Essentially what the title asks. For an argument $x$, how can I analytically acquire values for the function: $$ f(x)=\sum_{k=0}^{\infty}\frac{x^{2k+1}}{(2k+1)(k!)} $$ Again, it is important that I know how to do this analytically, as there are other series comparable to this one that I also wish to evaluate.
How to evaluate the infinite sum $\sum_{k=0}^{\infty}\frac{x^{2k+1}}{(2k+1)(k!)}$
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sequences-and-series
convergence
summation
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3try take derivative of $f(x)$ and see if you can spot something. – 2017-02-16
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0Try [this](https://en.wikipedia.org/wiki/Series_multisection) and try to detect a well-known power series. – 2017-02-16
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0Typically, you start with aseries you know, like a geometric series or some known Maclaurin series (exp, ln etc). Then you could see if integrating or differentiating gets you near to your series. Another possible experiment is to differentiate it and see if you can find a diferential equation satisfied by f. – 2017-02-16
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0I don't understand why this is being downvoted. – 2017-02-16
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2Question shows no effort maybe? – 2017-02-16
3 Answers
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$$f(x)=\int_{0}^{x}e^{t^2}\,dt =\int_{0}^{x}\exp\left(x^2+t^2-2tx\right)\,dt=xe^{x^2}\int_{0}^{1}e^{-t^2 x^2}\,dt = \frac{\sqrt{\pi}}{2} e^{x^2}\,\text{Erf}(x)$$ has the following continued fraction representation:
$$ f(z)=\sum_{n\geq 0}\frac{z^{2n+1}}{(2n+1)n!}=\frac{z}{1-\frac{2z^2}{3+\frac{4z^2}{5-\frac{6z^2}{7+\ldots}}}}. $$ For any $x\in(-1,1)$ the approximation $f(x)\approx e^{x^2}\arctan(x)$ is quite accurate.
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0Amazing typesetting of a continued fraction! (+1) – 2017-02-17
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0Thank you for this answer! However, I would like to know how you got $\int_{0}^{x}\exp (x^2+t^2-2tx)\ dt$ from the previous integral. Also, you put $dx$ at the end instead of $dt$ – 2017-02-17
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0@dsillman2000: in the first step it is enough to replace $t$ with $x-t$. In the second step $t$ is replaced by $xu$ so that $dt=x\,du$ and the integration range becomes $(0,1)$. Typos fixed. – 2017-02-18
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Hint: $$\sum_{k\ge0}\frac{x^{2k}}{k!}=\mathrm e^{x^2}.$$
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You ask for specific values. By noting that this is simply $$\sum_{k \ge 0} \int_0^x \frac{t^{2k}}{k!}dx= \int_0^x\sum_{k \ge 0} \frac{t^{2k}}{k!}dx= \int_0^xe^{t^2}dt = \frac{-i\sqrt{\pi}}{2}\operatorname{Erf}(ix)$$ All we have to do is look up known values of the Error Function. What we end up finding is this MSE question on the topic and this list of known values, which shows that there are no known, non-trivial, exact values of the Error Function.
Thus the only known values of $\operatorname{Erf}(x)$ are $x \in\{0,\pm\infty, \pm i\infty\}$
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0How can the integral of a real function on $(0,x)\subset\mathbb{R}$ be a imaginary number? – 2017-02-16
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0@JackD'Aurizio good catch. Meant $\operatorname{Erf}(\color{red}{i}x)$. – 2017-02-16