How do you use a Power Series to estimate the integral #int_0^0.01cos(sqrt(x))dx# ?

Question

Scarlett Allison · Accepted Answer

Since

#cosx=sum_{n=0}^infty(-1)^n{x^{2n}}/{(2n)!}#,

we have

#cos(sqrt{x})=sum_{n=0}^infty{(-1)^n(sqrt{x})^{2n}}/{(2n)!}=sum_{n=0}^infty{(-1)^nx^{n}}/{(2n)!}#

Now, consider the integral in question.

#int_0^{0.01}cos(sqrt{x})dx=int_0^{0.01}sum_{n=0}^infty{(-1)^nx^{n}}/{(2n+1)!}dx#

by integrating term by term,

#=sum_{n=0}^infty[{(-1)^nx^{n+1}}/{(2n)!(n+1)}]_0^{0.01}#

#={(0.01)^1}/{0!cdot1}-(0.01)^2/{2!cdot2}+{(0.01)^3}/{4!cdot3}-cdots#

By adding a few terms of the above series, we can approximate the value of the original definite integral.

Luke Alvarez · Answer

undefined To estimate the integral $\int_{0}^{0.01} \cos(\sqrt{x}) \, dx$ using a power series, we first need to represent $\cos(\sqrt{x})$ as a power series. One way to do this is by using the Maclaurin series expansion of $\cos(x)$ and then substituting $x$ with $\sqrt{x}$.

The Maclaurin series expansion of $\cos(x)$ is:

$$
\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}
$$

Substituting $x$ with $\sqrt{x}$, we get:

$$
\cos(\sqrt{x}) = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{(2n)!}
$$

Then, we integrate this power series term by term from $0$ to $0.01$:

$$
\int_{0}^{0.01} \cos(\sqrt{x}) \, dx = \int_{0}^{0.01} \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{(2n)!} \, dx
$$

Because the series representation is an infinite sum, integrating term by term can be computationally intensive. However, we can truncate the series to a finite number of terms based on the desired level of accuracy.

After integrating, we can compute the sum of the series for the given limits to estimate the value of the integral.