# Logarithm of an infinite series

I am curious about simplifying the following expression:

$$\log\bigg(\sum_{n=0}^{\infty} \frac{x^{\frac{n}{v}}{}}{n!}\bigg)^{v}, v>0, x>0$$

Is there any rule to simplify a summation inside the log?

Also, is there any way to derive the following expression:

$$\frac{d}{dx}\log\bigg(\sum_{n=0}^{\infty} \frac{x^{\frac{n}{v}}{}}{n!}\bigg)^{v}$$ or $$\frac{d}{dv}\log\bigg(\sum_{n=0}^{\infty} \frac{x^{\frac{n}{v}}{}}{n!}\bigg)^{v}$$

Thanks!

• Does the infinite series $\sum_0^\infty \frac{y^n}{n!}$ look familiar to you? If so, can you think of a way of rewriting your infinite summation so it's in this form? – K. A. Buhr Dec 27 '16 at 3:12

The exponential function $e^y$ is defined as

$$e^y:=\sum_{k = 0}^{\infty} {y^k \over k!}.$$

So your sum is simply (by setting $y=x^{\frac{1}{v}}$)

$$e^{x^{\frac{1}{v}}}.$$

Summary

• If you mean to calculate the logarithm of the sum raised to the power of $v$, log(sum^v):

\begin{align} \log \left(\left(\sum_{k = 0}^{\infty} {x^{\frac{k}{v}} \over k!}\right)^v\right)&=v\log \left(\sum_{k = 0}^{\infty} {x^{\frac{k}{v}} \over k!}\right)\\ &=v\log\left(e^{x^{\frac{1}{v}}}\right)\\ &=vx^{\frac{1}{v}}\log e \\ &=vx^{\frac{1}{v}}.\end{align}

• If you mean to calculate the logarithm of the sum all raised to the power of $v$, (log(sum))^v:

\begin{align} \left(\log \sum_{k = 0}^{\infty} {x^{\frac{k}{v}} \over k!}\right)^v&=\left(\log e^{x^{\frac{1}{v}}}\right)^v\\ &=\left(x^{\frac{1}{v}}\right)^v\\ &=x^{\frac{v}{v}}=x.\end{align}

• Thanks for your answer. I really appreciate your efforts. So, you are sure your answer is correct. – Rukna's Dec 27 '16 at 3:36
• You can check here en.m.wikipedia.org/wiki/Exponential_function and here mathworld.wolfram.com/ExponentialFunction.html – drzbir Dec 27 '16 at 3:39
• @gowarth Why this is incorrect? – drzbir Dec 27 '16 at 3:59
• @Azzo Then study the two answers below. – Juniven Dec 27 '16 at 4:00
• It's a good answer now, covering both possible interpretations. The way it's currently written in the question, it's certainly the first expression. But there's always a possibility that the OP may have misrepresented what they actually meant to ask... – zipirovich Dec 27 '16 at 5:27

The exponential function is defined by $$e^z = \sum\limits_{n=0}^\infty \frac{z^n}{n!}.$$

Hence we have that $$\log\left(\sum\limits_{n=0}^\infty \frac{x^{n/v}}{n!}\right)^v = \log\left(\sum\limits_{n=0}^\infty \frac{(x^{1/v})^n}{n!}\right)^v = \log\left(e^{x^{1/v}}\right)^v = (x^{1/v})^v = x.$$

Now that we've simplified the original expression, it should be easy to compute the derivatives.

• I upvote also this answer. – Juniven Dec 27 '16 at 4:07

Let's start over. As the other answers already worked it out, the expression inside the logarithm, or rather inside the big parentheses, simplifies because it's an instance of the series for the exponential function:

$$\sum_{n=0}^{\infty}\frac{x^{\frac{n}{v}}}{n!}=\sum_{n=0}^{\infty}\frac{\left(x^{1/v}\right)^n}{n!}=e^{x^{1/v}}.$$

But then remember what $a^{b^c}$ means: $a^{b^c}=a^{(b^c)}$, not $(a^b)^c$ (which would be equal to $a^{bc}$). So we get

$$\log\left(\sum_{n=0}^{\infty}\frac{x^{\frac{n}{v}}}{n!}\right)^v=\log\left(e^{x^{1/v}}\right)^v=v\log\left(e^{x^{1/v}}\right)=vx^{1/v}.$$

From that, you can take the derivative either with respect to $x$ or with respect to $y$. Let us know if you'd like to see that worked out.

• Thanks! Your solution is more appropriate! – Rukna's Dec 27 '16 at 14:41

We know that $$e^{x} =\sum_{k =0}^{\infty} \frac {x^k}{k!}$$ In our problem we can write our expression as $$\log(\sum_{n=0}^{\infty} \frac{x^{\frac{n}{v}}}{n!})^v =\log(\sum_{n=0}^{\infty} \frac {(x^{\frac {1}{v}})^n}{n!})^v$$ Thus we have, $$\log(e^{x^{\frac {1}{v}}})^v = (x^{\frac {1}{v}})^v=x$$ Hope it helps.

• I upvote coz this is correct – Juniven Dec 27 '16 at 4:04
• @juniven My heartfelt thanks. – Rohan Dec 27 '16 at 4:04
• How did $x^{\frac{1}{v}}$ become $\frac{x}{v}$??? Do you actually think that e.g. $16^{\frac{1}{2}}=\frac{16}{2}$??? – zipirovich Dec 27 '16 at 5:25