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Hi members,
Could any one help me with the problem following?
If x is a real number and P is a polynomial function, then
lim {P(x+3h)+P(x-3h)-2P(x)}/h^2
h->0
A)0
B)6P'(x)
C)3P''(x)
D)9P''(x)
E) 00
I guess D should be the answer, I need an explanation.
Thank You
arildno
Jun10-04, 01:56 AM
The easiest way is to use Taylor-series expansions of the terms P(x+3h), P(x-3h)
to verify your guess.
Then we have, for example:
P(x+3h)=P(x)+P'(x)3h+1/2P''(x)(3h)^2+O(h^3)
O(h^3) is a higher order term, i.e lim h->0 O(h^3)/h^2=0
Thank you, arildno
It couldn't be more wonderful solutions.
lobsterism
Nov12-10, 06:32 PM
It looks messy, but it's really no different from any other limit problem. What is typically the easiest way to find the limit of 0/0? L'Hopital's rule!
Differentiate top and bottom with respect to h (not x!) twice, using chain rule for top terms. So the first differentiation gives [3P'(x+3h)-3P'(x-3h)]/2h (notice the third term has no h, so drops out). The second round, you get [9P''(x+3h)+9P''(x-3h)]/2. Then setting h=0 gives the desired answer.
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