How do I do this the way I am supposed to? (without differentiating 9 times)

Any help would be greatly appreciated!

Thanks in advance!

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- May 16th 2010, 02:43 PMs3aCompute the 9th derivative
How do I do this the way I am supposed to? (without differentiating 9 times)

Any help would be greatly appreciated!

Thanks in advance! - May 16th 2010, 03:10 PMJester
Use the hint. The Taylor series for $\displaystyle \tan^{-1}x$ is

$\displaystyle

\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots.

$

Subs in for $\displaystyle x \rightarrow \frac{x^3}{4}$. You really only need to consider powers up to $\displaystyle x^9$. - May 16th 2010, 03:13 PMskeeter
- May 16th 2010, 03:49 PMs3a
For arctan(x) I get = sigma from 0 to inf of (-x^(2n)) which gives - x - x^3 /3 - x^5 /5 etc without the alternating. What did I do wrong so far? I did d/dx arctan(x) = 1/(1-(-x^2)) then integrate the sum.

- May 16th 2010, 03:56 PMskeeter
- May 16th 2010, 09:35 PMs3a
Ok I see how to get the summation of arctan(x^3 /4) but then I don't know what to do. Do I just write down the polynomials up to the x^9 power and do something there? What about f^(n) (a) x^n /n! formula ? What I was trying to do before my current attempt was to just do the summation of f^(n) (a) x^n /n! and to replace f^(n) with it's appropriate algebraic expression but it was tough to compute the second derivative and on.

Sorry for being stupid :( - May 17th 2010, 12:37 AMdrumist
So you're OK with this, correct?

$\displaystyle f(x) = \arctan \left(\frac{x^3}{4}\right) = \frac{x^3}{4} - \frac{x^9}{4^3 \cdot 3} + \frac{x^{15}}{4^5 \cdot 5} - \frac{x^{21}}{4^7 \cdot 7} + \cdots$

Now, consider what happens when you take 9 derivatives of this expression.

The first term $\displaystyle \frac{x^3}{4}$ will eventually become zero (beyond 3 derivatives).

The second term $\displaystyle -\frac{x^9}{4^3 \cdot 3}$ will become a constant. What constant does it become?

The remaining terms will turn into the form $\displaystyle C \cdot x^n$ where $\displaystyle C$ is a constant and $\displaystyle n$ is a positive integer. When you substitute $\displaystyle x=0$ into a term of this form, does it make sense that these terms will all become zero?

Therefore, there is only one nonzero term, which is the second term. - May 17th 2010, 09:03 AMs3a
OMG! Yes, I get it now, thanks!