- #1

Potatochip911

- 318

- 3

- TL;DR Summary
- Having trouble with math involving two series being multiplied together.

Hi, I've been following the derivation of wolfram mathworld for this problem and I'm running into some trouble regarding the summation indices. Currently I am at the step where we have found that (it's pretty much just binomial expansion and taylor series to get to this point)

$$ f(x) = x^n\left[\sum_{k=0}^{\infty}\binom{n}{k}(-1)^kx^{sk}\right ]\sum_{l=0}^\infty\binom{n+l-1}{l}x^l $$

Now we're interested in finding the coefficient for the term ##x^p## to find how many possibilities there are to obtain that exponent ##p##. This gives the equality $$p = n + sk + l$$ which we can use to put ##l## in terms of ##k## in the 2nd summation via $$l = p-n-sk$$

This gives

$$x^n\left[\sum_{k=0}^{\infty}\binom{n}{k}(-1)^kx^{sk}\right ]\sum_k \binom{p-sk-1}{p-sk-n}x^{p-n-sk} $$

For the new summation indices we have ##k = (p-n-l)/s##, plugging in ##l=0## and ##l=\infty## gives lower index ##(p-n)/s## and upper index ##-\infty## which doesn't make sense. On wolfram they are just claiming that the coefficient of ##x^p## is given by

$$c = \sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{p-sk-1}{p-sk-n}$$

I'm having trouble understanding this as they are simply multiplying the coefficients for the two different series together. If we consider as an example:

$$\sum_{k=0}^1\binom{3}{k}x^k\sum_{k=0}^1\binom{2}{k}x^k=(1+3x)(1+2x)=1+(3+2)x+6x^2 = 1+5x+6x^2$$

then multiplying the coefficients together only results in the correct coefficient for their proposed solution when there are no "interfering" (idk the proper math word) terms.

To summarize, I'm having trouble understanding how in their solution they don't change the limits of the sum by plugging in ##l##, and I'm also confused by the logic/math surrounding the coefficient of ##x^p##.

$$ f(x) = x^n\left[\sum_{k=0}^{\infty}\binom{n}{k}(-1)^kx^{sk}\right ]\sum_{l=0}^\infty\binom{n+l-1}{l}x^l $$

Now we're interested in finding the coefficient for the term ##x^p## to find how many possibilities there are to obtain that exponent ##p##. This gives the equality $$p = n + sk + l$$ which we can use to put ##l## in terms of ##k## in the 2nd summation via $$l = p-n-sk$$

This gives

$$x^n\left[\sum_{k=0}^{\infty}\binom{n}{k}(-1)^kx^{sk}\right ]\sum_k \binom{p-sk-1}{p-sk-n}x^{p-n-sk} $$

For the new summation indices we have ##k = (p-n-l)/s##, plugging in ##l=0## and ##l=\infty## gives lower index ##(p-n)/s## and upper index ##-\infty## which doesn't make sense. On wolfram they are just claiming that the coefficient of ##x^p## is given by

$$c = \sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{p-sk-1}{p-sk-n}$$

I'm having trouble understanding this as they are simply multiplying the coefficients for the two different series together. If we consider as an example:

$$\sum_{k=0}^1\binom{3}{k}x^k\sum_{k=0}^1\binom{2}{k}x^k=(1+3x)(1+2x)=1+(3+2)x+6x^2 = 1+5x+6x^2$$

then multiplying the coefficients together only results in the correct coefficient for their proposed solution when there are no "interfering" (idk the proper math word) terms.

To summarize, I'm having trouble understanding how in their solution they don't change the limits of the sum by plugging in ##l##, and I'm also confused by the logic/math surrounding the coefficient of ##x^p##.