I have a formula where both summation and product are involved \begin{align*} f(n-1)=\sum_{m=0}^{n-1}\sum_{j=0}^{m}\frac{\binom{m}{j}(-1)^j\mathrm{i}^m (a+k)^{2m-n+1}}{m!(2 a)^m} \prod_{p=1}^{n-1}(2m-2j-p+1). \end{align*} I want that the summation and product runs until $ n $ instead $ n-1 $. i.e., $ f(n)$. What changes will be made? I have no idea, how to do this.
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$\begingroup$ It seems you just want an expression for $f(n)$. If so, just plug in $n$ instead of $n-1$... Would you like me to write up what I mean? $\endgroup$– The CountCommented Jan 25, 2017 at 20:58
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$\begingroup$ yes....If possible. My main aim is to run the formula for $ n=0$ as well. In current situation, it can run from $ n=1$ $\endgroup$– shabbirCommented Jan 25, 2017 at 20:59
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$\begingroup$ I will write up a solution. One moment. $\endgroup$– The CountCommented Jan 25, 2017 at 20:59
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$\begingroup$ There. Helpful? $\endgroup$– The CountCommented Jan 25, 2017 at 21:08
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$\begingroup$ Yes....Thanks a lot :) $\endgroup$– shabbirCommented Jan 25, 2017 at 21:09
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1 Answer
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You already have this formula for $f(n-1)$:
$$f(n-1)=\sum_{m=0}^{n-1}\sum_{j=0}^{m}\frac{\binom{m}{j}(-1)^j\mathrm{i}^m (a+k)^{2m-n+1}}{m!(2 a)^m} \prod_{p=1}^{n-1}(2m-2j-p+1).$$
And since you want $f(n)$, we just need to replace every instance of $n-1$ with $n$. But this is a little tricky. There are three things to tackle. The first two are easy: the upper limits on the sum and product become $n$ instead of $n-1$. But on your factor of $(a+k)^{2m-n+1}$, we also have an $n-1$ hiding. The exponent needs to become $2m-(n+1)+1=2m-n$. Do you see why? In any case, our final expression is:
$$f(n)=\sum_{m=0}^{n}\sum_{j=0}^{m}\frac{\binom{m}{j}(-1)^j\mathrm{i}^m (a+k)^{2m-n}}{m!(2 a)^m} \prod_{p=1}^{n}(2m-2j-p+1).$$
answered Jan 25, 2017 at 21:04