# Taylor Series (again)Watch

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#1
Ok, i really have no idea on this... how do you even formulate the taylor series for q3?? Can someone explain this to me simply as if i'm an idiot - because at the moment i sure feel like one.

http://www.maths.ox.ac.uk/current-st...nalysis2-8.pdf

Thanks
0
13 years ago
#2
Let x be in [0, 2]. Then

f(0) = f(x) + (-x)f '(x) + (1/2)(-x)^2 f ''(k) for some k with 0 <= k <= x
f(2) = f(x) + (2 - x)f '(x) + (1/2)(2 - x)^2 f ''(K) for some K with x <= K <= 2

Subtracting,

f(2) - f(0) = 2f '(x) + (1/2)(2 - x)^2 f ''(K) - (1/2)(-x)^2 f ''(k)

Making f '(x) the subject,

f '(x) = (1/2)f(2) - (1/2)f(0) - (1/4)(2 - x)^2 f ''(K) + (1/4)(-x)^2 f ''(k)

|f '(x)|
<= (1/2)|f(2)| + (1/2)|f(0)| + (1/4)(2 - x)^2 |f ''(K)| + (1/4)(-x)^2 |f ''(k)|
<= (1/2) + (1/2) + (1/4)(2 - x)^2 + (1/4)(-x)^2
= 1 + (1/4)(4 - 4x + 2x^2)
= 1 + (1/2)(2 - 2x + x^2)
= 1 + (1/2)((x - 1)^2 + 1)
<= 2
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