1. ## Taylor expansion questions

Hello im new to this forum so im not sure if this is on the right forum. Im having trouble with taylors when 2 functions are combined. E.g. Expand ln(sinx) up to x^3 with a=pi/6. Could someone show me the steps to do this, thank you.

2. Do you know the formula?

3. [Fixed post based on SammyS's corrections below. Thanks SammyS.]

Just start from the beginning. It is the same as any function.
Take derivatives:
$f(x) = \ln(\sin(x))$
$f'(x) = \frac{1}{\sin(x)}\cos(x) = \cot(x)$
$f''(x) = -\csc^2(x)$
$f'''(x) = 2\csc^2(x)\cot(x)$

Evaluate derivatives:
$f(\frac{\pi}{6}) = \ln(\frac{1}{2})$
$f'(\frac{\pi}{6}) = \sqrt{3}$
$f''(\frac{\pi}{6}) = -4$
$f'''(\frac{\pi}{6}) = 8\sqrt{3}$

Now just write the Taylor expansion:
$
f(x)=f(\frac{\pi}{6}) + f'(\frac{\pi}{6})(x - a) +\displaystyle {{f''(\frac{\pi}{6})}\over{2!}}(x - a)^2 + {{f'''(\frac{\pi}{6})}\over{3!}}(x - a)^3 + ...
$

$
=
-\ln(2) + \sqrt{3}(x - \frac{\pi}{6}) - 2(x - \frac{\pi}{6})^2 + \displaystyle{{{4\sqrt{3}}\over{3}}}\textstyle (x - \frac{\pi}{6})^3 + ...
$

Disclaimer: Note this is just to give an idea (there may be mistakes).

4. Thanks i understand that method, but is there any way you can use the separate expansions of lnx and sinx and combine them to get the answer?

5. In this case, I don't think there is an "easy" way.

If it was ln(x)sin(x), then you can multiply two Taylor expansions, but ln(sin(x)) is nested.

6. Ah ok i see, thank you

7. Originally Posted by snowtea
Just start from the beginning. It is the same as any function.
Take derivatives:
$f(x) = \ln(\sin(x))$
$f'(x) = \frac{1}{\sin(x)}\cos(x) = \cot(x)$
$f''(x) = -\csc^2(x)$
$f'''(x) = 2\csc^2(x)\cot(x)$

Evaluate derivatives:
$f(\frac{\pi}{6}) = \ln(\frac{1}{2})$
$f'(\frac{\pi}{6}) = \sqrt{3}$
$f''(\frac{\pi}{6}) = -4$
$f'''(\frac{\pi}{6}) = 8\sqrt{3}$

Now just write the Taylor expansion:
$
f(\frac{\pi}{6}) + f'(\frac{\pi}{6})(x - a) + f''(\frac{\pi}{6})(x - a)^2 + f'''(\frac{\pi}{6})(x - a)^3 + ...
$

$
=
\ln(\frac{1}{2}) + \sqrt{3}(x - \frac{\pi}{6}) - 4(x - \frac{\pi}{6})^2 + 8(x - \frac{\pi}{6})^3 + ...
$

Disclaimer: Note this is just to give an idea (there may be mistakes).
I expect you're supposed to evaluate the derivatives at $\displaystyle \frac{\pi}{2}$ in order to get rational values for the coefficients. The whole point of a Taylor Series is to be able to evaluate the function at any value, it's pretty hard to do that when your coefficients can't even be evaluated...

8. Originally Posted by snowtea
...

Now just write the Taylor expansion:
$
f(\frac{\pi}{6}) + f'(\frac{\pi}{6})(x - a) + f''(\frac{\pi}{6})(x - a)^2 + f'''(\frac{\pi}{6})(x - a)^3 + ...
$

$
=
\ln(\frac{1}{2}) + \sqrt{3}(x - \frac{\pi}{6}) - 4(x - \frac{\pi}{6})^2 + 8(x - \frac{\pi}{6})^3 + ...
$

Disclaimer: Note this is just to give an idea (there may be mistakes).

You forgot the factorials !

Also dropped the $\displaystyle \sqrt3$ from the cubic term.

$
f(x)=f(\frac{\pi}{6}) + f'(\frac{\pi}{6})(x - a) +\displaystyle {{f''(\frac{\pi}{6})}\over{2!}}(x - a)^2 + {{f'''(\frac{\pi}{6})}\over{3!}}(x - a)^3 + ...
$

$
=
-\ln(2) + \sqrt{3}(x - \frac{\pi}{6}) - 2(x - \frac{\pi}{6})^2 + \displaystyle{{{4\sqrt{3}}\over{3}}}\textstyle (x - \frac{\pi}{6})^3 + ...
$