$$ \sum_{k=0}^\infty\ x^k$$

has radius of convergence R = 1

so we may define a function f : (-1,1) by $$ \sum_{k=0}^\infty\ r^k$$ for any r in (-1,1)

Let g(r) = r for all r in (-1,1).

I am asked to find a power series expansion for the function ${gf'}$ on (-1,1). Here ${gf'}$ is the product of g with the derivative of f .

My question is does power series expansion mean find the Taylor series or I guess Maclaurin series since this in this case it's centered at zero.

When I try to find the Maclaurin series for $x^k$ I get

$$f'(x)=k(x)^{k-1}$$ $$f''(x)=k(k-1)(x)^{k-2}$$ $$f'''(x)=k(k-1)(k-2)(x)^{k-3}$$ so I think $$f^n(x)=k(k-1)...(kn+1)(x)^{k-n}$$

But I don't know what is meant by ${gf'}$ on (-1,1). Here ${gf'}$ is the product of g with the derivative of f. If anyone could help me understand what the question is asking I would really appreciate it.

Thank you,


1 Answer 1


By the fact that we can do termwise differentiation of a power series, if

$$f(x) = \sum_{k=0}^\infty x^k,$$


$$f'(x) = \sum_{k=0}^\infty k x^{k-1}.$$

Then we have

$$(g f')(x) = x \sum_{k=0}^\infty kx^{k-1} = \sum_{k=0}^\infty kx^k.$$ Note that one can either prove directly that this has radius of convergence $R = 1$ (for instance using the Cauchy-Hadamard Theorem), or by the property that if two power series are both finite on the interval $(-1,1)$ then so two is their product.

An alternative derivation

This is a simple derivation, we can also check it using the fact that $$f(x) = \sum_{k=0}^\infty x^k = \frac{1}{1-x}.$$ Therefore

$$f'(x) = \frac{1}{(1-x)^2},$$ and

$$(gf')(x) = \frac{x}{(1-x)^2}.$$

You can now take the Taylor expansion of this and re-derive the above result.

A comment on your solution

A power series is really just a polynomial representation of a function (with technicalities about which functions have valid power series representations).

In your solution you try to compute a Maclaurin / Taylor expansion for the function $h(x) = x^k$. The fact is this is already a power series, so there is no need to do any calculations.

  • 2
    $\begingroup$ Indeed, when I was teaching Calculus, I would sometimes test my students’ understanding by asking them to write down the Maclaurin series for the polynomial $x^5-6x^4+17x^2+x+8$. The results were most discouraging: very few wrote down $8+x+17x^2-6x^4+x^5$ with no computation. $\endgroup$
    – Lubin
    Mar 24, 2018 at 19:13

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