MAT187 L13 - Integrating and Differentiating Power Series

#MAT187 PCE
The main benefit of using power series to approximate complicated functions arises when we try to integrate or differentiate them, as they are "infinite"-degree polynomials.

Building new Taylor series from other Taylor series

For functions that aren't easily infinitely differentiable (unlike $\cos (x)$ or $e^{x}$ ), we may notice that it gets very cumbersome to take successive derivatˊives for a Taylor series.

To circumvent this, we can consider a simpler version of the Taylor series that we know how to evaluate, and then substitute in some value of $x$ into it to make it equal to the function we want to evaluate. This works because of the explanation in L12.

Example: $\frac{1}{1 + x^{2}}$

We can find the Taylor series for a simpler version of this function, $\frac{1}{1 - x}$ , pretty easily. It is simply:

\frac{1}{1 - x} = lim_{n \to \infty} \sum_{i = 0}^{n - 1} x^{i}

... and we know that this is only valid on $x \in [- 1, 1]$ , as per L11.

Since $x$ can really be any value (as per L12), we can just substitute $- x^{2}$ into the Taylor series of $\frac{1}{1 - x}$ to get the function we want. So, the Taylor series of $\frac{1}{1 + x^{2}}$ is:

\begin{aligned} \frac{1}{1 + x^{2}} & = \frac{1}{1 - (- x^{2})} \\ = lim_{n \to \infty} \sum_{i = 0}^{n - 1} (- x^{2})^{i} \end{aligned}

... and the bounds must also be changed to account for this substitution. $| - x^{2} | < 1 ⟹ | x | < 1$ . Coincidentally (or maybe not, idrk), this is the same restriction as before.

Differentiating and integrating Taylor series

We can now use this tool to more easily differentiate and integrate Taylor series.

Theorem: Differentiating and Integrating Taylor Series

On its own interval of convergence, a power series (Taylor series) can be differentiated and integrated term by term.

Example: expressing $\arctan (x)$ as a Taylor series

We can use the property that $\int_{0}^{x} \frac{1}{1 + t^{2}} d t = \arctan (x) + C$ to express $\arctan (x)$ as a Taylor series, since we know the Taylor series of $\frac{1}{1 + t^{2}}$ from the above example.

\begin{aligned} \int_{0}^{x} \frac{1}{1 + t^{2}} d t & = \int_{0}^{x} lim_{n \to \infty} \sum_{i = 0}^{n - 1} (- t^{2})^{i} \\ = lim_{n \to \infty} \sum_{i = 0}^{n - 1} ((- 1)^{i} \frac{1}{2 i + 1} t^{2 i + 1}) |_{0}^{x} \end{aligned}

And since $C$ in this case equals the Taylor series evaluated at $0$ ,

\begin{aligned} \arctan (x) & = \int_{0}^{x} \frac{1}{1 + t^{2}} d t \\ = lim_{n \to \infty} \sum_{i = 0}^{n - 1} ((- 1)^{i} \frac{1}{2 i + 1} x^{2 i + 1}) \end{aligned}

This is useful (well, "useful") since it allows us to approximate $\arctan (x)$ for some value of $x$ without having to use a calculator.

Building new Taylor series from other Taylor series

Example: 11+x2

Differentiating and integrating Taylor series

Example: expressing arctan⁡(x) as a Taylor series

Example: $\frac{1}{1 + x^{2}}$

Example: expressing $\arctan (x)$ as a Taylor series