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Taylor series / 2nd deriv test

  1. Jan 7, 2008 #1
    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?
    [tex]\lim_{x \to a} a\right0 + a\right1 (x-a) + a\right2(x-a)^2 ... > a\right0 [/tex] ?
    Last edited: Jan 7, 2008
  2. jcsd
  3. Jan 7, 2008 #2


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    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.
  4. Jan 7, 2008 #3


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    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 [itex]a_0+ a_2(x- a)^2+[/itex] 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
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