## 8.4. Taylor Series

### Example 8.4.18 (e): Finding Taylor Series by Multiplication

Multiply two known series together until a pattern emerges.

Let's start with ### Example

- Find the Taylor series centered at zero for
*f(x) = x*.^{2}e^{2x} - Find the first 3 terms of the Taylor series centered at zero for
*g(x) = sin(2x) e*^{x2}

*f(x) = x*. We know the series for the exponential function

^{2}e^{2x}*e*, so by substitution we also know

^{x}*e*. Then we multiply through by

^{2x}*x*to get the answer:

^{2}x^{2}e^{(2x)}= x^{2}1/n! (2x)^{n}=

= x^{2}(1 + (2x) + 1/2! (2x)^{2}+ 1/3! (2x)^{3}+ 1/4! (2x)^{4}+ ... ) =

= x^{2}(1 + 2 x + 1/2! 2^{2}x^{2}+ 1/3! 2^{3}x^{3}+ 1/4! 2^{4}x^{4}+ ... ) =

= x^{2}1/n! 2^{n}x^{n}=

= 2^{n}/n! x^{n+2}

Now let's move on to *g(x) = sin(2x) e ^{x2}*. We
could, of course, take derivatives at

*x = 0*, but because of product and chain rule they quickly get out of hand. Therefore, we will try another method: multiplication with an "infinite" distributive law. We know:

sin(x) =

e^{x}=

Substituting and multiplying we therefore get:

sin(2x) e^{x2}= =

= =

= =

= =

=

This approximation is not bad as you can see when you overlay the two functions. The red curve below shows the original function, the blue one shows the first three terms from the Taylor serires.

Of course the polynomial is *much* simpler than the original
function, yet both functions are pretty close to each other if
*x* is small.