I'm trying to calculate the error involved in using $x$ to approximate $\sin({1^\circ})$. The Taylor polynomial is centered around $x=0$ (i.e. $x_0=0$), by the way. Since I have to use $x$ as the approximation for $\sin({x}),$ I'm guessing the Taylor polynomial will have to be of degree one:

$$f(x)=e^x=P_1(x)+R_1(x)$$ since the polynomial $P_1(x)$ does result in $f(x_0)+f'(x_0)(x-x_0)=\sin({0})+\cos({0})x=x$.

When calculating the remainder (i.e. $R_1(x)$), I get that it is given by: $$\frac{f^{(1+1)}(z)}{(n+1)!}(x-x_0)^2$$ where $z$ is a real number between $x$ and $x_0$. Plugging in the information for this case, then, $$R_1(x)=\frac{f''(z)}{2!}x^2$$ Since $f(z)=\sin({z})$, $f''(z)=-\sin({z})$. The error is then given by $|R_1(x)|=\lvert-\frac{1}{2}x^2 \sin{(z)}\rvert $.

Trying to find the boundaries for the error specifically when $x=1$ (since the question asks for the error in using $x$ to approximate $\sin(1^\circ)$) gets me that the lower bound is zero (if the estimation were perfectly accurate, which it isn't). I'm having trouble finding the upper boundary, though. The maximum value for |$\sin{(z)}$| is 1 (since the absolute value of the sine function oscillates between 0 and one). Since I'm being asked to calculate the error for when $x=1$, I think $x$ is supposed to be $1$. This would mean that the upper bound would be given by: |$-\frac{1}{2}(1)^2 $|$=\frac{1}{2}$. The given answer, though, is that the error is approximately $8.86 \cdot 10^{-7}$.

What did I do wrong? It always struck me that $x$ would be an unsuitable approximation for $\sin{(1^\circ)}$ since, at $x=1$, $\sin({1^\circ})$ is very small, certainly less than one (the maximum value for the sine function). I don't understand how the error can be so relatively small.

  • $\begingroup$ Two comments: Did you convert degrees into radians? And, more subtly, because $\sin$ is an odd function, the Taylor polynomial of degree $1$ at $0$ is in fact the polynomial of degree $2$, so you get a higher-order error estimate. $\endgroup$ – Ted Shifrin Jun 7 '18 at 1:17
  • $\begingroup$ Would I need to convert degrees into radians for the $x$ function? Or would I only need to do that for the sine function? $\endgroup$ – A. Lieber Jun 7 '18 at 1:18

The error is that you are plugging in $x=1$, instead of $x=1^\circ$. A degree is $\frac{\pi}{180}$ radians, so you should set $x=\frac{\pi}{180}$ instead.

(The right way to think about this is that the sine function is just defined by $\sin x=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$. When $x$ is an angle measure in radians, then $\sin x$ defined in this way happens to also coincide with the geometric "sine" function you learn about in trigonometry. But if you measure $x$ in degrees it does not.)

Also, you can get a much better upper bound for the $\sin(z)$ part. Since $z$ is between $0$ and $1^\circ$ and sine is increasing on this interval, the most that $\sin(z)$ could be is $\sin 1^\circ=\sin\frac{\pi}{180}$. Of course, you are trying to estimate $\sin\frac{\pi}{180}$, so you don't yet have an exact value for this that you can plug in. But you can use this to iteratively get a better bound for the error: once you have one bound for the error by saying $|\sin(z)|\leq 1$, you can then plug in the bound on $\sin \frac{\pi}{180}$ you got as a new bound on $|\sin z|$ to get a better estimate of the error.

  • $\begingroup$ So, then, would the upper bound be |$-\frac{1}{2}(\frac{\pi}{180})\sin\frac{\pi}{180}$|? $\endgroup$ – A. Lieber Jun 7 '18 at 1:24
  • $\begingroup$ Yes, modulo some obvious typos you seem to have. $\endgroup$ – Eric Wofsey Jun 7 '18 at 1:31
  • $\begingroup$ Ah. I see now. Thank you, Eric Wolfsey! $\endgroup$ – A. Lieber Jun 7 '18 at 1:31
  • $\begingroup$ Follow up question, why is it obligatory to use |$R_2(x)$| when calculating the error here (i.e. abs[$\frac{1}{3!}(-\cos{\frac{\pi}{180}})(\frac{\pi}{180})^3$)]$\approx 8.86 \cdot 10^{-7}$? I was told above that since "sine is an odd function, the Taylor polynomial of degree 1 at 0 is in fact the polynomial of degree 2". Is it just customary to always use the highest possible degree? $\endgroup$ – A. Lieber Jun 7 '18 at 1:38
  • $\begingroup$ You might as well use the highest possible degree, since it will give a better bound on the error. $\endgroup$ – Eric Wofsey Jun 7 '18 at 2:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.