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I Mathematical series in physics - Why and when do we need them?

  1. Oct 9, 2018 #1
    Hi,

    Before I post my question, let me admit that my foundation on mathematics is poor. I am trying to work on it, specifically on the application part.

    When I came through the following image, I was stuck to understand why I will need one like Taylor's series in a simple case like "F+ΔF = F (x* + Δx)"?

    Why don't we live with F (x* + Δx) instead of expanding it as a series? What exactly is the benefit?

    upload_2018-10-9_12-53-28.png


    Thanks.
     
  2. jcsd
  3. Oct 9, 2018 #2

    PeroK

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    If you don't know the value of ##F(x + \Delta x)## then the Taylor series gives you an approximation in terms of ##F(x)##, ##F'(x)## and ##\Delta x ##.

    It is, in fact, one of the most widely used mathematical tricks.
     
  4. Oct 9, 2018 #3
    Well, relationship between F and x* are known. Thus, if ΔF is known Δx also can be found. Why to use Taylor's series then?
     
  5. Oct 9, 2018 #4

    PeroK

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    You may not even know what function ##F## is. For example, if we have a local minimum of ##F## at a point ##x = 0##, say. We know that ##F'(0) =0## and ##F''(0) >0##, at a local minimum. So, for small ##x##, we have the approximation:
    ##F(x)= F(0) + kx^2##, where ##k## is a positive constant.

    That is then a general result for any function ##F## around a local minimum.
     
  6. Oct 9, 2018 #5
    The expression ## F+\Delta F = F(x+\Delta x)## is true only for very small values of ## \Delta F## (as they say: an incremental force). A Taylor expansion is true for its range of convergence, which can be over all numbers or over a smaller range like [-1,1]. So we use a Taylor expansion because it represents the function exactly in some larger neighborhood of x. As they show, for small displacement values of a harmonically oscillating system, we can forget about the nonlinear effects and only keep the first term. For larger displacement or force values, you have to take an extra term in the Taylor series to accurately describe the physical motion. This flexibility makes Taylor series very important for many mathematical derivations. It is also important because we can estimate beforehand how large the error is going to be for our approximation by looking at the largest of the truncated terms.
     
  7. Oct 9, 2018 #6

    sophiecentaur

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    Taylor is only one of many series for approximating functions. On a practical note, your function "F" can be very complicated and very difficult / impossible to work out manually. Transendental functions like Sin and Cos and all the others you come across are only calculable using a series approximation. In the old days we had books of 'tables' of all sorts of functions for a range of arguments. To calculate a value for a general input argument, it was often quicker and easier to use the Taylor series to give a result, based on a 'known' start value and use the slope, curvature etc. (higher and higher order terms of the Taylor series) for other values Nowadays people use computers for most calculations and the computer algorithms for functions do exactly the same thing with the best chosen series for whatever function you call for. Taylor works very well for many functions BUT you can only use it when the function it is used to describe behaves 'nicely' i.e. it is continuous (no gaps) and also, the derivatives are also continuous over the range you want to use. It doesn't work for functions with gaps and rapid changes of direction etc..
     
  8. Oct 9, 2018 #7
    Thanks, mfig.

    1. For simplicity, can I take this way: In the equation referred in my original post, if I substitute ΔF and assume relationship between F and x to be linear, the error is going to be huge (courtesy - My assumption). However, Taylor's expansion gives me a flexibility to find out an accurate answer - The amount of accuracy is chosen by me by truncating at my convenient place.

    2. Can I not find out the Δx by a simple differentiation - Like finding out what dF/dx is?

    Edit:
    "2. Can I not find out the Δx by a simple differentiation - Like finding out what dF/dx is?"

    Cannot. Because dF/dx finds only the slope but the locus.

    Thus only an expansion can help here. Doubt cleared gentlemen. Thanks.
     
    Last edited: Oct 9, 2018
  9. Oct 9, 2018 #8

    Dale

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    How?
     
  10. Oct 9, 2018 #9
    Dale I was missing the point that the primary assumption was x varies "non-linear" with F
     
  11. Oct 9, 2018 #10
    A few additional comments on this interesting discussion.

    1. These linear approximations are used all over the place in physics, especially introductory physics classes, to reduce difficult equations to usable ones. And they are used in many practical applications as well. Here are a few off the top of my head.

    a. The potential energy of an object of mass m at height h relative to the surface of earth is ##mgh##. No, actually it's not. The exact expression is the difference between ##GMm/R## and ##GMm/(R+h)## where R is the radius of the earth, G is the gravitational constant, and M is the mass of the earth. When ##(h/R)## is small, the linear terms of the Taylor series give you the approximation ##mgh##.
    b. In a pendulum, the restoring force is proportional to the angular displacement ##\theta##, which gives you the equations for simple harmonic motion. No, actually it's not. The restoring force is proportional to ##\sin\theta##. We use the first term of the Taylor series for sine, ##\sin\theta \approx \theta##, which is pretty good for small ##\theta##.
    c. The horizon when you are at height h above the surface is at ##\sqrt{2Rh}## where R is the radius of the earth. (This is one of the practical ones I mentioned, it has a lot of usage in calculations dealing with the earth. I used it many times on the job). No, actually, it's not. That's an approximation of the exact formula when ##(h/R)## is small. The exact formula, which comes from the Pythagorean Theorem, is ##\sqrt{(R+h)^2 - R^2} = \sqrt{2Rh + h^2}##.

    2. When you want more accuracy, practical applications often still use series, but they keep more terms in the series. An example again from my job history is the gravitational field of the earth. For introductory physics, the earth is a uniform sphere of mass, ##F = GMm/r^2##. For more advanced physics courses, they might introduce the variation of the density with depth and the fact that the earth is a flattened spheroid, not a sphere. But for high-accuracy prediction of satellite motions, you need to represent the irregularities that vary over the surface. The biggest are that mountain ranges represent a high concentration of mass, and deep ocean trenches represent a lower than average concentration of mass. In those applications, the earth is represented by a Taylor series going up to several higher powers of ##1/r##, and the coefficients of that series are derived from experimental measurements. We don't have any "original function" that the Taylor series is approximating, all we have is the series.

    3. You'll also notice that the vague term "small" is used when introducing these approximations, and nobody ever defines it. How small is small enough to use the approximation? That depends on the original function, as the theory of Taylor series tells you the error after n terms depends on the n-th derivative of ##f(x)##. It also depends on how accurate an answer you need. But for many applications requiring only 2-3 significant figures, 0.1 is usually "small enough" and 0.01 is almost always "small enough".
     
  12. Oct 9, 2018 #11

    Dale

    Staff: Mentor

    Even if F is linear in x then to find ##\Delta x## requires the computation above.
     
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