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Establish an infinite series expansion for the function $y=e^{-x}\cos(x)$ from just the known series expansions of $e^x$ and $\cos(x)$. Include terms up to the sixth power.

I know that the expansions for $e^x$ and $\cos(x)$ are

$$e^{-x} = 1+x+ \frac{x^2}{2!}+ \frac{x^3}{3!}+ \frac{x^4}{4!}+ \frac{x^5}{5!}+ \frac{x^6}{6!},$$


$$\cos(x) = 1- \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}.$$

But I don't understand how to combine them.

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If all you want is up to the $6$-th power, just multiply, discarding (or not computing) terms of degree $\gt 6$. Note that your series for $e^{-x}$ is wrong, it should be $1-x+x^2/2!-x^3/3!+x^4/4!-x^5/5!+x^6/6!$. – André Nicolas Jan 4 '12 at 14:50
second it. With the given truncated formulae for $e^{-x}$ and $cos x$, just multiply them and discard terms with degree >$6$. – NikBels Jan 4 '12 at 14:53
you can multiply power series very similarly to the way you multiply polynomials, in other words – Zarrax Jan 4 '12 at 14:55
What you want here is a Cauchy product; i.e., a convolution. – J. M. Jan 4 '12 at 15:03
up vote 3 down vote accepted

You are meant to just simply multiply the two series (in an informal manner) and keep the terms that have a power less than or equal to $6$.

It should be evident that you can consider the product $$ \tag{1} \color{maroon}{ \Bigl( 1-x+ \frac{x^2}{2!}- \frac{x^3}{3!}+ \frac{x^4}{4!}- \frac{x^5}{5!}+ \frac{x^6}{6!}\Bigr)}\color{darkgreen}{ \Bigl( 1- \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}\Bigr)}; $$ because, for example, if you included the term $x^7\over7!$ of the series for $e^x$, then after doing the multiplication, that term would introduce powers of $x$ that are larger than 6, and you don't care about those.

So, let's do the multiplication. One way would be just to distribute the $\color{maroon}{\text{left}}$ factor of (1) across the $\color{darkgreen}{right}$, then keep distributing until no multiplications remain, and finally collect like terms.

But, a better method is available. You know you'll wind up with an expression of the form $$\tag{2} a_0+a_1x^1+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6. $$ We just need to figure out what the $a_i$ are.

Now, to obtain the product (1), you can do the following: select one term from the $\color{maroon}{\text{left}}$ factor and one term from the $\color{darkgreen}{right}$ factor and then take their product. Do this for every such choice of selections and add them all up.

So what we will do is, fix a value of $i$, and figure out the coefficient $a_i$ for the $x^i$ term appearing in the product (2).

So, how do you get $a_0$?

Well, there is only one way to obtain a constant by selecting terms from (1), and that is to select the "1" in both factors.

So $a_0=1$.

How do you get $a_1$?

Well, there is only one way to obtain a $x$ term by selecting terms from (1), and that is to select the "$\color{maroon}{-x}$" term from the $\color{maroon}{\text{left}}$ factor and the $\color{darkgreen}{1}$ from the $\color{darkgreen}{right}$.

So $a_1=1$.

How do you get $a_2$?

Here it's more interesting. There are exactly two ways:

$\ \ \ \ \bullet$select $\color{maroon}{1}$ from the $\color{maroon}{\text{left}}$ factor and $\color{darkgreen}{-{x^2\over 2!}}$ from the $\color{darkgreen}{right}$

$\ \ \ \ \bullet$select $\color{\maroon}{x^2\over 2!}$ from the $\color{maroon}{\text{left}}$ factor and $\color{darkgreen}{1}$ from the $\color{darkgreen}{right}$

Multiplying and adding these together will give $$ 1\cdot{-x^2\over2!}+{x^2\over2!}\cdot1=0 $$

So $a_2=0$.

Continuing in this manner:

$$ \eqalign{ a_3&= \underbrace{-1\over2!}_{\color{maroon}{-x}\cdot \color{darkgreen}{x^2\over2!} } + \underbrace{-1\over3!}_{ \color{maroon}{-x^3\over3!}\cdot\color{darkgreen}1 } \cr a_4&= \underbrace{1\over4!}_{\color{maroon}1\cdot\color{darkgreen}{x^4\over4!}} -\underbrace{1\over2!2!}_{\color{maroon}{x^2\over2!}\cdot\color{darkgreen}{-x^2\over2!}} +\underbrace{1\over4!}_{\color{maroon}{x^4\over4!}\cdot\color{darkgreen}1} \cr a_5&= \underbrace{-1\over 4!}_{\color{maroon}{-x}\cdot\color{darkgreen}{x^4\over4!}}+ \underbrace{1\over3!2!}_{\color{maroon}{-x^3\over3!}\cdot\color{darkgreen}{-x^2\over2!}}+ \underbrace{-1\over5!}_{\color{maroon}{-x^5\over5!}\cdot\color{darkgreen}1} \cr a_6&= \underbrace{-1\over 6!}_{\color{maroon}1\cdot\color{darkgreen}{-1\over6!}}+ \underbrace{1\over2!4!}_{\color{maroon}{x^2\over2!}\cdot\color{darkgreen}{x^4\over4!}}- \underbrace{1\over4!2!}_{\color{maroon}{x^4\over4!}\cdot\color{darkgreen}{-x^2\over2!}} + \underbrace{1\over6!}_{ \color{maroon}{x^6\over6!}\cdot\color{darkgreen}1} \cr } $$

Now you know the values of all the $a_i$ and you can form the required polynomial.

Note that, as Andre points out in the comments, your series for $e^{-x}$ is incorrect. The signs should be alternating: $e^{-x}=1-x+{x^2\over2!}-{x^3\over3!}+\cdots$.

I used the incorrect form in my answer before this edit; but have corrected it...

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Write $\cos x = \frac{e^{ix}+e^{-ix}}{2}$ which gives $e^{-x}\cos x = \frac{e^{x(-1+i)} + e^{x(-1-i)}}2$

then $$e^{-x}\cos x = \sum_{n=0}^\infty \frac{a_nx^n}{n!}$$

where $a_n = \frac{(-1+i)^n + (-1-i)^n} 2 = \Re{(-1+i)^n}$

(Changing rest of argument using the comment from J.M. below)

Now, use that $-1+i = \sqrt{2}e^{i\frac{3\pi}{4}}$, we see that the real part of $(-1+i)^n$ is $(\sqrt{2})^n \cos\left(\frac{3\pi n}4\right)$

So, finally, we get:

$$e^{-x}\cos x = \sum_{n=0}^\infty 2^{\frac n 2}\cos\left(\frac{3\pi n}4\right)\frac{x^n}{n!}$$

Note that the value $a_n=2^{\frac{n}2}\cos(\frac{3\pi n}4)$ is always an integer since it is the real part of $(-1+i)^n$. Specifically, $a_0=1$, $a_1=-1$, $a_2=0$, and $a_3=2$. Since $(-1+i)^4 = -4$, we see that $a_{n+4} = -4 a_n$

So this gives the terms $a_4=-4, a_5=4, a_6=0$.

So the sixth-degree Taylor sum is:

$$1 - x + 2\frac{x^3}{3!} - 4 \frac{x^4}{4!} + 4\frac{x^5}{5!} = 1-x + \frac{x^3}{3} - \frac{x^4}{6} + \frac{x^5}{30}$$

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$$\exp(-x)\cos\,x=\sum_{j=0}^\infty \cos\left(\frac{3j\pi}{4}\right)\frac{\left(x\sqrt{2}\right)^j}{j!}$$ – J. M. Jan 4 '12 at 15:11
@J.M. Thanks, changed argument to reflect this easier view. – Thomas Andrews Jan 4 '12 at 15:57

As André Nicolas and Nikhil Bellarykar have commented, you can do this just looking at terms up to $x^6$, for example

$$(1-x+ \tfrac{x^2}{2!}- \tfrac{x^3}{3!}+ \tfrac{x^4}{4!}- \tfrac{x^5}{5!}+ \tfrac{x^6}{6!} - \cdots)(1- \tfrac{x^2}{2!}+ \tfrac{x^4}{4!}- \tfrac{x^6}{6!}- \cdots)$$

$$=(1-x+ \tfrac{x^2}{2!}- \tfrac{x^3}{3!}+ \tfrac{x^4}{4!}- \tfrac{x^5}{5!}+ \tfrac{x^6}{6!} - \cdots) - \tfrac{x^2}{2!}(1-x+ \tfrac{x^2}{2!}- \tfrac{x^3}{3!}+ \tfrac{x^4}{4!}- \cdots)$$

$$+ \tfrac{x^4}{4!} (1-x+ \tfrac{x^2}{2!}- - \cdots) - \tfrac{x^6}{6!}(1- \cdots)$$

$$ = 1 +x(-1) + x^2(\tfrac{1}{2!} + \tfrac{-1}{2!}) + x^3(\tfrac{-1}{3!} + \tfrac{1}{2!}) + x^4(\tfrac{1}{4!} + \tfrac{-1}{2!2!} + \tfrac{1}{4!}) \cdots \text{etc.} $$

and you can do the rest

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Since $e^x=\sum \frac{x^n}{n!}$, we have $e^{-x}=\sum (-1)^n \frac{x^n}{n!}$. It is also well known that $\cos(x)=\sum (-1)^n \frac{x^{2n}}{(2n)!}$. Thus, $$\left(\sum_{i=0}^\infty (-1)^i \frac{x^i}{i!}\right)\left(\sum_{j=0}^\infty (-1)^j \frac{x^{2j}}{(2j)!}\right) = \sum_{k=0}^\infty c_k x^k$$ where $$c_k = \sum_{i+2j=k} (-1)^{i+j}\frac{1}{i!\cdot (2j)!}=\begin{cases} \sum_{j=0}^{k/2} (-1)^{k-j} \frac{1}{(k-2j)!(2j)!}&, \text{ if } k \text{ is even,} \\ \sum_{j=0}^{(k-1)/2} (-1)^{k-j} \frac{1}{(k-2j)!(2j)!} &, \text{ if } k \text{ is odd}. \end{cases}$$

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