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I want to do the following derivative

$$ \frac{d}{d(x+y)}\sum_{r=0}^\infty(x+y)^{\alpha +r}$$

and I guessed I must begin from the following statement:

$$ \frac{d}{d(x+y)}= \frac{d}{dx}\sum_{r=0}^\infty(x+y)^{\alpha +r}+ \frac{d}{dy}\sum_{r=0}^\infty(x+y)^{\alpha +r}$$

I want to know, is it right or not?

Blue
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Habib
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  • where did you see this? – Praphulla Koushik Jan 03 '20 at 12:11
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    For one thing $\frac{1}{a+ b}$ is not equal to $\frac{1}{a}+ \frac{1}{b}$ so you can't "separate" the derivative like that. Instead, let u= x+ y so that the derivative becomes $\frac{d}{du}\sum_{r=0}^\infty u^{n+r}= \sum_{r=0}^\infty \frac{du^{n+r}}{du}$ – user247327 Jan 03 '20 at 12:19

2 Answers2

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No, your first attempt is not correct. (Or rather, I'd be astonished if it were correct; I haven't actually proved it incorrect.)

Hint: substitute $z = x + y$ to avoid confusion.

Patrick Stevens
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ASSUMPTION: $\left|x+y\right| \lt 1$. To avoid confusion lets substitute $(x+y)=t$ Following this substitution, we have : $$\frac{d}{dt} \sum_{r=0}^{\infty} \left(t\right)^{\alpha +r}$$ To be evaluated. $\sum _{r=0}^{\infty}t^{\alpha +r} $ can be written as : $$t^{\alpha}+t^{\alpha +1}+ \cdots + t^{\alpha +n} + \cdots$$ taking $t^{\alpha}$ common from all terms: $$t^{\alpha} (1+t+t^2+ \cdots + t^n +\cdots)$$ Applying sum of infinite geometric progression for common ratio $\in (-1,1)$ we get : $$ t^{\alpha} \left(\frac{1}{1-t}\right)$$ $\therefore$ we have to finally evaluate : $$\frac {d}{dt} \frac{t^{\alpha}}{1-t}$$ This can be easily evaluated by the quotient rule and we finally end up in: $$ \frac{\alpha t^{\alpha -1} (1-t) +t^{\alpha }}{(1-t)^2}$$

sai-kartik
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  • Thanks sai-kartik, your comment is so useful but the summation is vanished in your method while i don't want it. – Habib Jan 03 '20 at 17:13
  • I don't think the derivative will be easy to simply without resolving the summation first..I have made the assumption that $|x+y| \lt 1$ because if it wasn't, the summation would become infinite and differentiation would not be possible – sai-kartik Jan 03 '20 at 17:50