Find the Maclaurin Series for tanx

[turkcyclone]

Homework Statement

Find the terms through x^5 in the Maclaurin series for f(x)

f(x)=tanx

Homework Equations

tanx=sinx/cosx

Maclaurin Series for:

sinx=x-x^3/3!+x^5/5!-x^7/7!...
cosx=1-x^2/2!+x^4/4!-X6/6!...

The Attempt at a Solution

I have done tanx=sinx/cosx

So I need to solve for (x-x^3/3!+x^5/5!-x^7/7!+x^9/9!)/(1-x^2/2!+x^4/4!-X6/6!+x^8/8!),
but I don't know how to solve this. Can someone help me with my division? Thanks!

 

[micromass]

Cant you just find the Maclaurin series the direct way? By using the formula

$$\sum_{n=0}^{+\infty}{\frac{f^{(n)}(0)}{n!}x^n}$$

So to find the term at $$x^5$$, you'll need to derive 5 times.

 

[turkcyclone]

Cant you just find the Maclaurin series the direct way? By using the formula

$$\sum_{n=0}^{+\infty}{\frac{f^{(n)}(0)}{n!}x^n}$$

So to find the term at $$x^5$$, you'll need to derive 5 times.

No, for this problem I am suppose to use the way I am trying, those were the directions.

 

[micromass]

So you'll need to find a power series $$\sum_{n=0}^{+\infty}{a_nx^n}$$ such that

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

Perform the product (up to $$x^5$$) and compare the coefficients...

 

[hgfalling]

Well, if you're not allowed to just expand the series with derivatives, you can do this:

$$\tan x = \frac{\sin x}{\cos x}$$
$$\sum_{n=0}^\infty a_n x^n = \frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}{1-\frac{x^2}{2!}+\frac{x^4}{4!}-...}$$
$$(1-\frac{x^2}{2!}+\frac{x^4}{4!}-... )\sum_{n=0}^\infty a_n x^n = x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$$

Now multiply out the first few terms of the left side, up to terms of order $x^5$, and generate a set of linear equations in $a_n$ (since the coefficients of the powers on either side must be the same). Solve those and you have your answer.

 

[jackmell]

Can someone help me with my division? Thanks!

Can you follow this and continue the long division to get the remaining terms?

$$\begin{array}{rc@{}c} & \multicolumn{2}{l}{\, \, \, x} \vspace*{0.12cm} \\ \cline{2-3} \multicolumn{1}{r}{1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dotsb \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} x-\frac{x^3}{2}+\frac{x^5}{24}+\dotsb} \\ & \multicolumn{2}{l}{\, \,\,\hspace{-18pt} -(x-\frac{x^3}{2}+\frac{x^5}{4!}+\dotsb} \\ \vspace{1mm} \cline{2-3} \multicolumn{2}{1}{\,\,\,\hspace{150pt} 1/3 x^3+\dotsb} \\ \vspace{1mm} \end{array}$$

 

[turkcyclone]

Can you follow this and continue the long division to get the remaining terms?

$$\begin{array}{rc@{}c} & \multicolumn{2}{l}{\, \, \, x} \vspace*{0.12cm} \\ \cline{2-3} \multicolumn{1}{r}{1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dotsb \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} x-\frac{x^3}{2}+\frac{x^5}{24}+\dotsb} \\ & \multicolumn{2}{l}{\, \,\,\hspace{-18pt} -(x-\frac{x^3}{2}+\frac{x^5}{4!}+\dotsb} \\ \vspace{1mm} \cline{2-3} \multicolumn{2}{1}{\,\,\,\hspace{150pt} 1/3 x^3+\dotsb} \\ \vspace{1mm} \end{array}$$

I actually started using that method and got stuck after the (x^3)/3.

 

[jackmell]

Ok, the arithmetic is a mess. I got the third one that way but wouldn't want to get any more that way. Here's the latex for fun:

$$\begin{array}{rc@{}c} & \multicolumn{2}{l}{\, \, \, x+1/3 x^3} \vspace*{0.12cm} \\ \cline{2-3} \multicolumn{1}{r}{1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dotsb \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} x-\frac{x^3}{2}+\frac{x^5}{24}+\dotsb} \\ & \multicolumn{2}{l}{\, \,\,\hspace{-12pt} -(x-\frac{x^3}{2}+\frac{x^5}{4!}+\dotsb} \\ \vspace{1mm} \cline{2-3} \multicolumn{2}{1}{\,\,\,\hspace{180pt} 1/3 x^3-1/30 x^5+\frac{6}{7!} x^7+\dotsb}\vspace{1mm} \\ \multicolumn{2}{1}{\,\,\,\hspace{170pt} -(1/3 x^3-x^5/6+x^7/72+\dotsb} \\ \cline{2-3}\\ \multicolumn{2}{1}{\,\,\,\hspace{170pt} 2/15 x^5+\dotsb} \\ \end{array}$$

 

[turkcyclone]

Ok, the arithmetic is a mess. I got the third one that way but wouldn't want to get any more that way. Here's the latex for fun:

$$\begin{array}{rc@{}c} & \multicolumn{2}{l}{\, \, \, x+1/3 x^3} \vspace*{0.12cm} \\ \cline{2-3} \multicolumn{1}{r}{1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dotsb \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} x-\frac{x^3}{2}+\frac{x^5}{24}+\dotsb} \\ & \multicolumn{2}{l}{\, \,\,\hspace{-12pt} -(x-\frac{x^3}{2}+\frac{x^5}{4!}+\dotsb} \\ \vspace{1mm} \cline{2-3} \multicolumn{2}{1}{\,\,\,\hspace{180pt} 1/3 x^3-1/30 x^5+\frac{6}{7!} x^7+\dotsb}\vspace{1mm} \\ \multicolumn{2}{1}{\,\,\,\hspace{170pt} -(1/3 x^3-x^5/6+x^7/72+\dotsb} \\ \cline{2-3}\\ \multicolumn{2}{1}{\,\,\,\hspace{170pt} 2/15 x^5+\dotsb} \\ \end{array}$$

I tried it and you are right, it is a mess. I think I am going to talk to my instructor on and see if I can use the other method. Thanks for your help!

 

[jackmell]

I think I am going to talk to my instructor on and see if I can use the other method.

Wait a minute dude. It ain't messy enough to do that. He may say, "auh, poor dawg, what, you can't handle a lil' bit of arithmetic?" I mean, it's only two more. :)

 

[turkcyclone]

Wait a minute dude. It ain't messy enough to do that. He may say, "auh, poor dawg, what, you can't handle a lil' bit of arithmetic?" I mean, it's only two more. :)

Haha, okay I will try it. We just started the Maclaurin Series and this was the first problem, so I was frustrated when I couldn't get it and the solutions manual did a TERRIBLE job explaining how to get it. But I did a couple of more and starting to get the hang of it, so I will go back and try it again. Thanks again though!