Finding Double Roots for f(x;p) = cos x - 0.8 + px^2 using Python

[member 428835]

Homework Statement

What value of $p$ gives a double root for $f(x;p) = \cos x - 0.8+px^2$? I'm using python.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

I was thinking about choosing a window $p\in[a,b]$ such that $a=0.3$ yields 4 roots and $b=0.4$ yields 0 roots. Then cut this interval in half at $p = b-0.5(a+b)$ and evaluate the number of roots of $f(x;p)=0$. If no root is found then let $b = p$ and reiterate. If a root is found then let $a = p$ and reiterate.

The issue is, every root finder I see doesn't use a good value when no root is found. It returns an error message with an arbitrary value. I need a way to distinguish when a root is found or not...any ideas?

 

[fresh_42]

How about setting $f(x;p)=p(x-x_0)^2$ and examination of the consequences either by comparison of the terms or differentiation? Not sure how far you get with this approach, but it's an idea.

 

[member 587159]

Could you define double root please?

 

[member 428835]

Could you define double root please?

Sorry, I can see I've been somewhat inaccurate. A double root implies the multiplicity of a root is 2. If a double root occurs at $x=a$ for a given $p$, then $f(a;p) = f'(a;p) = 0$. This is a necessary condition...

So I just solved it! Following your question I applied the above two equation equalities and out came a good answer!