Ariawan's Ratio

Level 2

Given $\mathrm{A}=\lim_{N \to \infty}\left|\frac{\sum\limits_{\alpha=0}^N \frac{(1 + i\pi)^\alpha}{\alpha!}}{\sum\limits_{\alpha=0}^N \frac{1}{\alpha!}}\right|,$ then determine the value of $11^\mathrm{A}$ ?

The answer is 11.

1 solution

Michael Tong
Jan 25, 2014

Note that the taylor series of $e^x$ is equal to $\displaystyle \sum{a = 0}^{\infty} \frac{x^a}{a!}$ . Thus $A = \frac{e^{1 + i\pi}}{e}$ . Furthermore, note that $e^{i \pi} = -1$ . Thus, $A = | \frac{-e}{e}| = 1$ , so $11^A = 11$ .

This is a perfect example of a problem that looks difficult at first, but is actually quite simple.

ugh that first line.... what happened

$\displaystyle \sum_{a=0}^{\infty} \frac{x^a}{a!}$ is what it's supposed to say.

Michael Tong - 7 years, 4 months ago

Well done Michael... :) Requote: "This is a perfect example of a problem that looks difficult at first, but it is actually quite simple".

Tunk-Fey Ariawan - 7 years, 4 months ago

I don't know why I was missing the Absolute Value sign. :-/

Avijit Sarker - 7 years, 2 months ago

Ariawan's Ratio

The answer is 11.

1 solution

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