Taylor Series

Final exam: Wednesday, March 22, 7-10pm in PCYNH 109. Bring ID!
Last Quiz 4: This Friday
Next: 11.10 Taylor and Maclaurin series
Next: 11.12 Applications of Taylor Polynomials
Midterm Letters:
A, 32-38
B, 26-31
C, 20-25
D, 14-19
Mean: 23.4, Standard Deviation: 7.8, High: 38, Low: 6.

Example 6.6.1   Suppose we have a degree-$3$ (cubic) polynomial $p$ and we know that $p(0) = 4$, $p'(0)=3$, $p''(0)=4$, and $p'''(0)=6$. Can we determine $p$? Answer: Yes! We have

$$p(x) = a + bx + cx^2 + dx^3$$

$$p'(x) = b + 2cx + 3dx^2$$

$$p''(x) = 2c + 6dx$$

$$p'''(x) = 6d$$

From what we mentioned above, we have:

$$a = p(0) = 4$$

$$b = p'(0) = 3$$

$$c = \frac{p''(0)}{2} = 2$$

$$d = \frac{p'''(0)}{6} = 1$$

Thus

$$p(x) = 4 + 3x + 2x^2 + x^3.$$

Amazingly, we can use the idea of Example  to compute power series expansions of functions. E.g., we will show below that

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$

Convergent series are determined by the values of their derivatives.

Consider a general power series

$$f(x) = \sum_{n=0}^{\oo } c_n (x-a)^n = c_0 + c_1 (x-a) + c_2 (x-a)^2 + \cdots$$

We have

$$c_0 = f(a)$$

$$c_1 = f'(a)$$

$$c_2 = \frac{f''(a)}{2}$$

$$\cdots$$

$$c_n = \frac{f^{(n)}(a)}{n!},$$

where for the last equality we use that

$$f^{(n)}(x) = n! c_n + (x-a)(\cdots + \cdots)$$

Remark 6.6.2   The definition of $0!$ is $1$ (it's the empty product). The empty sum is 0 and the empty product is $1$.

Theorem 6.6.3 (Taylor Series)   If $f(x)$ is a function that equals a power series centered about $a$, then that power series expansion is

$$f(x) = \sum_{n=0}^{\oo } \frac{f^{(n)}(a)}{n!} (x-a)^n$$

$$= f(a) + f'(a)(x-a) + \frac{f''(a)}{2} (x-a)^2 + \cdots$$

Remark 6.6.4   WARNING: There are functions that have all derivatives defined, but do not equal their Taylor expansion. E.g., $f(x) = e^{-1/x^2}$ for $x\neq 0$ and $f(0)=0$. It's Taylor expansion is the 0 series (which converges everywhere), but it is not the 0 function.

Definition 6.6.5 (Maclaurin Series)   A Maclaurin series is just a Taylor series with $a=0$. I will not use the term ``Maclaurin series'' ever again (it's common in textbooks).

Example 6.6.6   Find the Taylor series for $f(x)=e^x$ about $a=0$. We have $f^{(n)}(x) = e^x$. Thus $f^{(n)}(0) = 1$ for all $n$. Hence

$$e^x = \sum_{n=0}^{\oo } \frac{1}{n!} x^n = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$$

What is the radius of convergence? Use the ratio test:

$$\lim_{n\to\infty} \left\vert \frac{ \frac{1}{(n+1)!} x^{n+1}}{ \frac{1}{n!} x^n} \right\vert = \lim_{n\to\infty} \frac{n!}{(n+1)!}\vert x\vert$$

$$= \lim_{n\to\infty} \frac{\vert x\vert}{n+1} = 0, x .$$

Thus the radius of convergence is $\oo$.

Example 6.6.7   Find the Taylor series of $f(x)=\sin(x)$ about $x=\frac{\pi}{2}$.^6.1 We have

$$f(x) = \sum_{n=0}^{\oo } \frac{f^{(n)}\left(\frac{\pi}{2}\right)}{n!} \left(x - \frac{\pi}{2}\right)^n.$$

To do this we have to puzzle out a pattern:

$$f(x) = \sin(x)$$

$$f'(x) = \cos(x)$$

$$f''(x) = -\sin(x)$$

$$f'''(x) = -\cos(x)$$

$$f^{(4)}(x) = \sin(x)$$

First notice how the signs behave. For $n=2m$ even,

$$f^{(n)}(x) = f^{(2m)}(x) = (-1)^{n/2} \sin(x)$$

and for $n=2m+1$ odd,

$$f^{(n)}(x) = f^{(2m+1)}(x) = (-1)^{m} \cos(x) = (-1)^{(n-1)/2} \cos(x)$$

For $n=2m$ even we have

$$f^{(n)}(\pi/2) = f^{(2m)}\left(\frac{\pi}{2}\right) = (-1)^m.$$

and for $n=2m+1$ odd we have

$$f^{(n)}(\pi/2) = f^{(2m+1)}\left(\frac{\pi}{2}\right) = (-1)^m\cos(\pi/2) = 0.$$

Finally,

$$\sin(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(\pi/2)}{n!}(x-\pi/2)^n$$

$$= \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(2m)!} \left(x - \frac{\pi}{2}\right)^{2m}.$$

Next we use the ratio test to compute the radius of convergence. We have

$$\lim_{m\to\infty} \frac{\displaystyle \left\vert \frac{(-1)^{m+1}... ...ft\vert \frac{(-1)^{m}}{(2m)!} \left(x - \frac{\pi}{2}\right)^{2m} \right\vert} = \lim_{m\to\infty} \frac{(2m)!}{(2m+2)!} \left(x-\frac{\pi}{2}\right)^2$$

$$= \lim_{m\to\infty} \frac{\left(x-\frac{\pi}{2}\right)^2}{(2m+2)(2m+1)}$$

which converges for each $x$. Hence $R=\infty$.

Example 6.6.8   Find the Taylor series for $\cos(x)$ about $a=0$. We have $\cos(x) = \sin\left(x+\frac{\pi}{2}\right)$. Thus from Example  (with infinite radius of convergence) and that the Taylor expansion is unique, we have

$$\cos(x) = \sin\left(x+\frac{\pi}{2}\right)$$

$$= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!} \left(x + \frac{\pi}{2} - \frac{\pi}{2}\right)^{2n}$$

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