Taylor expansion: Complex, composite functions

**MathCrusader** · January 24th 2013, 04:33 PM

I am to determine $f^{(3)}(0)$ by using Taylor expansion about $x= 0$ for

$f(x) = \sin^3 (\ln (1+x)) \, .$

In our toolkit, we have

$\ln (1+x) = x - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4 + \text{O}(x^5)\, , \ \ x \rightarrow 0 \, .$
$\sin (x) = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + \text{O}(x^9)\, , \ \ x \rightarrow 0 \, .$

Now, plugging in the polynomial for $\ln (1+x)$ into the polynomial for $\sin (x)$ and then cubing(!!!) it all sounds not like something fun at all to do, but I have done similar problems for and that is really the best(?) way to go.

Do you guys have any tips for how I can make the polynomial multiplication easier and quickly obtain the sought coefficient, instead of having to really multiply out too many terms and add upp their coefficients etc.

**Prove It** · January 24th 2013, 05:05 PM

If you're evaluating the third derivative at 0, then surely you would need to differentiate this function three times first...

**Prove It** · January 24th 2013, 05:31 PM

Originally Posted by MathCrusader

I am to determine $f^{(3)}(0)$ by using Taylor expansion about $x= 0$ for

$f(x) = \sin^3 (\ln (1+x)) \, .$

In our toolkit, we have

$\ln (1+x) = x - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4 + \text{O}(x^5)\, , \ \ x \rightarrow 0 \, .$
$\sin (x) = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + \text{O}(x^9)\, , \ \ x \rightarrow 0 \, .$

Now, plugging in the polynomial for $\ln (1+x)$ into the polynomial for $\sin (x)$ and then cubing(!!!) it all sounds not like something fun at all to do, but I have done similar problems for and that is really the best(?) way to go.

Do you guys have any tips for how I can make the polynomial multiplication easier and quickly obtain the sought coefficient, instead of having to really multiply out too many terms and add upp their coefficients etc.

On second thought, remembering that the Taylor Expansion of a function is $\displaystyle \begin{align*} f(x) = f(a) + f'(a) (x - a) + \frac{f''(a)}{2}(x-a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \dots \end{align*}$ , you should be able to read off what $\displaystyle \begin{align*} f'''(0) \end{align*}$ is if you have the cube term.

Luckily since the sine function only has a linear term and a cubic term to worry about, the expansions won't be difficult.

$\displaystyle \begin{align*} \ln{(1 + x)} &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \\ \\ \sin{(x)} &= x - \frac{x^3}{3!} + \dots \\ \\ \sin{\left[ \ln{(1 + x)} \right]} &= \left( x - \frac{x^2}{2} + \frac{x^3}{3} + \dots \right) - \frac{\left( x - \frac{x^2}{2} + \frac{x^3}{3} + \dots \right)^3}{3!} + \dots \end{align*}$

Notice that when you expand $\displaystyle \begin{align*} \left( x - \frac{x^2}{2} + \frac{x^3}{3} \right)^3 \end{align*}$ , only the first term will be needed, since all other terms will be of higher order than $\displaystyle \begin{align*} x^3 \end{align*}$ , so really we can write

$\displaystyle \begin{align*} \sin{\left[ \ln{ \left( 1 + x \right) } \right]} &= x - \frac{x^2}{2} + \frac{x^3}{3} + \dots - \frac{x^3}{3!} + \dots \\ &= x - \frac{x^2}{2} + \frac{2x^3}{3!} + \dots - \frac{x^3}{3!} + \dots \\ &= x - \frac{x^2}{2} + \frac{x^3}{3!} + \dots \\ &= 0 \left( x - 0 \right)^0 + 1 \left( x - 0 \right) - \frac{1}{2}\left( x - 0 \right)^2 + \frac{1}{3!} \left( x - 0 \right)^3 + \dots \end{align*}$

So it should be clear that $\displaystyle \begin{align*} f'''(0) = 1 \end{align*}$ .

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