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We know that the small-angle approximation says that:

$$\sin(x) \approx x$$ $$\cos(x) \approx 1-\frac{x^2}{2}$$

I'm trying to understand, using the Taylor series and Lagrange error, why the first term of the $\sin(x)$ Maclaurin is so much better an approximation than $\cos(x)$ Maclaurin. It must have something to do with the fact that the first nonzero term for $\cos(x)$ is zero-degree, while the first nonzero term for $\sin(x)$ is first-degree. But the issue is that the next term in both cases has a derivative of $0$ at $x=0$, so I thought they'd be similarly good approximations. Why is this the case? And could the difference be shown clearly using Lagrange error bound? This is not homework, I'm just curious.

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    $\begingroup$ Very loosely speaking, the relation between $\sin$ and $\cos$ (which likely prompted your question to begin with) is quadratic $\sin^2 x + \cos^2 x = 1\,$. If you look at those Taylor series, $\sin^2 x \sim x^2$ and $\cos^2 x \sim 1 - x^2\,$ are of the same order. $\endgroup$ – dxiv May 7 '17 at 7:33
  • $\begingroup$ That's so interesting! Thank you! $\endgroup$ – rb612 May 7 '17 at 8:10
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The following diagram depicts the absolute difference between $\cos(x)$ and $1$ (pink), $\sin(x)$ and $x$ (blue), and $\cos(x)$ and $1-\frac12x^2$ (yellow).

enter image description here

It is apparent that $1$ is a quite bad approximation for for $\cos(x)$, and $1-\frac12x^2$ is a better approximation for $\cos(x)$ than $x$ is for $\sin(x).$ However, this is only a relative quality issue...

Further to this qualitative analysis, one has to consult the theory of the remainder term to the Taylor approximations.

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  • $\begingroup$ Thank you for your answer. I would like to see how the theory of the remainder term relates exactly to this problem as stated in my question. $\endgroup$ – rb612 May 7 '17 at 7:57
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I don't know what would constitute an actual answer to this question. Merely restating various points about Taylor series doesn't seem like it says anything since the question seems to be why the cosine approximation, term for term, is worse than the sine approximation (near 0). This is not intuitive, since $\sin$ and $\cos$ are just shifted versions of each other—shouldn't their approximations be as good?

For comparison, consider one of the popular interpolation methods: cubic splines. We like cubic splines because 1) they hit every point 2) the slopes match at the points 3) the curvature matches at the points.

What would it mean to constitute a good approximation? We often like to approximate with low-order polynomials (e.g. "everything is approximately a line if you zoom in enough", Simpson's rule uses a parabola to estimate, etc.). Since the derivative of a line is a constant, this kind of approximation should be alright if the derivative isn't changing very much (the derivative of a line doesn't change at all). This is by analogy to the point (3) of cubic splines. Since the rate of change of the derivative of $\sin(x)$ is (ignoring the sign) just $\sin(x)$ and at 0 the sine is 0, this means our small approximation should be alright: it's not a line, but close to 0 it sure acts like one . If you compare this to cosine the situation is reversed and we are at the maximum rate of change of the derivative, so our small approximation should have more error. It's not a line, and it isn't acting like one, either.

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  • $\begingroup$ This definitely gets at what I'm thinking about, especially the last few sentences. When you say, "If you compare this to cosine the situation is reversed and we are at the maximum rate of change of the derivative, so our small approximation should have more error." -- this is exactly what I'm trying to explore in terms of bounding the error using Lagrange error. With Lagrange error, in this case, I believe you'd be looking at the max value of the derivative on the interval in question, but really, it should be the max second derivative from what you're saying. $\endgroup$ – rb612 May 7 '17 at 8:03
  • $\begingroup$ @rb612 It's not so much that you should look to the second derivative, you should look at the way your approximating function behaves compared to the function being approximated. We can approximate anything with a line, but our linear approximations should be best when the second derivative is zero, since the second derivative of a line is 0. If we are approximating with a quadratic the story is not the same. $\endgroup$ – law-of-fives May 9 '17 at 14:14

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