Infinite(simal) Sum

Have fun!

Find a formula for $\sum_{i=0}^\infty i^{2014}x^{i}$. In other words, find $0^{2014}x^{0}+1^{2014}x^{1}+2^{2014}x^{2}+3^{2014}x^{3}+4^{2014}x^{4}+...$.

Hint: Try a form of recursion(not with numbers, but with a formula).

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

We can get a recurrence relation in the following manner:

For positive integers k we define $T_{k}(x)=\sum_{i=0}^{\infty }i^kx^i$ $\forall x \in (0,1)$

We know $T_{1}(x)=\frac{x}{(1-x)^2}$

Also $T_{k+1}(x)=x\frac{dT_{k}(x)}{dx}$

Sambit Senapati - 7 years, 2 months ago

To be precise, is there a way of expanding your recursion formula?

Tristan Shin - 7 years, 2 months ago

Sorry, I didnt get you. What do you exactly mean by expanding the recursion formula? Did you mean this-

$T_{k+1}(x)=T_{k}(x)+x^2T_{k-1}''(x)$

Sambit Senapati - 7 years, 2 months ago

@Sambit Senapati – I found a simpler recursion formula that only reaches back one term.

For a start, I labeled $T_{k}=\frac {f(x)}{(1-x)^{k+1}}$ . Then, try working around with that.

Tristan Shin - 7 years, 2 months ago

Just a note, I know how to do it, but it would take a very long time(probably over a month straight) of working out some functions. I'm not really looking for an exact solution(don't actually take the time to do it), but if anyone can discover the method I'm using, that would be wonderful.

Tristan Shin - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$