approximate cos(1/10) with three correct decimals around origin with Taylor polynomial

Question

Using the Maclaurin with two derivatives I get the following. Substituting x = 1/10 gives approx 0.9800, but the answers say it should be 0.995.

Answer 1

A Maclaurin series is a Taylor series expansion of a function about $0$,

$$f(x)=f(0)+f'(0)\cdot x+\frac{f''(0)}{2!}\cdot x^2+\frac{f'''(0)}{3!}\cdot x^3+\frac{f''''(0)}{4!}\cdot x^4+\dots\tag1$$

For $f(x)=\cos(x)$, we get:

$$\cos(x)=1+0-\frac{x^2}{2}+0+\frac{x^4}{24}+\dots=1-\frac{x^2}{2}+\frac{x^4}{24}+\dots\tag2$$

So, when $x=\frac{1}{10}$, we get:

$$\cos\left(\frac{1}{10}\right)=1-\frac{\left(\frac{1}{10}\right)^2}{2}+\frac{\left(\frac{1}{10}\right)^4}{24}+\dots=\frac{238801}{240000}+\dots\tag3$$

And:

$$\frac{238801}{240000}\approx0.995004166666\tag4$$

Note: the error term is given by the terms we did not put into our approximation. The more terms you pick the more accurate it will become.

Answer 2

The second term of the Taylor series is $f'(0)(x-0)$. Because the sine is $0$ at $0$ the term is zero. The third term is $\frac 1{2!}f''(0)(x-0)^2=-\frac 12(0.1)^2=-0.005$. Adding this to $1$ gives $0.995$. The series is alternating, so the error is of the sign of and smaller than the first neglected term. The next term is $\frac 1{4!}\cdot 1 \cdot 0.1^4 \ll .001$, so our three places are correct.

Answer 3

In your series, you write $\cos x$ and $\sin x$ when you should be writing $\cos 0 $ and $\sin 0$. The reason you stop after the 2nd-order term is that this is an alternating series. The error is bounded by the next term, $(0.1)^4/4!$, so two non-zero terms are sufficient.