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MacLaurin and Taylor series

  1. Sep 12, 2007 #1
    Hi I have some questions. If you're doing a MacLaurin expansion on a function say sinx or whatever, if you take an infinite number of terms in your series will it be 100% accurate? So will the MacLaurin series then be perfectly equal to the thing you're expanding?

    Also I don't really understand why, in a Taylor series, if you expand a function at x=5 (say) that it's more accurate for x near 5. Why is it different to an expansion of x=0? I can vaguely see that your chosen point of expansion, the value is perfect and it gets worse as you go away from it but is that good enough?

    I don't really understand what the remainder term does, can someone just explain it? Thanks. I promise I have looked on the net but I couldnt' find anything.
  2. jcsd
  3. Sep 12, 2007 #2
    A MacLaurin series is just a special case of a Taylor series, centred about zero. To understand the motivation, notice that around any point of a (continuous) function, the value of the function is approximately the same. Thus, we can use that value as an approximation near that point. We can get a better approximation by using the tangent to that point -- it's more closely approximated over a larger interval. Repeating, we can add a quadratic term, a cubic term, etc. and get successively better approximations, but all they do is to extend the range over which the approximation is "good". I suggest you draw the graph of sin x, and it's first 5 or so truncated taylor expansions, to see this for yourself. Under some circumstances, the range over which the expansion is accurate extends to the entire real line, and in that case, the Taylor series is "perfectly" equal to the original function.

    Caveats: all of this needs to be made far more precise, and lots of quite difficult issues turn up. Properly founding all this will easily take you up to post-grad mathematics (depends on the course and where you're taking it).
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