Find the Maclaurin Series for tanx

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Homework Help Overview

The discussion revolves around finding the Maclaurin series for the function f(x) = tan(x) up to the x^5 term. Participants explore various methods to derive the series, including using the definitions of sine and cosine functions and their respective series expansions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants suggest using the series expansions for sine and cosine to derive the series for tan(x) through division. Others propose using the direct formula for Maclaurin series, which involves taking derivatives of the function.

Discussion Status

Participants are actively engaging with different methods to approach the problem. Some express confusion about the division process, while others are attempting to clarify the steps involved in generating the series. There is no explicit consensus on the best method to use, but several lines of reasoning are being explored.

Contextual Notes

Some participants mention that they are required to follow specific directions for the problem, which may limit their approach to finding the series. There is also a recognition of the challenges faced with arithmetic in the division process.

turkcyclone
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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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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.
 
micromass said:
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.
 
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...
 
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.
 
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turkcyclone said:
Can someone help me with my division? Thanks!

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

<br /> \begin{array}{rc@{}c}<br /> &amp; \multicolumn{2}{l}{\, \, \, x} \vspace*{0.12cm} \\<br /> \cline{2-3}<br /> \multicolumn{1}{r}{1-\frac{x^2}{2}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dotsb \hspace*{-4.8pt}} &amp; \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} x-\frac{x^3}{2}+\frac{x^5}{24}+\dotsb} \\<br /> &amp; \multicolumn{2}{l}{\, \,\,\hspace{-18pt} -(x-\frac{x^3}{2}+\frac{x^5}{4!}+\dotsb}<br /> \\<br /> \vspace{1mm}<br /> \cline{2-3}<br /> \multicolumn{2}{1}{\,\,\,\hspace{150pt} 1/3 x^3+\dotsb} \\<br /> \vspace{1mm}<br /> \end{array}<br />
 
jackmell said:
Can you follow this and continue the long division to get the remaining terms?

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

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

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

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

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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