Integrating $\exp(-x^2)$ using power series

integrationpower series

I am aware of the technique to evaluate $\displaystyle\int_{-\infty}^{+\infty} \exp(-x^2)dx$ using polar coordinates.

I recently wondered, why is it not possible to evaluate this integral using power series representation? If you express the integral using a power series, its interval of convergence is $(-\infty,+\infty)$. Why cannot we get a value closer and closer to $\sqrt{\pi}$ by using a power series?

Other functions can be integrated using a power series solution, with good approximation and precision (depending on the number of terms), in the absence of a closed form solution. Why is this integral different?

Best Answer

$$e^z=\sum_{n=0}^\infty \frac{z^n}{n!}\qquad\forall z\in\mathbb{C}$$ and so: $$e^{-x^2}=\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{n!}$$ the problem is that the terms alternate between positive and negative, so if you cut the series off at any arbitrary point (which is going to be small compared to the infinite number of terms). To demonstrate this point look at the graph below: The red line represents the actual $\exp(-x^2)$ function whilst blue is $n\in[0,5]$ and green is $n\in[0,10]$ and for your integral you want to integrate this series and then take the limit for $n\to\infty$ whilst will just accentuate any "small" error you see here

Related Solutions

[Math] Understanding and Integrating/Differentiating Power Series

Are these the notes you took in class? ie.. do you just want some explanation on some of the details? Well for your first example, in the last step the $$1\over 5$$ is multiplied into the summation giving the term $$x^{n+1}\over {5^{n+1}(n+1)}$$. Then the power series is shifted by replacing n with n-1 (which explains the term you have written in the last step). However, in doing this, the series now starts at $n=1$ not $n=0$. Was that a typo? Clearly the term to the right of the last summation is undefined at $n=0$

For the second example (and I guess the first one too) the term $C$ is an arbitrary constant of integration. So to evaluate the definite integral in your second question, you can choose $C$ to be zero as any antiderivative can be used in the fundamental theorem of calculus.

In general, power series are nice because polynomials are by far the easiest functions to work with, and a power series allows you approximate a much harder function as a polynomial. There are only a few well known power series in elementary calculus (namely, $sin(x)$, $cos(x)$, $e^x$, and $1\over {x-1}$ ) but by differentiating and integrating you can produce power series for many, many functions. That's the whole point of this.

[Math] Understanding power series and their representation of functions

It's not that your understanding is wrong, it's more that the wording of the question suggests that the definitions, properties, applications, and "cultural context" of power series are a bit tangled.

For concreteness and definiteness, let's work over the real numbers. If $x_{0}$ is a real number, and if $(a_{k})_{k=0}^{\infty}$ is an arbitrary sequence of real numbers, the associated power series with coefficients $(a_{k})$ and center $x_{0}$ is the expression $$ \sum_{k=0}^{\infty} a_{k} (x - x_{0})^{k}. \tag{1} $$

Power series make sense as purely algebraic entities. For example, two power series (with the same center) can be added termwise and multiplied using the "Cauchy product": \begin{gather*} \sum_{k=0}^{\infty} a_{k} (x - x_{0})^{k} + \sum_{k=0}^{\infty} b_{k} (x - x_{0})^{k} = \sum_{k=0}^{\infty} (a_{k} + b_{k}) (x - x_{0})^{k}, \\ \left(\sum_{k=0}^{\infty} a_{k} (x - x_{0})^{k}\right) \left(\sum_{k=0}^{\infty} b_{k} (x - x_{0})^{k}\right) = \sum_{k=0}^{\infty} \left(\sum_{j=0}^{k} a_{j} b_{k - j}\right) (x - x_{0})^{k}. \end{gather*} These definitions make sense whether or not the individual series converge (next item). The crucial point is, each coefficient of the sum or product is a finite algebraic expression in the coefficients of the "operands".
For each real number $x$, the power series (1) is an infinite series of real numbers, which may converge (the sequence of partial sums $$ s_{n}(x) = \sum_{k=0}^{n} a_{k} (x - x_{0})^{k} $$ converges to a finite limit) or diverge (otherwise). Clearly, (1) converges for $x = x_{0}$. It's not difficult to show that: a. If (1) converges for some $x$ with $|x - x_{0}| = r$, then (1) converges for every $x$ with $|x - x_{0}| < r$;

b. If (1) diverges for some $x$ with $|x - x_{0}| = r$, then (1) diverges for every $x$ with $|x - x_{0}| > r$.

It follows that for every power series (1), there exists an extended non-negative real number $R$ (i.e., $0 \leq R \leq \infty$) such that (1) converges for $|x - x_{0}| < R$ and diverges for $|x - x_{0}| > R$. (If $R = 0$, the former condition is empty; if $R = \infty$, the latter is empty.) This $R$ is called the radius of the power series (1). The power series $$ \sum_{k=0}^{\infty} k!\, x^{k},\qquad \sum_{k=0}^{\infty} \frac{x^{k}}{R^{k}},\qquad \sum_{k=0}^{\infty} \frac{x^{k}}{k!} $$ have radii $0$, $R$, and $\infty$, respectively.
If (1) has positive radius (i.e., $0 < R$), the sum of the series defines a function $$ p(x) = \sum_{k=0}^{\infty} a_{k} (x - x_{0})^{k} $$ whose domain is the set of $x$ with $|x - x_{0}| < R$. In this region, $p$ is infinitely differentiable, and its derivatives are found by termwise differentiation, e.g., $$ p'(x) = \sum_{k=1}^{\infty} ka_{k} (x - x_{0})^{k-1} = \sum_{k=0}^{\infty} (k + 1) a_{k+1} (x - x_{0})^{k}. $$ Each derivative is itself a power series, convergent in the same region.

A function $f$ is analytic at $x_{0}$ if $f$ is represented by a convergent power series in some open interval about $x_{0}$, and is analytic if $f$ is analytic at $x_{0}$ for every interior point $x_{0}$ of the domain of $f$.
If $f$ is infinitely-differentiable at some point $x_{0}$, the Taylor series of $f$ at $x_{0}$ is the power series $$ \sum_{k=0}^{\infty} \frac{f^{(k)}(x_{0})}{k!} (x - x_{0})^{k}. $$ As is well-known, a function can be infinitely differentiable at $x_{0}$ without the Taylor series converging to $f$ in any open interval of $x_{0}$. The standard example is $f(x) = e^{-1/x^{2}}$ if $x \neq 0$, and $f(0) = 0$, whose Taylor series is identically zero.
A transcendental function is real-analytic by definition. As Paramanand Singh notes, the defining property of a transcendental function is not "requires an infinite series" (i.e., "not a polynomial"), but "does not satisfy a polynomial equation in two variables" (i.e., "is not an algebraic function").
Power series are important for many reasons. The most common rationale for introducing them in calculus is to obtain algebraic/analytic expressions for the exponential function (and the closely related circular and hyperbolic functions, and power functions with non-integer exponents), etc., etc. For example, from the exponential power series, one obtains $$ e = \exp(1) = \sum_{k=0}^{\infty} \frac{1}{k!}, $$ from which one can easily show $e$ is irrational.

Another application is to obtain power series solutions of linear ordinary differential equations with non-constant coefficients. Power series are by no means the only infinite series useful for studying functions; Fourier series and wavelets come immediately to mind.
As for generalizations: Power series with complex coefficients (and complex centers) make perfect sense; the wordings and notation above are chosen to minimize the modifications required to consider complex power series. There are useful notions of matrix-valued power series, operator-valued power series, etc.

Complex power series exhibit interesting behavior (such as monodromy) not encountered over the reals, because a connected open set in the plane need not be simply-connected. Further, every holomorphic (i.e., complex-differentiable) function of one complex variable is automatically analytic. The logical strength of being "once complex-differentiable" gives complex analysis a completely different flavor from real analysis.

Best Answer

Related Solutions

[Math] Understanding and Integrating/Differentiating Power Series

[Math] Understanding power series and their representation of functions

Related Question