Having problem finding an error that is less than what is given.

Question

so I have a problem with a problem that has to do with Taylor polynomials.

Here is what they want me to do: "Find a value do cos(1/10) with means of Taylor polynomial method, with an error less than 10^(-5)."

Now, I did the whole Taylor thing (chose Maclaurin, a=0) with a degree of n = 2 with f(x)=cos(x). And in order to find the error, I differentiated the third time and got the third derivative to be sin(c) with c being between 0 and 1/10.

The largest error I got was 1/6000 which in itself is greater than 1/100000 (0r 10^(-5)). I can't unfortunately post a picture in this forum it seems like, but here is how I computed my calculations.

Third degree of the derivative of the function f(x)=cos(x) is sin(x).

The error is absolut value of sin(c) multiplied by (x-a)^(3) and this being divided with 6 (n! of n=3). What I got was absolut value of sin(c)/6000. And since we want this error to be as big as possible, I will choose sin(c) to be 1 (c is between 0 and 1/10 so this is allowed). So my error is 1/6000 with is greater AND not less than 10^(-5).

I have this exam that I have to do again in Calculus in 4 weeks and I remember I solved this problem the same way as I have done now, but I got the right answer but I can't seem to get it now.

As far as the theory goes, please don't even waste your time on this. I just want the problem to be solved. So if you could attach a paper with the solution the same way as I did would be prefferable or just describe why I get an error being greater than (10^5)?

Thanks

Answer 1

The bound $0\le\sin c\le 1$ can be impoved by the bound $0\le \sin c\le c\le \frac1{10}$, since $c\in(0,\frac1{10})$. Therefore $0\le\cos \frac1{10}-P_2(\frac1{10})\le\frac1{60000}$.

Answer 2

We can use Taylor-Lagrange inequality: for $f \in \mathcal{C}^{n+1}$ with $f^{(n+1)} \le M$, $$ |f(b) - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(b-a)^k | \le \frac{M}{(n+1)!}(b-a)^{n+1} $$ Here we want the error to be less then $10^{-5}$, so applying this formula to $x \longmapsto \cos(x)$ for $b = (1/10)$, $a=0$ and $n=3$, we have $$ \frac{(1/10)^4}{4!} = 0.42 \times 10^{-5} < 10^{-5} $$ We can therefore compute our approximation $$ \cos(1/10) \approx 1 - \frac{(1/10)^2}{2} = 0.995 $$

Answer 3

Oh ok, so in other words, when the error is NOT less than what they want it to be, you just raise the degree of the Taylor polynomial (and obviously then the error term)? Beacause the higher the degree n the smaller the error, I kind of remember this from the lectures. Am I right?