How to find the inverse of a complicated function?

[ColdheartedGod]

Homework Statement

Find the inverse function for a given function.

Relevant Equations

f(x) =(x+(x^2+1)) ^(1/3)+(x-(x^2+1))^(1/3)

 

[QuantumQuest]

In order to be helped in a useful way you must first put your own efforts. So, what have you tried / come up with, so far?

 

[BvU]

Hello Cold one,

Read the guidelines to find out why we are not allowed to help and why your thread will probably be deleted and end up in the PF black hole

Your text and picture don't agree !
Read and check what you post

1. Learn to post using $\LaTeX$
2. post an attempt at solution

 

[ColdheartedGod]

I'm so sorry for not reading the guidelines carefully!
Here's my current attempt.

However I cannot go any further. I've tried using sin(x) to the power of three to replace sin(x) a second time but it doesn't go well.
I really would like to know if I'm on the right track or if there's other approach to the solution.

 

[BvU]

other approach

Did you try to write out $y^3$ ?

 

[BvU]

Never mind, doesn't look that good either

 

[BvU]

Other tack: try $x = \cosh(z)$

[edit] I mean the other one: sinh !

 

[BvU]

Concerning

I'm so sorry

And

1. Learn to post using $\LaTeX$

Example: (right-click on formula to show math as $\TeX$ comands)
$$y = \sqrt[\leftroot{-3}\uproot{12}3]{x+\sqrt{x^2+1}} + \sqrt[\leftroot{-3}\uproot{12}3]{x-\sqrt{x^2+1}} +$$
Or simpler: your f(x) =(x+(x^2+1)) ^{(1/3)}+(x-(x^2+1))^{(1/3)} -- adorned with {} to group the (1/3) and enclosed in double $$ becomes
$$f(x) =(x+(x^2+1)) ^{1/3}+(x-(x^2+1))^{1/3} $$
when it should have been
$$f(x) =\left(x+(x^2+1)^{1/2}\right ) ^{1/3}+\left(x-(x^2+1)^{1/2}\right)^{1/3} $$
Enough showing off; back to the exercise at hand
Any light at the end of the tunnel ?

 

[ColdheartedGod]

Concerning

And
Example: (right-click on formula to show math as $\TeX$ comands)
$$y = \sqrt[\leftroot{-3}\uproot{12}3]{x+\sqrt{x^2+1}} + \sqrt[\leftroot{-3}\uproot{12}3]{x-\sqrt{x^2+1}} +$$
Or simpler: your f(x) =(x+(x^2+1)) ^{(1/3)}+(x-(x^2+1))^{(1/3)} -- adorned with {} to group the (1/3) and enclosed in double $$ becomes
$$f(x) =(x+(x^2+1)) ^{1/3}+(x-(x^2+1))^{1/3} $$
when it should have been
$$f(x) =\left(x+(x^2+1)^{1/2}\right ) ^{1/3}+\left(x-(x^2+1)^{1/2}\right)^{1/3} $$
Enough showing off; back to the exercise at hand
Any light at the end of the tunnel ?

Thanks for the explanations!
Well, it seems that my current knowledge in trigonometry isn't good enough to carry that part so I'm now struggling to find out more about sinh functions and how to apply it in this case.

 

[PeroK]

Thanks for the explanations!
Well, it seems that my current knowledge in trigonometry isn't good enough to carry that part so I'm now struggling to find out more about sinh functions and how to apply it in this case.

The most important hyperbolic trig identity is $\sinh^2 \theta + 1 = \cosh^2 \theta$.

You can get hyperbolic multiple angle formulas quite easily using the definitions:

$\cosh \theta = \frac{e^\theta + e^{-\theta}}{2}, \ \ \sinh \theta = \frac{e^\theta - e^{-\theta}}{2}$

In this case you can get simple formulas for $\sinh \theta \pm \cosh \theta$. And:

$\sinh^3 \theta = \frac 1 8 ( e^\theta - e^{-\theta})^3 = \frac 1 8 (e^{3\theta} - 3e^\theta + 3e^{-\theta} - e^{-3\theta}) = \dots$

This problem looks like a very good exercise in the use of the hyperbolic trig functions.

 

[epenguin]

Not that anything is guaranteed before you try, but have you tried straight algebra?

In order to not get over-long formulae I would write this as

y = (A^⅓ + B^⅓) at first, then look at y³. (Even if, as I see, this has been disadvised.)

Note that inverses are not necessarily unique, but if you even get one that would be something.

 

[ehild]

Not that anything is guaranteed before you try, but have you tried straight algebra?

In order to not get over-long formulae I would write this as

y = (A^⅓ + B^⅓) at first, then look at y³. (Even if, as I see, this has been disadvised.)

It is worth trying!

 

[vela]

Not that anything is guaranteed before you try, but have you tried straight algebra?

I'll third this recommendation. I didn't work it out, but I think I see how it goes. You get some nice simplification.

 

[epenguin]

Note that inverses are not necessarily unique, but if you even get one that would be something.

Not to mention that an inverse does not necessarily exist, in the reals, for all y. However to see into that in this case I think you need to make progress in the elaboration along the lines indicated.
You can easily see the function f(x) itself does not exist in the reals for x between -1 and 1.

 

[PeroK]

Not to mention that an inverse does not necessarily exist, in the reals, for all y. However to see into that in this case I think you need to make progress in the elaboration along the lines indicated.
You can easily see the function f(x) itself does not exist in the reals for x between -1 and 1.

$f(0) = 0$

What's the problem?

 

[PeroK]

I'll third this recommendation. I didn't work it out, but I think I see how it goes. You get some nice simplification.

What's the point learning about hyperbolic trig functions if you don't use them?

 

[Vanadium 50]

You can easily see the function f(x) itself does not exist in the reals for x between -1 and 1.

Which suggests, if you are going to go the substitution route, to pick a function with similar restrictions, e.g. sec, csc , cosh (maybe) or coth.

 

[epenguin]

$f(0) = 0$

What's the problem?

The problem is that I mistranscribed the function and thought it had a √(x² - 1) in it - but it showed that that case could arise, even if not here.

Happy Christmas everybody!

 

[WWGD]

Maybe you can follow recommendation of cubing by first leaving one of the radical expressions alone in one side before doing the cubing to avoid a mess of an expression.

 

[ehild]

Maybe you can follow recommendation of cubing by first leaving one of the radical expressions alone in one side before doing the cubing to avoid a mess of an expression.

It is not a good idea. Just take the cube of both sides as they are. The cube of $y=a+b$ is $y^3=a^3+b^3+3a^2b+3ab^2$ which can be written as $y^3=a^3+b^3+3(ab)(a+b)$. By substitution, it becomes very simple and x can be easily isolated.

 

[willem2]

It is not a good idea. Just take the cube of both sides as they are. The cube of $y=a+b$ is $y^3=a^3+b^3+3a^2b+3ab^2$ which can be written as $y^3=a^3+b^3+3(ab)(a+b)$. By substitution, it becomes very simple and x can be easily isolated.

You indeed get massive cancellation, but there's still a cube root left over. I get a cubic in x, lightly sprinkled with powers of y^3. I could now use cardano's formula, and maybe there's even more cancellation?
There is however an easier way. If you look at the original expression, you can see that it looks just like the outcome of cardano's formula. There is a depressed cubic in y with x as one of the parameters, which will have your expression as the real root. Solving this for x will be trivial.

 

[BvU]

Other tack: try $x = \cosh(z)$

[edit] I mean the other one: sinh !

Haven't seen much follow-up for this one. even though $\sinh + \cosh$ and $\sinh - \cosh$ yield such nice expressions ...

 

[archaic]

Haven't seen much follow-up for this one. even though $\sinh + \cosh$ and $\sinh - \cosh$ yield such nice expressions ...

I solved using that substitution, got my answer deleted though, as per the rules

 

[ehild]

I solved using that substitution, got my answer deleted though, as per the rules

I saw it, but it was a bit complicated. The solution is quite simple just a polynomial of y.

 

[ehild]

You indeed get massive cancellation, but there's still a cube root left over. I get a cubic in x, lightly sprinkled with powers of y^3. I could now use cardano's formula, and maybe there's even more cancellation?
There is however an easier way. If you look at the original expression, you can see that it looks just like the outcome of cardano's formula. There is a depressed cubic in y with x as one of the parameters, which will have your expression as the real root. Solving this for x will be trivial.

No need for anything complicated, like Cardano formula. If you follow the hint in #Post 20, the only cube roots remaining are in a+b, which is identical with y.