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Trigonometric and hyper. functions approx in large small argumnt

  1. Mar 2, 2014 #1
    hello guys ,

    i'm looking for approximation of trigonometric and hyperbolic functions for small and large argument, is it correct to say sin(x)=x and tg(x)=x and tgh(x)=x and cos(x) = 1 and cosh(x)=1 and coth(x)=1/x for small x what about large x ? what can we say about exponential function in large and small argument ?
     
  2. jcsd
  3. Mar 3, 2014 #2

    micromass

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    The key notion here is Taylor series. I don't think you've heard of that since you posted in precalc. But I think the answer to your question is going to be difficult without this notion.

    The idea is that ##\sin(x)\sim x## for small values of ##x##. Now, if ##x## becomes larger, then you need to add more and more error terms in order to still have a good approximation

    For example, for somewhat larger ##x##, we have

    [tex]\sin(x)\sim x - \frac{x^3}{3!}[/tex]

    as a better approximation

    See here for a list of approximations: http://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions
    The more terms you use, the better the approximation. If you use infinite terms, then the approximation is usually exact.
     
  4. Mar 3, 2014 #3

    HallsofIvy

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    Slightly simpler than Taylor's series here, but useful for these questions, is the "tangent line approximation". The derivative of sin(x) is cos(x) which is 1 at x= 0. Since sin(0)= 0 and its derivative at x= 0 is 1, the tangent line to the graph of y= sin(x) is y= 0+ (1)x= x. For small x, sin(x) is approximately equal to x.

    Similarly tan(0)= 0, the derivative of tan(x) is sec^2(x), and sec(0)= 1 so the tangent approximation to y= tan(x) is also y= 0+ (1)x= x. (which, of course, means that for small x, tan(x) is approximately equal to sin(x).)

    cos(0)= 1, the derivative of cos(x) is -sin(x), and -sin(0)= 0 so the tangent approximation to y= cos(x) is y= 1+ (0)x= 1.

    Similarly, cosh(0)= 1, the derivative of cosh(x) is sinh(x), and sinh(0)= 0 so the tangent approximation to y= cosh(x) is y= 1+ (0)x= 1.

    "1/x" is, of course, not a power series so not a Taylor's series. But it is true that tanh(0)= 1, the derivative of tanh(x) is sech^2(x), and sech(0)= 0 so that the tangent approximation to y= tanh(x) is y= x. Since coth(x)= 1/tanh(x), we can approximate y= coth(x) near x= 0 with 1/x.

    Finally, for e^x, the derivative is e^x so both function and derivative are e^0= 1 at x= 0. For small x, y= e^x can be approximated by y= 1+ x.

    e^x "goes to infinity" as x goes to infinity so we cannot say anything about an approximation to y= e^x for "large x".
     
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