Find the Maclaurin Series for tanx

  • #1
turkcyclone
9
0

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!
 
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  • #2
Cant you just find the Maclaurin series the direct way? By using the formula

[tex]\sum_{n=0}^{+\infty}{\frac{f^{(n)}(0)}{n!}x^n}[/tex]

So to find the term at [tex]x^5[/tex], you'll need to derive 5 times.
 
  • #3
micromass said:
Cant you just find the Maclaurin series the direct way? By using the formula

[tex]\sum_{n=0}^{+\infty}{\frac{f^{(n)}(0)}{n!}x^n}[/tex]

So to find the term at [tex]x^5[/tex], 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.
 
  • #4
So you'll need to find a power series [tex]\sum_{n=0}^{+\infty}{a_nx^n}[/tex] such that

[tex]\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}}[/tex]

Perform the product (up to [tex]x^5[/tex]) and compare the coefficients...
 
  • #5
Well, if you're not allowed to just expand the series with derivatives, you can do this:

[tex] \tan x = \frac{\sin x}{\cos x} [/tex]
[tex] \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!}-...} [/tex]
[tex] (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!}-...[/tex]

Now multiply out the first few terms of the left side, up to terms of order [itex]x^5[/itex], and generate a set of linear equations in [itex]a_n[/itex] (since the coefficients of the powers on either side must be the same). Solve those and you have your answer.
 
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  • #6
turkcyclone said:
Can someone help me with my division? Thanks!

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

[tex]
\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}
[/tex]
 
  • #7
jackmell said:
Can you follow this and continue the long division to get the remaining terms?

[tex]
\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}
[/tex]

I actually started using that method and got stuck after the (x^3)/3.
 
  • #8
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:

[tex]
\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}
[/tex]
 
  • #9
jackmell said:
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:

[tex]
\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}
[/tex]

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!
 
  • #10
turkcyclone said:
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. :)
 
  • #11
jackmell said:
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!
 

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