# Taylor series / 2nd deriv test

1. Jan 7, 2008

### jesuslovesu

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

Use the Taylor series about x = a to verify the second derivative test for a max or min. Show if f'(a) = 0 then f''(a) > 0 implies a min point at x = a ... Hint for a min point you must show that f(x) > f(a) for all x near enough to a.

2. Relevant equations

3. The attempt at a solution
f(x) = a0 + a1(x-a) + a2(x-a)^2 + ....
f'(x) = a1 + 2a2(x-a) + ...
f''(x) = 2a2

f'(a) = a1
if a1 = 0 then it's either a max or min
but I don't quite know what I should do to show that if f''(a) > 0 or < 0 that the point will be a max or min.
Should I do a limit?
$$\lim_{x \to a} a\right0 + a\right1 (x-a) + a\right2(x-a)^2 ... > a\right0$$ ?

Last edited: Jan 7, 2008
2. Jan 7, 2008

### Dick

At x=a your Taylor series has no a1 term, since f'(a)=0. So your series is just f(a)+(f''(a)/2)*(x-a)^2+.... You can ignore the higher order terms if x is 'near enough' to a.

3. Jan 7, 2008

### HallsofIvy

Staff Emeritus
Yes, if a is a max or min for f(x), then f'(a)= 0. That means that the taylor's series for f(x) is $a_0+ a_2(x- a)^2+$ higher order terms. Close to a, (x-a)3[/sup is very small compared to (x-a)2 (if x- a= 0.001, (x-a)2= (0.001)2= 0.000001 and (x-a)3= (0.001)3= 0.000000001. (x-a)2 is 1000 times as large as (x-a)3). Ignoring higher power terms, f(x)= f(a)+ a_2(x-a)2. Of course, (x-a)2 is never negative and for x not equal to a is not 0. Whether f(x) is f(a)+ a positive number or f(x)= f(a)- a positive number depends entirely upon whether a2 is positive or negative-which in turn depends upon whether f"(a) is positive or negative.

Last edited: Jan 7, 2008