# Taylor expansion

#### talolard

1. The problem statement, all variables and given/known data
Let f be differentiable on [a,b] and f'(a)=f'(b)=0. Prove that if f'' exists then there exists a point c in (a,b) such that
$$test$$
$$|f''(c)| \geq \frac{4}{(b-a)^2}|f(b)-f(a)|$$

2. Relevant equations

All of the equations are supposed to be in absolute value but I had trouble getting tha tto apear so please assume that everything is bsolute value. I didnt use any inequalities so it shouldnt make much difference.

3. The attempt at a solution
Assume f'' exists
Via Taylor we get that
1. $$f(x) (at a) = f(a) + f'(a)x + \frac{f''(c)}{2}(x-a)^2 = f(a)+\frac{f''(c)}{2}(x-a)^2$$
2. $$f(x) (at b) = f(b) + f'(b)x + \frac{f''(d)}{2}(x-b)^2 = f(a)+\frac{f''(d)}{2}(x-b)^2$$
Then Via 1 we get that
3. $$|f(b)|= |f(a)+\frac{f''(c)}{2}(b-a)^2| \iff |\frac{2(f(b) -f(a))}{(b-a)^2}| = | f''(c)|$$
via 2 we get that
4. $$|f(a)|= |f(b)+\frac{f''(d)}{2}(b-a)^2| \iff |\frac{2(f(b) -f(a))}{(b-a)^2}| = | f''(d)|$$
adding equation 3 and 4 together we get
$$|f''(c)|+|f''(d)|=|frac{4(f(b) -f(a))}{(b-a)^2}|$$
then $$\frac |{f''(c)+f''(d)}{2}|=|f''(e)| =| \frac{2(f(b) -f(a))}{(b-a)^2}|$$

At this point I am stuck. I'm thinking that I can show that c and d are actualy the same point.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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"Taylor expansion"

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