Deriving Maclaurin series for tanx

In summary: The Maclaurin series for tanx is:In summary, the Maclaurin series for tanx is the sum of the series for sinx and cosx, where each term is the product of the previous term and a Bernoulli number.
  • #1
Lucy Yeats
117
0

Homework Statement



State the Maclaurin series for sinx and cosx. Hence derive the Maclaurin series for tanx.


Homework Equations



sin(x) = x - x3/3! + x5/5! - x7/7!...
cos(x) = 1 - x2/2! + x4/4! - x6/6!...

The Attempt at a Solution



I know you divide the series for sinx by the series for cosx. However, can you do polynomial long division to get the answer? I've only done long division with polynomials starting with the highest power, so I'm confused.
 
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  • #2
One way you can do it is to write
\begin{align*}
\tan x &= \frac{x - \frac{x^3}{3!}+\frac{x^5}{5!}-\cdots}{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots} \\
&= \frac{x - \frac{x^3}{3!}+\frac{x^5}{5!}-\cdots}{1 - \left(\frac{x^2}{2!}-\frac{x^4}{4!}+\cdots\right)} \\
&= \left( x - \frac{x^3}{3!}+\frac{x^5}{5!} -\cdots \right) \left[ 1 + \left(\frac{x^2}{2!}-\frac{x^4}{4!}+\cdots\right) + \left(\frac{x^2}{2!}-\frac{x^4}{4!} + \cdots\right)^2+\cdots\right]
\end{align*}What you'd want to do is figure out what terms contribute to each order of x.
 
  • #3
Sorry for being slow, but how did you get that last line?
 
  • #4
Oh, sorry. I meant to explain that. It's using the series for 1/(1-z):
[tex]\frac{1}{1-z} = 1 + z + z^2 + \cdots[/tex]In this case, you have
[tex]z = \frac{x^2}{2!} - \frac{x^4}{4!} + \cdots[/tex]

You might want to see if you can figure out how to adapt long division to solve this problem. Try starting from the low-degree end and see how it works out.
 
  • #5
Thanks, that's a clever method.
 
  • #7
Notice that [itex]\tan{x}[/itex] is an odd function. Therefore, its Maclaurin series expansion contains only odd powers:
[tex]
\tan{x} = \sum_{n= 0}^{\infty}{T_{n} \, x^{2 n + 1}}
[/tex]

You are left with finding the coefficients [itex]\lbrace T_{n} \rbrace[/itex]. Notice that:
[tex]
\tan{x} = \frac{\sin{x}}{\cos{x}} \Rightarrow \sin{x} = \cos{x} \, \tan{x}
[/tex]
Although you had stated the first few terms of the sine and the cosine series expansions, you need an analytic form for the whole series:
[tex]
\sin{x} = \sum_{n = 0}^{\infty}{\frac{(-1)^{n}}{(2 n + 1)!} \, x^{2 n + 1}}
[/tex]
[tex]
\cos{x} = \sum_{n = 0}^{\infty}{\frac{(-1)^{n}}{(2 n)!} \, x^{2 n}}
[/tex]

If you plug all these expansions in, and you use the product[/itex] to find the general term of the product of the series, and you compare them term by term, you should get an equation for the coefficients [itex]\lbrace T_{n} \rbrace[/itex]. If you compare it with the equation that defines the Bernoulli numbers, you might find a connection.
 
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  • #8
Thanks everyone!
 

1. What is a Maclaurin series?

A Maclaurin series is a mathematical representation of a function as an infinite sum of terms. It is named after the Scottish mathematician Colin Maclaurin, who first introduced this concept.

2. How do you derive the Maclaurin series for tanx?

To derive the Maclaurin series for tanx, you must use the Taylor series expansion of the function. This involves taking the derivatives of the function at a specific value (in this case, x = 0) and plugging them into the general formula for the Taylor series.

3. What is the general formula for the Maclaurin series of tanx?

The general formula for the Maclaurin series of tanx is: tanx = x + (x^3)/3 + (2x^5)/15 + (17x^7)/315 + ... This pattern continues, with each term having an odd power of x in the numerator and a corresponding factorial in the denominator.

4. Why is the Maclaurin series important for tanx?

The Maclaurin series allows us to approximate the value of tanx for any given value of x. This is especially useful in calculus and other areas of mathematics where finding the exact value of a function may be difficult or impossible.

5. Are there any limitations to using the Maclaurin series for tanx?

Yes, there are limitations to using the Maclaurin series for tanx. The series only converges (approaches the true value) for values of x between -π/2 and π/2. Outside of this range, the series diverges and becomes increasingly less accurate as x gets farther from 0.

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