Can YOU Simplify This?

Algebra Level 3

Evaluate

$2+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+...$

2 \frac{8}{11}

2.5

\pi

e

1 solution

Jason Simmons
Jan 3, 2016

The Maclaurin series of $e^x$ is as follows :

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

In this case, we can see (by inspection) that the base number is in fact $e$ because

$e \approx \sum_{n=0}^{5} \frac{1}{n!}= 2+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}$

The ellipsis in the problem indicates that the series continues forever so we can safely say that the series equals Euler's number!

There seems to be an issue with this, $\ln(-1)$ is undefined.

Thus $e^{\ln(-1)}$ would also be undefined?

Isaac Buckley - 5 years, 5 months ago

$Log(-1)=\pi i$ , which leads to Euler's equation ${e}^{\pi i}=-1$

Michael Mendrin - 5 years, 5 months ago

Correct! Nice Job!

Jason Simmons - 5 years, 5 months ago

@Jason Simmons – I always like to tell people that if they want to write the exact value for $\pi$ , write $-i Log(-1)$ . Real handy not to have to write out all those decimals!

Michael Mendrin - 5 years, 5 months ago

@Michael Mendrin – Yes! In high school I was always told that there are no values for the natural logarithm of negative numbers, but then I read up on Complex Analysis and I found that this isn't true. I guess that for high school purposes we don't need to worry about how you can find $\ln (-1)$ , but I was intrigued when I furthered my knowledge of Complex Analysis.

Jason Simmons - 5 years, 5 months ago

$\ln (-1)$ is an imaginary number. The natural logarithm can have negative numbers as well as complex numbers; however, these numbers (complex or real-negative) will come out to be complex themselves. Another example is that $\ln i = \frac{i \pi}{2}$ .

Read over my wiki and you'll gain a better understanding of how $\ln x$ is not limited to $x > 0$ .

With that said, the natural logarithm can not be evaluated at zero, so $\ln 0 = \textbf{undefined}$ but $\ln(-1) = i \pi$ .

Jason Simmons - 5 years, 5 months ago

I'm well aware that we can generalise the logarithm. For me it's a definition issue.

I take the logarithm to be a function that maps the positive reals to the reals ( $\log : \mathbb{R}^+ \to \mathbb{R}$ ).

I see the complex logarithm as an extension that can make sense of otherwise ill-defined expressions.

Isaac Buckley - 5 years, 5 months ago

@Isaac Buckley – I guess the thing is they have the same notation so the most general function should always be assumed?

Isaac Buckley - 5 years, 5 months ago

@Isaac Buckley – Call out the attorneys. This subject is messier still, because $Log(-1)$ is actually $(2n+1)\pi i$ where $n$ is an integer. Functions of complex numbers are usually multi-valued. However, this doesn't change the answer to this problem.

Michael Mendrin - 5 years, 5 months ago

@Isaac Buckley – Hmmm, I'm not too sure about that. Where's Calvin when you need him haha? I don't really know how to contact him. I'm sorry but I don't really have an answer to your question. I guess I think of it this way: a square is a kind of rhombus with four congruent sides. Similarly, the function $\ln (x) \; : \; \mathbb{R}^+ \rightarrow \mathbb{R}$ is actually within the complex logarithm. Essentially I'm restating that

$\mathbb{R} \subset \mathbb{C}$

So, I guess I'm saying that you could fit the "real" natural logarithm into the complex logarithm and they become one. I hope that makes sense. I wish I could somehow get Calvin to join our conversation, but I'm fairly new to Brilliant...

Jason Simmons - 5 years, 5 months ago

@Jason Simmons – Calvin is working on a machine that will generate clones of himself so that they can be of more help to Brilliant members.

Michael Mendrin - 5 years, 5 months ago

@Jason Simmons – @Calvin Lin What do you think Lord Calvin?

Isaac Buckley - 5 years, 5 months ago

@Isaac Buckley – His Lordship thinks that the problem should be rephrased for clarity.

Calvin Lin Staff - 5 years, 5 months ago

@Calvin Lin – If you were to rephrase this problem, how would you go about doing so? I don't like to have equivocal problems.

Jason Simmons - 5 years, 5 months ago

@Jason Simmons – I suggest removing the $\ln \ldots$ term completely, and just ask for the value of that summation. If you are in agreement, I can update the answer to $e$ , and change some of the options too.

Calvin Lin Staff - 5 years, 5 months ago

@Calvin Lin – Maybe we should just always assume $\ln$ denotes the complex logarithm at all times? Would that cause any problems?

It's similar to assuming that the Gamma function is the extension of the factorial function. It has nice properties but it's not the only interpolation. Although according to Bohr-Mollerup theorem it's the only one that satisfies certain properties for $x>0$ .

Here is a nice example of another extension: Hadamards Gamma function .

Is there a reason most people assume the regular Gamma function as the extension?

Is the Gamma function the only one which you can analytically continue to the whole complex plane?

Isaac Buckley - 5 years, 5 months ago

@Isaac Buckley – It could introduce issues/complexities, because the complex logarithm is multi-valued, and I prefer not to have to explain that every time.

With regards to the Gamma function, there are many ways that we can analytically extend the factorial function, because it is only defined on the integers. For example, we could add $\sin (\pi x )$ to the Gamma function, and still obtain a smooth function which is equal to the factorial function on the integers. However, once we define the function on a set which has an accumulation point, then the analytic continuation is uniquely determined.

Calvin Lin Staff - 5 years, 5 months ago

@Calvin Lin – Calvin I agree to you changing it however you see fit. Thank you.

Jason Simmons - 5 years, 5 months ago

@Jason Simmons – Thanks! I've edited the problem. Let me know if you want to suggest any changes.

Calvin Lin Staff - 5 years, 5 months ago

Can YOU Simplify This?

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