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

AI Thread Summary
To find the inverse function for the approximation given by a = cos(x) / (3x^2 - π^2), the Newton-Raphson method is suggested as a viable approach. The original function is not a bijection in the specified range, resulting in two distinct inverse functions near a ≈ -0.086 ± 0.01. Using different initial values in the Newton-Raphson method can yield the two separate solutions. The discussion emphasizes the importance of numerical methods and the potential for power series to express the inverse functions. Overall, applying these techniques can help in accurately determining x(a) for the given approximation.
Theia
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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. :confused:

Thank you!
 
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Theia said:
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. :confused:

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:
$$x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} = x_k - \frac{a(3x_k^2 - \pi^2) - \cos x_k}{6ax_k + \sin x_k}$$
 
I like Serena said:
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.
 
Theia said:
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}
 
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