Taylor series

2011 Paper 1 Question 11

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(a)

Given it is for 4 marks, I suspect they desire the remainder term including.

f(x) = f(a) + f^{(1)}(a)(x-a) + f^{(2)}(a)\frac{(x-a)^2}{2!}+f^{(3)}(\xi)\frac{(x-a)^3}{3!}

where a < \xi < x.

(b)

For x=0, the Taylor series becomes

f(x) = f(0) + f^{(1)}(0)x + f^{(2)}(0)\frac{x^2}{2!}+f^{(3)}(0)\frac{x^3}{3!}+\cdots

(i)

f(x) = \frac{1}{(x^2+9)^{1/2}}

f(0) = 1/3

Note f(x) is an even function so we expect f^{(1)}(0) =f^{(3)}(0) =0 but we still need to differentiate to work out f^{(2)}(0).

\begin{aligned}f^{(1)}(x) &= (-1/2)(x^2+9)^{-3/2}\,2x \\ &= -\frac{x}{(x^2+9)^{3/2}} \\ &= -x\,f^3 \end{aligned}

As expected f^{(1)}(0)=0

Notice also the very useful device here of expressing the derivative in terms of the original function. It can save a lot of effort sometimes.

\begin{aligned} -f^{(2)}(x) &= x (3)f^2 \, f^{(1)}(x) + f^3 \\ \Rightarrow -f^{(2)}(0) &=\left( f(0) \right)^3 \\ &= 1/27 \end{aligned}

Putting it all together we obtain

{f(x) = 1/3 - 1/54 \, x^2 }