# Linear approximation higher order terms

1. Feb 14, 2013

### bacon

My questions are from lecture 9, MIT OCW SV Calculus, Jerison, 2009;
At 27:50 he is deriving the linear approximation for the function
e^(-3x)(1+x)^(-1/2)≈(1-3x)(1-1/2x)≈1-3x-1/2x+3/2x^2≈1-7/2x, for x near 0.
In the last step he drops the x squared term since it is negligible(no questions so far). He then says that dropping the quadratic term is ok since in an earlier analysis, the quadratic and higher terms were ignored and that methodology is just being consistently applied. However, that was not done, at least explicitly. He points to an earlier approximation done, e^x≈1+x.
This was done using the expression f(x)≈f(0) + f'(0)x where f(x)=e^x. However, no non linear terms were generated and and therefore none were dropped. Although I believe him, I don't see the how the dropped terms are generated in the earlier analysis. Does anyone know how one would generate those terms? I'm not trying to needlessly complicate things, just fully understand what he was saying.
My other question concerns the example at 29:50 in the same lecture. He shows how the time dilation expression for a GPS satellite is given by the lorentz transformation approximated by a fairly simple linear expression. The approximation is what was used by the engineers who designed the satellite. My question here is really about the motivation of the engineers. Is it simply easier to write the code for a linear calculation, instead of the original expression? Does it use less memory? Any Ideas?

2. Feb 14, 2013

### jbunniii

Regarding the higher-order terms for approximating e^x, use however many terms of the Taylor series you need:
$$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$
Thus for example a quadratic approximation you would use
$$e^x \approx 1 + x + \frac{x^2}{2}$$
and for a cubic you would use
$$e^x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$
etc.

I'm not sure specifically about the GPS relativistic correction, but in general algorithmic complexity is not much of an issue, as the calculations are not performed on board the satellite. Ground stations upload the (time-varying) orbital information needed to compute the correction, and the satellite simply echoes this information to receivers. The receivers then calculate the correction based on the orbital parameters and the current time and user position.

The bottleneck is the craptacular 50 bits per second data communication speed from the satellite to the receiver, so there's only so much orbital information that can be sent, and this is further limited by the fact that this information must be repeated frequently (every 30 seconds) because a receiver may "tune in" at any time. This limitation means that only first and second order coefficients are used for most parameters, each represented by as few data bits as possible. This in turn means that the ground stations have to broadcast updated orbital parameters fairly often, every few hours.

Last edited: Feb 14, 2013
3. Feb 15, 2013

### bacon

Thank you, jbunniii, for your response. The expansion of e^x certainly shows the terms I was looking for, the terms do not come from the expression, f(x)≈f(0) + f'(0)x, I thought maybe I missed something in the lecture.
Thanks also for giving me more depth on GPS operations, there is quite a lot going there behind the scene.