Approximate the graph of $e^{x^2}$ While solving the integrals related to $e^{x^2}$, we try to approximate it. My question is there some function of which the graph is approximately like that of $e^{x^2}$?
 A: Using the power series representation we have
$$e^{x^2}=\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}=1+x^2+\frac{x^4}{2!}+...$$
Taking more terms will give a better and better approximation.
A: The Taylor series for $e^{x^2}$ is $$1 + x^2 + \frac{x^4}{2} + O(x^6),$$ so that any function that you sum $1$ to an even power of $x$ will ressemble approximately the function you mention. Take a look at its behaviour in Wolfram Alpha.
A: Here is more fun way to find an approximation using these representations with varying convergence. Note the Modified Bessel Function of the First Kind. Note the Pi-Product notation:
$$e^{x^2}=\sum_{n=0}^\infty \frac{x^{4n}(x^2+2n+1)}{(2n+1)!}= \sum_{n=0}^\infty \frac{x^{4n-2}(x^2+2n)}{(2n)!}= \sum_{n\in\Bbb Z} \text I_n(x^2)=  \sum_{n\in\Bbb Z} (-1)^n\text I_n(-x^2)=\cosh(x^2)+\sinh(x^2)=\sum_{n=0}^\infty \left(\frac{x^{4n}}{(2n)!}+ \frac{x^{4n-2}}{(2n-1)!} \right)=\text I_0(x^2)+2\sum_{n=1}^\infty \text I_{2n(x^2)}+2\sum_{n=0}^\infty \text I_{2n+1}(x^2)=\prod_{n=1}^\infty \left(\frac{4x^4}{\pi^2(2n-1)^2}+1\right)+x^2\prod_{n=1}^\infty\left(\frac{x^4}{\pi^2n^2}\right)=\pmb …$$
There are many non summation representations as well. Please correct me and give me feedback!
