How do i show the Taylor Series of inverse sine?

chinkmeista · #1 $Old$ November 7th, 2009, 07:42 PM

Hiya, i've got to show that the Taylor series for arcsine(x) = x + (x^3/6) + (3x^5/40) +....

I've tried using the general formula where i set it out as a table of derivatives i.e f'(a), f''(a), f'''(a) etc and when a=0 but it doesn't work and i only get x. The rest turns out as 0 for the coeffiecent when i substitute a=0 in to the higher derivatives.

Does anyone have any ideas? Many thanks.

Chinkmeista.

xxlvh · #2 $Old$ November 7th, 2009, 08:54 PM

Maybe the derivatives were incorrect?
$f(x) = sin^{-1}x$
$f(0) = sin^{-1}0 = 0$

$f'(x) = \frac{1}{\sqrt{1-x^{2}}}$

$f'(0) = 1$

$f''(x) = \frac{x}{(1-x^{2})^{\frac{3}{2}}}$

$f''(0)= 0$

$f'''(x) = \frac{2x^{2}+1}{(1-x^{2})^{\frac{5}{2}}}$

$f'''(0) = 1$

$f^{(4)}(x) = \frac{2x^{2}+1}{(1-x^{2})^{\frac{5}{2}}}$

$f^{(4)}(0) = 0$

$f^{(5)}(x) = \frac{24x^{4}+12x^{2}+9}{(1-x^{2})^{\frac{9}{2}}}$

$f^{(5)}(0) = 9$

This gives $\sin^{-1}x = x+\frac{x^{3}}{3!}+\frac{9x^{5}}{5!}+... = x+\frac{x^{3}}{6}+\frac{3x^{5}}{40}+...$

redsoxfan325 · #3 $Old$ November 7th, 2009, 08:58 PM

Quote:

Originally Posted by chinkmeista $View Post$

Hiya, i've got to show that the Taylor series for arcsine(x) = x + (x^3/6) + (3x^5/40) +....

I've tried using the general formula where i set it out as a table of derivatives i.e f'(a), f''(a), f'''(a) etc and when a=0 but it doesn't work and i only get x. The rest turns out as 0 for the coeffiecent when i substitute a=0 in to the higher derivatives.

Does anyone have any ideas? Many thanks.

Chinkmeista.

It works fine just taking the derivatives, though it is pretty tedious. The third derivative is $\frac{2x^2+1}{(x^2-1)^2\sqrt{1-x^2}}$ and the fifth is $\frac{24x^4+72x^2+9}{(x^2-1)^4\sqrt{1-x^2}}$ .

Note a pattern: The $(n+1)$ derivative of inverse sine is $\frac{(-1)^n(n!x^n+...)}{(x^2-1)^n\sqrt{1-x^2}}$

The numerator is some polynomial, but the only obvious pattern to it was the factorial coefficient for the highest power (though there is undoubtedly some pattern for the rest of the terms).

chinkmeista · #4 $Old$ November 7th, 2009, 10:14 PM

Quote:

Originally Posted by xxlvh $View Post$

Maybe the derivatives were incorrect?
$f(x) = sin^{-1}x$
$f(0) = sin^{-1}0 = 0$

$f'(x) = \frac{1}{\sqrt{1-x^{2}}}$

$f'(0) = 1$

$f''(x) = \frac{x}{(1-x^{2})^{\frac{3}{2}}}$

$f''(0)= 0$

$f'''(x) = \frac{2x^{2}+1}{(1-x^{2})^{\frac{5}{2}}}$

$f'''(0) = 1$

$f^{(4)}(x) = \frac{2x^{2}+1}{(1-x^{2})^{\frac{5}{2}}}$

$f^{(4)}(0) = 0$

$f^{(5)}(x) = \frac{24x^{4}+12x^{2}+9}{(1-x^{2})^{\frac{9}{2}}}$

$f^{(5)}(0) = 9$

This gives $\sin^{-1}x = x+\frac{x^{3}}{3!}+\frac{9x^{5}}{5!}+... = x+\frac{x^{3}}{6}+\frac{3x^{5}}{40}+...$

Ah ok, but how does the 3rd derivative equal the 4th one: $f'''(x) = \frac{2x^{2}+1}{(1-x^{2})^{\frac{5}{2}}}$ and $f^{(4)}(x) = \frac{2x^{2}+1}{(1-x^{2})^{\frac{5}{2}}}$ ?

And also, for the 3rd derivative, how did you get the 2x+1 on top? did you use the quotient rule?

redsoxfan325 · #5 $Old$ November 7th, 2009, 10:17 PM

Quote:

Originally Posted by chinkmeista $View Post$

Ah ok, but how does the 3rd derivative equal the 4th one: $f'''(x) = \frac{2x^{2}+1}{(1-x^{2})^{\frac{5}{2}}}$ and $f^{(4)}(x) = \frac{2x^{2}+1}{(1-x^{2})^{\frac{5}{2}}}$ ?

It's supposed to be $\frac{6x^3+9x}{(1-x^2)^{7/2}}$ , so $f^{(4)}(0)=0$ .

xxlvh · #6 $Old$ November 7th, 2009, 11:07 PM

Typing error, my apologies I was quickly trying to recopy those results from a program I used to find the derivatives. However they're still slightly off from the ones that redsoxfan325 has provided so it is best to disregard the ones I posted, it doesn't seem to always work for inverse trig functions..lucky coincidence it resulted with the same coefficients.

redsoxfan325 · #7 $Old$ November 7th, 2009, 11:11 PM

I was doing them on my TI-89. The only difference was the numerator for the $f^{(5)}(x)$ . Everything else was the same, though we were representing our denominators differently.