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Uniqueness of the Taylor series


The purpose of this lab is to acquaint you with several useful techniques for generating Taylor series. These techniques include substitution and multiplication and division of series by simple polynomials.


The general form of the Taylor series representation with base point x=a of a function f(x) is given by the following

Definition 347

Suppose that f(x) is a smooth function in some open interval containing x=a and that for every fixed value of x in this interval, it can be shown that the Taylor polynomial remainder Rn(x) goes to zero as $n\rightarrow\infty$. Then the Taylor series representation with base point x=a of the function f(x) is given by

f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x-a)^k \end{displaymath}

In class we will derive the following four important Taylor series,

\exp(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!} \mbox{, for all $x$} \end{displaymath}

\cos(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(2k)!} \mbox{,
for all $x$} \end{displaymath}

\sin(x) = \sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!}
\mbox{, for all $x$} \end{displaymath}

\frac{1}{1+x} = \sum_{k=0}^{\infty} (-1)^k x^k \mbox{, for $\mid x
\mid < 1$} \end{displaymath}

where the last series is really the formula for the sum of a geometric series in disguise.

Once you have a Taylor series representation for a function, it can be used in several ways to generate Taylor series representations of related functions. This is because of the following theorem.

Theorem 358

Suppose f satisfies

f(x) = c_0 + c_1 (x-a) + c_2 (x-a)^2 + c_3 (x-a)^3 + \cdots \end{displaymath}

for all x in some interval around a. Then,

c_n = \frac{f^{(n)}(a)}{n!} \end{displaymath}

Thus a function cannot have more than one power series in x-a that represents it.

The rest of the Background describes several different techniques for generating Taylor series of functions that are related to Taylor series that are already known. The four techniques are substitution, multiplication and division, integration, and differentiation. We have already seen examples of integration and differentiation with Taylor polynomials, but we haven't talked about the first two techniques yet.


To get a Taylor series for $\exp(2x)$, you could go through the standard procedure of differentiating and substituting into the general formula, but an easier (and also correct) procedure is to take the series for $\exp(x)$ and substitute 2x for x, obtaining

\exp(2x) = \sum_{k=0}^{\infty} \frac{(2x)^k}{k!}. \end{displaymath}

Writing out the first few terms gives

\exp(2x) = 1 + 2x +2x^2 + \frac{4 x^3}{3} + \ldots \end{displaymath}

By the uniqueness theorem, this must be the Taylor series for $\exp(2x)$.

The most commonly useful Taylor series have base point x=0 and that is what we will focus on in this lab. The technique of substitution is most useful if the substitution is of the form axn where a is a constant and n is a positive integer. For example, the series for $\cos(x^2)$ is easy to obtain as

\cos(x^2) = \sum_{k=0}^{\infty} (-1)^k \frac{(x^2)^{2k}}{(2k)!} \end{displaymath}

but it is not clear at all if the following substitution

\exp(x^2+x) = \sum_{k=0}^{\infty} \frac{(x^2+x)^k}{k!} \end{displaymath}

produces a useful result. This is because you would have to do a lot of work expanding powers of x2+x and collecting terms to recover a power series in x.

Even if you use a substitution of the form axn, you have to be careful if the series is only valid for a finite interval about the base point. For example, suppose you wanted to find the Taylor series with base point t=0 for the function

f(t) = \frac{1}{1+2t}\end{displaymath}

You can obtain the desired series by substitution as

\frac{1}{1+2t} = \sum_{k=0}^{\infty} (-1)^k (2t)^k \end{displaymath}

but you have to be careful because this formula is not valid for all values of t. In fact this formula is only valid if $\mid t \mid < 1/2$. The reason for this is that the series for 1/(1+x) is only valid if $\mid x \mid < 1$ and when we substitute 2t for x, the formula only makes sense if $\mid
2t \mid < 1$.

Multiplication and Division

If you have the Taylor series for f(x), and you want the Taylor series for something like x2 f(x), you just multiply each term of the series for f(x) by x2. If the leading term for the Taylor series of g(x) is xk for some integer k > 0, you can use division to obtain the Taylor series for g(x)/xn for any integer $n
\leq k$. Some examples follow.

x\sin(x) = x^2 - \frac{x^4}{3!} + \frac{x^6}{5!} + \ldots \end{displaymath}

x^2 \exp(x) = x^2 + x^3 + \frac{x^4}{2} + \ldots \end{displaymath}

\frac{\sin(x)}{x} = 1-{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}} +
\ldots \end{displaymath}

Term-by-term integration and differentiation

One of the nice properties of Taylor series is that they can be integrated and differentiated term-by-term. Here is the formal theorem, but the procedures are pretty straightforward.

Theorem 406

If a function f(x) has a Taylor series representation

f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x-a)^k \end{displaymath}

then the derivative of f(x) has the Taylor series representation

f'(x) = \sum_{k=1}^{\infty} \frac{f^{(k)}(a)}{(k-1)!} (x-a)^{k-1}\end{displaymath}

obtained by differentiating each term in the series for f(x), and the series

F(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{(k+1)!} (x-a)^{k+1} \end{displaymath}

obtained by integrating each term in the Taylor series for f(x), is an antiderivative of f(x).


Verify that substitution and multiplication work as described above to generate Taylor series (with base point a=0) for the following functions. That is, compare the Taylor polynomials for various orders obtained directly with those obtained by substitution, multiplication, or division.
$f(x) = \exp(4x)$.
$f(x) = x \sin(x^2)$.
$f(x) = (\cos(x)-1)/x^2$.
f(x) = x/(1+2x).
f(x) = x/(1-x2).
Use integration followed by substitution to generate the first four terms in the Taylor series with base point x=0 for $\ln(1+x^2)$. Start with the series for 1/(1+x).

Can you find the sum of the series

\sum_{n=2}^{\infty}n(n-1)r^{n-2} \end{displaymath}

if $\mid r \mid < 1$?

In the background section we only considered multiplication of series by polynomials. Suppose you wanted to generate the Taylor polynomial of order ten with base point x=0 for the function $h(x)=\cos(x)\sin(x)$. Can you do this by multiplying Taylor polynomials for $\cos(x)$ and $\sin(x)$? Can you think of another way to do this using the trig identity $\sin(2x)=2\sin(x)\cos(x)$?

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Jane E Bouchard