Equation involving inverse trigonometric function

[phymath7]

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.

 

[BvU]

For x close to 1 your lefthand side does not exceed the 1.5 from the righthand side. But arcsin does ($\pi/2$), so your approach doesn't work...

 

[phymath7]

For x close to 1 your lefthand side does not exceed the 1.5 from the righthand side. But arcsin does ($\pi/2$), so your approach doesn't work...

What is x?The equation doesn't contain x.Moreover,k can't exceed 1,so your assumption(taking k as ($\pi/2$))is wrong.Perhaps you didn't notice it's an inverse trig function.

 

[Mark44]

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

The left side is not the Maclaurin series expansion for the inverse sine function. It's the first couple of nonzero terms for the expansion of the sine function.

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.

Or maybe your setup for the area calculation was incorrect. The equation you're asked to solve doesn't seem to have a solution other than by numerical means, so I would suggest going back to the original problem to see if you've made a mistake there.

 

[Frabjous]

sin(1.5)=.997
I would take the sine of both sides and do a taylor series around 1.

 

[phymath7]

The left side is not the Maclaurin series expansion for the inverse sine function. It's the first couple of nonzero terms for the expansion of the sine function.
Or maybe your setup for the area calculation was incorrect. The equation you're asked to solve doesn't seem to have a solution other than by numerical means, so I would suggest going back to the original problem to see if you've made a mistake there.

The left side is the maclaurin series for inverse sine function.Have a look in here:https://mathworld.wolfram.com/MaclaurinSeries.html.

Or maybe your setup for the area calculation was incorrect. The equation you're asked to solve doesn't seem to have a solution other than by numerical means, so I would suggest going back to the original problem to see if you've made a mistake there.

There's no error in the setup(checked from the solution manual)

 

[phymath7]

sin(1.5)=.997

Where does this assumption come from?How does that help?

I would take the sine of both sides and do a taylor series around 1.

Taking sine of both sides leave us with:
$$k=sin( \frac {k^2}{2} +1)$$.How do we proceed?

 

[Mark44]

The left side is the maclaurin series for inverse sine function.Have a look in here:https://mathworld.wolfram.com/MaclaurinSeries.html.

You're right. The first couple of terms of the Maclaurin series for sin(x) looks similar -- $x - \frac {x^3}6 \pm \dots$, and that fooled me into thinking that you had used the series for sine rather than arcsine.

 

[BvU]

What is x?The equation doesn't contain x.Moreover,k can't exceed 1,so your assumption(taking k as ($\pi/2$))is wrong.Perhaps you didn't notice it's an inverse trig function.

Sorry, old habit to use $x$ as independent variable on horizontal axis. Your $k$, to be sure.
And I did not say $k = \pi/2$ but the value of the $\arcsin$ function near $k=1$.A simple plot clarifies things. Of course, you had made it already ?

Clearly the solution is within the red circle. So $k$ should be close to 1, and any series expansion should be around 1, not around zero.

$\$

 

[Frabjous]

Where does this assumption come from?How does that help?

Taking sine of both sides leave us with:
$$k=sin( \frac {k^2}{2} +1)$$.How do we proceed?

I then plugged in the bounds, 0 and 1. Since k=1 is close to the solution, it is preferable to take series expansion there because it will require less terms for the given number of significant digits required.

 

[phymath7]

I then plugged in the bounds, 0 and 1. Since k=1 is close to the solution, it is preferable to take series expansion there because it will require less terms for a given number of significant digits required.

It' clear from graphing utility that k is close to 1,but how do we use that guess?Mainly,I am asking how my Maclaurin series expansion is wrong?Do I need to take more terms?If I do ,then things get more complicated(due to terms with higher power).How to overcome this?

 

[Frabjous]

It' clear from graphing utility that k is close to 1,but how do we use that guess?Mainly,I am asking how my Maclaurin series expansion is wrong?Do I need to take more terms?If I do ,then things get more complicated(due to terms with higher power).How to overcome this?

Why are you wedded to the Maclaurin series? It is preferrable to do series expansions as close as possible to the solution so that you can minimize the number of terms that you need.

 

[phymath7]

It is preferrable to do series expansions as close as possible to the solution so that you can minimize the number of terms that you need.

Do I take a=1 in the Taylor series to do the series expansion?That's what you mean?

 

[BvU]

Yes. I take it you can accept a numerical solution ?

$\$

 

[Frabjous]

Do I take a=1 in the Taylor series to do the series expansion?That's what you mean?

Yes. You are expanding in powers of k-a. For a=0 (maclaurin) you are doing powers near 1. For a=1 you are doing powers near 0 which has much better convergence.

 

[BvU]

There's a small problem: $f'(1)$ doesn't exist for $f=\arcsin$.
A numerical solution, then ?

 

[Frabjous]

There's a small problem: $f'(1)$ doesn't exist for $f=\arcsin$.
A numerical solution, then ?

View attachment 324618

Which is why I said to take the sine of both sides of the equation in post #5.

 

[BvU]

Works like a dream !