Proof of tan^(-1z) maclauren series

In summary: Now substituting $\displaystyle \begin{align*} x = z \end{align*}$, the series becomes$\displaystyle \begin{align*} \arctan{(z)} = \sum_{n = 0}^{\infty} \frac{\left( -1 \right) ^n z^{2n + 1}}{2n + 1} \end{align*}$ if $\displaystyle \begin{align*} |z| < 1 \end{align*}$.In summary, we can prove that $\displaystyle \begin{align*} \arctan{(z)} = z - \frac{z^3}{3} + \frac
  • #1
Stumped1
9
0
prove that
\(\displaystyle tan^{-1}z=z-z^3/3 + z^5/5 -z^7/7 + ...\) for \(\displaystyle |z|<1\)

I know of a proof for this that takes the derivative, does long division, then integrates.

I would like a proof of this using the known Maclaurin series for e^z, cosz, or sinz.Is there a way to do this using these?

Thanks for any help!
 
Physics news on Phys.org
  • #2
Stumped said:
prove that
\(\displaystyle tan^{-1}z=z-z^3/3 + z^5/5 -z^7/7 + ...\) for \(\displaystyle |z|<1\)

I know of a proof for this that takes the derivative, does long division, then integrates.

I would like a proof of this using the known Maclaurin series for e^z, cosz, or sinz.Is there a way to do this using these?

Thanks for any help!

The easiest way is remember that $\displaystyle \begin{align*} \frac{d}{dx} \left[ \arctan{(x)} \right] = \frac{1}{1 + x^2} \end{align*}$.

Now notice $\displaystyle \begin{align*} \frac{1}{1 + x^2} = \frac{1}{1 - \left( -x^2 \right) } \end{align*}$, and if we recall that a geometric series has $\displaystyle \begin{align*} \sum_{n = 0}^{\infty} r^{n} = \frac{1}{1 - r} \end{align*}$ if $\displaystyle \begin{align*} |r| < 1 \end{align*}$, that means $\displaystyle \begin{align*} \frac{1}{1 - \left( - x ^2 \right) } = \sum_{n = 0}^{\infty} { \left( -x^2 \right) ^n } \end{align*}$ if $\displaystyle \begin{align*} \left| -x ^2 \right| < 1 \end{align*}$. Thus

$\displaystyle \begin{align*} \frac{1}{1 + x^2} &= \sum_{n = 0}^{\infty}{ \left( -1 \right) ^n x^{2n} } \textrm{ if } |x| < 1 \\ \int{ \frac{1}{1 + x^2} \, dx} &= \int{ \sum_{n = 0}^{\infty}{ \left( -1 \right) ^n x^{2n} }\,dx} \\ \arctan{(x)} + C &= \sum_{n = 0}^{\infty} \frac{ \left( -1 \right) ^n x^{2n + 1} }{2n + 1} \end{align*}$

and by substituting $\displaystyle \begin{align*} x = 0 \end{align*}$ it can be seen that $\displaystyle \begin{align*}C = 0 \end{align*}$.

Thus $\displaystyle \begin{align*} \arctan{(x)} = \sum_{n = 0}^{\infty} \frac{\left( -1 \right) ^n x^{2n + 1}}{2n + 1} \end{align*}$ if $\displaystyle \begin{align*} |x| < 1 \end{align*}$.
 
Last edited:

FAQ: Proof of tan^(-1z) maclauren series

What is the Maclauren series for "tan^(-1)z"?

The Maclauren series for "tan^(-1)z" is given by:
1/z + (1/3)z^3 + (2/15)z^5 + (17/315)z^7 + (62/2835)z^9 + ...

How is the Maclauren series for "tan^(-1)z" derived?

The Maclauren series for "tan^(-1)z" is derived using the Taylor series expansion of the arctangent function. By substituting z for x, we can obtain the series representation for "tan^(-1)z".

What is the purpose of the Maclauren series for "tan^(-1)z"?

The Maclauren series for "tan^(-1)z" is used to approximate the arctangent function for small values of z. It can also be used to simplify calculations involving the arctangent function.

Is the Maclauren series for "tan^(-1)z" accurate for all values of z?

No, the Maclauren series for "tan^(-1)z" is only accurate for small values of z. As z increases, the accuracy of the series decreases.

Can the Maclauren series for "tan^(-1)z" be used to find the exact value of the arctangent function?

No, the Maclauren series for "tan^(-1)z" is an approximation and cannot be used to find the exact value of the arctangent function. However, it can provide a close approximation for small values of z.

Similar threads

Replies
9
Views
2K
Replies
4
Views
2K
Replies
2
Views
1K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
2
Views
2K
Replies
10
Views
2K
Back
Top