

[night berry]

Civil Service Pigeon

[faint wharf]

you can divide the numerator and denominator by $x^2$, and setting $x=0$ yields the limit being equal to $\frac{1/3}{1/2}=2/3$

[night berry]

Civil Service Pigeon

[sand topaz]

So I never needed like the (-1)^r+1… part?

[faint wharf]

well you do, I just wrote it explicitly cause it's easier to see that way

the $(-1)^{r+1}$ is why the signs of successive terms alternate

[night berry]

Civil Service Pigeon

[sand topaz]

So first 3 terms over the first 3 terms simplify by dividing everything by x^2 set x=0 and you get your answer

[faint wharf]

So first 3 terms

The series extend beyond this, hence the ..., but you don't need to consider the following terms since even after being divided by x^2, they are still equal to 0 when x=0
over the first 3 terms simplify by dividing everything by x^2
dividing numerator and denominator by smallest exponent
set x=0 and you get your answer
yeah, that's how a bunch of stuff equals 0

[sand topaz]

Ok thank you for the help and thanks for writing that out

Actually can I ask you something else?

Not related to series

The last part of that question is there any other way to solve other then differentiation

A and B co ordinates

[faint wharf]

Maybe, but I don't see one immediately.

[sand topaz]

I’ve asked several people but everyone has said differentiation but was wondering cause work was about logs and exponentials

[faint wharf]

I mean mixing concepts is a thing

🤷‍♂️

[sand topaz]

Yea but thing is we’ve never done differentiation

So that’s why I was wondering

[faint wharf]

what the ...

ok maybe there's some rlly smart way I'm just not seeing lol

idk my only idea rn is to take this as a quadratic in e^x somehow and use the discriminant, but I have no idea of how to materialise that

[sand topaz]

Don’t matter that much anyway I was able to do the differentiation with a bit of help from my brother

I assume no one else in the class has done it tho