# 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? Commented Dec 27, 2016 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}

• You can check here en.m.wikipedia.org/wiki/Exponential_function and here mathworld.wolfram.com/ExponentialFunction.html Commented Dec 27, 2016 at 3:39
• @gowarth Why this is incorrect? Commented Dec 27, 2016 at 3:59
• @Azzo Then study the two answers below. Commented Dec 27, 2016 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... Commented Dec 27, 2016 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. Commented Dec 27, 2016 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! Commented Dec 27, 2016 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 Commented Dec 27, 2016 at 4:04
• @juniven My heartfelt thanks.
– user371838
Commented Dec 27, 2016 at 4:04
• How did $x^{\frac{1}{v}}$ become $\frac{x}{v}$??? Do you actually think that e.g. $16^{\frac{1}{2}}=\frac{16}{2}$??? Commented Dec 27, 2016 at 5:25