- #1

phymath7

- 48

- 4

- Homework Statement
- Solve the following equation:

$$\sin^{-1}(k)=\frac {k^2}{2} +1$$ where 0<k<1.

- Relevant Equations
- Maclaurin series expansion for ##sin^{-1}(k)##

I came across the mentioned equation aftet doing a integral for an area related problem.Doing the maclaurin series expansion for the inverse sine function,I considered the first two terms(as the latter terms involved higher power of the argument divided by factorial of higher numbers),doing so the equation then became like the following:

$$k+\frac {k^3}{6}=\frac{k^2}{2} +1$$

But this equation gives one real solution and two complex solution.And the real solution is greater than 1 which doesn't satisfies the condition.So what's wrong?Do I need to consider more terms in the series?But doing so makes the equation much more hardr to solve which I guess the author hasn't intended in the problem.

$$k+\frac {k^3}{6}=\frac{k^2}{2} +1$$

But this equation gives one real solution and two complex solution.And the real solution is greater than 1 which doesn't satisfies the condition.So what's wrong?Do I need to consider more terms in the series?But doing so makes the equation much more hardr to solve which I guess the author hasn't intended in the problem.