1. ## Find Function

Find all function $f: \mathbb{R} \longrightarrow [-1,1]$ verifying $\forall x \in \mathbb{R} , f(2x)=2\left(f(x)\right)^2 -1$ , such that $f(0)=1$ and $\lim_{x\rightarrow 0}\frac{1-f(x)}{x^2} = a$ , $a \in \mathbb{R}^{+}$

2. Actually cos[(2a^1/2)x] will work for any a in R+

3. hello,

hi Calculus26 , I know it's $\cos(x\sqrt{2a})$ but I need proof ...

4. If you need to verify:

f(0) =1 is obvious

using the double angle formula cos(2t) = 2cos^2(t)-1 for t = (2a)^1/2 x

which is precisely 2f(x)^2 -1

double application of L'Hopital's Rule verifies the limit

Now if you need to come up with cos((2a)^1/2 x) from first principles then that's another story. Do we have any other conditions such as differentiability ?

My thought process was simply

1. The condition f:R - [-1,1] led me to cosines and sines

2. f(0) =1 narrowed this to cos(bx)

3. the limit led to b^2 = 2a

Let me know if this not sufficient and I'll consider it further

5. ## Integers identity

Hello,

thanks claculus26