How can I find the inverse function for a given approximation?

[Theia]

Hello all

I was doing some approximation to solve another problem, but got stuck when trying to figure out a suitable inverse functions for this:

$$a = \frac{\cos x}{3x^2 - \pi^2}$$, where $$0 \le x \le \pi$$.

What I need is the two functions $$x(a)$$ at least near $$a \approx -0.086 \pm 0.01$$ but I'm not quite sure how to do it well.

Thus far I tried some sort of numerical way, meaning that I put $$a = $$ something and made a table of the results, then tried to fit some sort of simple polynomial (quadratic and cubic), but the results were not very convincing. So now I'm wondering should I still continue playing with the numerics and try to find some good function shape to fit, or should I try some other way, e.g. write the functions $$x(a)$$ in terms of power serie.

Thank you!

 

[I like Serena]

Hello all

I was doing some approximation to solve another problem, but got stuck when trying to figure out a suitable inverse functions for this:

$$a = \frac{\cos x}{3x^2 - \pi^2}$$, where $$0 \le x \le \pi$$.

What I need is the two functions $$x(a)$$ at least near $$a \approx -0.086 \pm 0.01$$ but I'm not quite sure how to do it well.

Thus far I tried some sort of numerical way, meaning that I put $$a = $$ something and made a table of the results, then tried to fit some sort of simple polynomial (quadratic and cubic), but the results were not very convincing. So now I'm wondering should I still continue playing with the numerics and try to find some good function shape to fit, or should I try some other way, e.g. write the functions $$x(a)$$ in terms of power serie.

Thank you!

Hey Theia! ;)

The first notion that springs to my mind is to apply Newton-Raphson.

What are you thinking btw with "two" functions?
The graph is smooth near $$a \approx -0.086 \pm 0.01$$.

To apply Newton-Raphson, we can for instance define:
$$f(x) = a(3x^2-\pi^2) - \cos(x)$$
To find the inverse $x(a)$, we can now apply:
$$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} = x_k - \frac{a(3x_k^2 - \pi^2) - \cos x_k}{6ax_k + \sin x_k}$$

 

[Theia]

Hey Theia! ;)

The first notion that springs to my mind is to apply Newton-Raphson.

What are you thinking btw with "two" functions?
The graph is smooth near $$a \approx -0.086 \pm 0.01$$.

To apply Newton-Raphson, we can for instance define:
$$f(x) = a(3x^2-\pi^2) - \cos(x)$$
To find the inverse $x(a)$, we can now apply:
$$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} = x_k - \frac{a(3x_k^2 - \pi^2) - \cos x_k}{6ax_k + \sin x_k}$$

Thank you for your reply. And thank you for reminding me that method. I think I've even used it in some problem but it seems I simply can't remember all usefull things... *oops*

As for your question why two functions, there are two distinct inverse functions for $$a \approx -0.086 \pm 0.01$$ because the original function is not a bijection in that region. Somewhat similar situation as in case of inverting $$y = x^2$$, but not as simple as that one.

 

[I like Serena]

Thank you for your reply. And thank you for reminding me that method. I think I've even used it in some problem but it seems I simply can't remember all usefull things... *oops*

That's why we have sites like MHB to remind us of the things we're not thinking of, or even worse, have forgotten. ;)

As for your question why two functions, there are two distinct inverse functions for $$a \approx -0.086 \pm 0.01$$ because the original function is not a bijection in that region. Somewhat similar situation as in case of inverting $$y = x^2$$, but not as simple as that one.

Ah, of course. (Nod)
If we pick a negative initial value, say $x_0=-1$, for Newton-Raphson, we'll get the one, and with $x_0=+1$ we'll get the other.Oh, oh, and now that we have TikZ pictures, here's how the graph looks - just because we can:
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[xmin=-1.5,xmax=1.5,samples=101]
\addplot[blue, ultra thick] (x,{cos(deg(x)) / (3*x*x - 10)});
\end{axis}
\end{tikzpicture}