Dealing with Domain Problems in Plotting: A Mathematician's Dilemma

[DivGradCurl]

I'm having a bit of difficulty plotting

$$y=\left( \frac{3}{2} \right) ^{1/3} \arcsin ^{2/3} x \qquad -1 < x < 1$$

I don't know how to work around the problem associated with the domain of $$\arcsin x$$. Is there any way to do it?

Any help is highly appreciated.

 

[whozum]

Sorry I was thinking arctan.

 

[whozum]

The arcsin function's domain fits in that domain doesn't it?

 

[DivGradCurl]

For x<0, y gives imaginary nos. That seems to be the problem. Is there any mathematical trick to work around it (probably no, I think... )?

 

[dextercioby]

Why?It's perfectly reasonable to extract root of 3-rd order from any real number.In your case,y>=0 and is very real...

Daniel.

 

[DivGradCurl]

Well, so the problem must be with me!

I've evaluated this with my TI-89 and Mathematica 5. I get imaginary nos. for x < 0. I double-checked my syntax, but it looks fine. Then, I must be doing something else wrong. I just don't have a clue.

 

[Data]

That's why you shouldn't trust your calculator. If I ask Maple what $(-2)^{1/3}$ is, it gives me a complex answer, because it's not perfect either. Of course, the answer it gives me is a cube coot of $-2$, but I want the real one. It doesn't know that.

Maybe this'll help:

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

Do you just need to sketch this, or plot it in detail?

 

[whozum]

Correct me if I am wrong, but squaring arcsin would result in a set of positive values, for which a cuberoot would always be positive, no?

 

[Data]

indeed, you can apply the square or the cube root first with no difference, as long as you're agreed that you want the real cube root.

 

[xanthym]

For x<0, y gives imaginary nos. That seems to be the problem. Is there any mathematical trick to work around it (probably no, I think... )?

Although there's no "actual" problem here with obtaining the (2/3) or even the Real (1/3) power of the negative argument for "arcsin()" in {-π ≤ arcsin() ≤ π}, if you still need to avoid it for some reason, the trick is to restrict the "arcsin()" range to {0 ≤ arcsin() ≤ 2*π} by adding "2*π" to your "arcsin()" function prior to evaluating the power:
(arcsin() + 2*π)^(2/3)


~~

 

[dextercioby]

Enjoy.It has no negative values,because of the square.


Daniel.

 

[DivGradCurl]

Enjoy.It has no negative values,because of the square.

I get the same picture here (attached). I agree with you that $$y\geq 0$$. The problem is that my textbook says $$y$$ is defined in the interval $$-1 < x < 1$$, while we both get $$0 \leq x \leq 1$$. That's what puzzles me.

About the imaginary numbers, I only get them when I evaluate a negative no. in my calculator. As Data said, I just shouldn't trust it. Unfortunately, they don't become real if I add "2*π" to my "arcsin()" function prior to evaluating the power. It's the nature of the beast!

Anyway, chances are my book has a mistake and the function is defined just in the interval $$0 \leq x \leq 1$$. Thus, everything else (imaginary nos.) could be disregarded. What do you guys think?

Thanks for your input

 

[xanthym]

I get the same picture here (attached). I agree with you that $$y\geq 0$$. The problem is that my textbook says $$y$$ is defined on the interval $$-1 < x < 1$$, while we both get $$0 \leq x \leq 1$$. That's what puzzles me.

About the imaginary numbers, I only get them when I evaluate a negative no. in my calculator. As Data said, I just shouldn't trust it. Unfortunately, they don't become real if I add "2*π" to my "arcsin()" function prior to evaluating the power. It's the nature of the beast!

Anyway, chances are my book has a mistake and the function is defined just in the interval $$0 \leq x \leq 1$$. Thus, everything else (imaginary nos.) could be disregarded. What do you guys think?

Thanks for your input

Something is very strange. The value {arcsin(x) + 2*π} should never be negative for {-1 ≤ x ≤ 1} with most implementations of "arcsin()".
ALSO -----> Make sure your calculator is set to RADIANS (not Degrees). Otherwise, if you still want to use Degrees, change method to {arcsin(x) + 360}

~~

 

[DivGradCurl]

Hey xanthym,

I've never said I got negative values as a result of the evaluation of the function. In fact, I said that whenever I evaluate anything out of the interval $$0 \leq x \leq 1$$ (i.e. x < 0 and x > 1) I get an imaginary result. Furthermore, I did not say the method you suggest is wrong.

This question came up as I looked up the answer of a differential equation (initial-value problem) in my textbook. I got the solution right, but the domain in which the solution is defined is said to be $$-1 < x < 1$$. I got in touch with you guys so that I know what is going on there.

In other words, it is possible that my book has a mistake. As far as I can see, the domain of

$$y=\left( \frac{3}{2} \right) ^{1/3} \arcsin ^{2/3} x$$

is defined for the real set of numbers only in the interval $$0 \leq x \leq 1$$. Thus, everything else (imaginary nos.) could be disregarded. However, I just can't say that what I've got here is correct. I need some feedback.

Thanks

 

[DivGradCurl]

I don't know whether it helps or not, but here is the problem exactly as I have here in the textbook:

"20. $$y^2 \left( 1 - x^2 \right) ^{1/2} \: dy = \arcsin x \: dx, \qquad y(0) = 0$$

(a) Find the solution of the given initial value problem in the explicit form.
(b) Plot the graph of the solution.
(c) Determine (at least approximately) the interval in which the solution is defined."

Just in case you're wondering...

Regards

 

[dextercioby]

I know my Maple was a bonehead,but this is ridiculous...

Daniel.

 

[dextercioby]

Since the solution is

$$y(x)=\left(\frac{3}{2}\right)^{\frac{1}{3}} \left(\arcsin^{2} x\right)^{\frac{1}{3}}$$

,i'd say the domain is [-1,+1],as if it were only $\arcsin$.

Daniel.

 

[DivGradCurl]

I've just found out what I was doing wrong. Both Mathematica and my TI-89 interpret

$$y=\left( \frac{3}{2} \right) ^{1/3} \arcsin ^{2/3} x \qquad (1)$$

and

$$y=\left( \frac{3}{2} \right) ^{1/3} \left[ \left( \arcsin x \right) ^2 \right] ^{1/3} \qquad (2)$$

differently. Only (2) displays both branches and has the correct domain $$[-1,1]$$.

Compared to $$[0,1]$$, it is not a subtle difference. That was my point all along, and I think it was worth taking the time to find out.

Thank you

 

[Data]

Like I said, the solution is to not trust your calculator. It will lie to you whenever possible (last week, I had to explain to my sister why sometimes it would give her the wrong answer when she took arcsines! hahah).

 

[dextercioby]

Thiago,it happened to my Maple,too.It's definitely something weird.That's why i posted both graphs to show that mathematical software is not 100% reliable

Daniel.