Question is express 1/(1-x)^2 as a power series

[stable sparrow]

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[dusty peak]

I don’t get this at all

Like what the point of derivatetive if ur not even using it

And that is a representation of 1/(1-x) not even the original question??

[manic shuttle]

you're asked for a series representation of 1/(1-x)^2, which is just the derivative of 1/(1-x) by chain rule

you know the series representation of 1/(1-x), and can show summation and differentiation can be freely swapped, hence you get a series representation of 1/(1-x)^2

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

[zinc sphinx]

Omegabet_

[dusty peak]

Why do they use 1/(1-x)

Am I suppose to always get it into a 1/(1-r) form

[manic shuttle]

I mean, you use stuff you know

like how do you find the Taylor expansion of 1/(1-x^2)?

[dusty peak]

Idk I’m not on that

[manic shuttle]

you are

you're doing taylor expansions rn

[dusty peak]

I’m doing representing power series by differentiation and integration

Is that same

[manic shuttle]

I mean same idea

you use the basic power series you know to build new ones

1/(1-x) is the ubiquitous example of a power series you should know

[dusty peak]

So basically the function 1/(1-x)^2 looks similar to this so I use that

Is that how it works

[manic shuttle]

again, from calc 1 there's a clear link between 1/(1-x) and 1/(1-x)^2

the latter is the derivative of the prior

to which it follows

[dusty peak]

So I’m suppose to always find the anti derivative then? What

[manic shuttle]

no, you use logic

you want 1/(1-x)^2

you know from calc 1 that 1/(1-x)^2 = d/dx 1/(1-x)

you know the power series of 1/(1-x)

the rest is just computation

[elfin pendant]

Because 1/(1 - x) is a convenient expression with a known series expansion.

[elfin pendant]

Look, let's just take this from the top. We want to find a series expansion of 1/(1 - x)^2. We don't know that, but we do know that the series expansion for 1/(1 - x) = sum(n = 0, inf) x^n. We also know that d/dx(1/(1 - x)) = 1/(1 - x)^2. We therefore deduce that d/dx(sum(n = 0, inf) x^n) is the series expansion for 1/(1 - x)^2.

Is there any particular step you're having trouble with here?

[manic shuttle]

If only I had said all this already

[digital creek]

@dusty peak

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[golden zinc]

Hm, this is an interesting problem lol