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Inverse functions

  1. Dec 30, 2005 #1
    find the inverse of y= x + sinx
    i got as far as x=y+siny..but how do i get to y= ____?
    thanks
     
  2. jcsd
  3. Dec 31, 2005 #2

    arildno

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    How did you get as far as that??
    It doesn't seem right.

    In the neighbourhood of (x,y)=(0,0), you might use the technique of successive substitutions to gain a fairly accurate approximation:

    1. First, expand your right hand-side in a power series, and get:
    [tex]y=2x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}....[/tex]

    2. Next, assume that x can be written as some power series in y:
    [tex]x=\sum_{n=1}^{\infty}a_{n}y^{n}[/tex]

    3. Plug in that power series in your equation:
    [tex]y=2a_{1}y+2a_{2}y^{2}+2a_{3}y^{3}-\frac{a_{1}^{3}y^{3}}{3!}++++[/tex]

    4. Thus, by comparing coefficients, you get:
    [itex]a_{1}=\frac{1}{2}, a_{2}=0, a_{3}=\frac{1}{96}[/tex]

    5. Hence, to the three lowest orders, you've got:
    [tex]x=\frac{y}{2}+\frac{y^{3}}{96}[/tex]
     
    Last edited: Dec 31, 2005
  4. Dec 31, 2005 #3

    HallsofIvy

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    If y= x+ sin x then x is NOT y+ sin y- in particular, if [itex]x= \frac{\pi}{2}[/itex] then [itex]y= \frac{\pi}{2}+1[/itex] but [itex]\frac{\pi}{2}+1+ sin(\frac{\pi}{2}+1) \ne \frac{\pi}{2}[/itex]. Simply subtracting sin x from both sides gives x= y- sin x, not y+ sin y. Essentially, you need to solve the equation
    y= x+ sin x for x- and that cannot be done algebraically.
     
  5. Dec 31, 2005 #4
    You could simply use a 2 by 2 matrix that reflects the function in the line y = x, that will be the same as the inverse.
     
  6. Dec 31, 2005 #5
    I think he's using the method we all learned in high school: to get an inverse function, replace x with y and vice-versa, then solve for y (where y, in this case, is really [itex]y^{-1}(x)[/itex] ). For instance:
    [tex]y = e^{x}[/tex]
    Repleace x with y
    [tex]x = e^y[/tex]
    Solve for y
    [tex]\ln{x} = \ln{e^y} = y[/tex]
    Therefore, the inverse of [itex]y=e^x[/itex] is [itex]y=\ln{x}[/itex]
    I doubt he's up to the point of using Power Series expansions (most people learn that after calculus, and this is the precalculus forum) or matrices.
    That said, this one is tricky and I don't recall how you'd do it aside from the more complicated methods mentioned here.
     
  7. Dec 31, 2005 #6
    since it seems like that finding the inverse is beyond high school calculus..maybe my approach to the probelm was incorrect

    the probelms states f(x) = x + sin(x), g(x) is the inverse of f(x) so f(g(x)) = x

    a) write an expression for f(g(x)) in terms of g(x) and
    b) use f(g(x))= x to find g'(x) in terms of g(x)

    i do not understand what it means by writing an expression interms of g(x) when i have no idea how to find the inverse..
     
  8. Dec 31, 2005 #7

    LeonhardEuler

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    I think they mean to write an expression that has "g(x)" in it. In other words just substitute g(x) for x in f(x)=x+sin(x).

    For the second part I would just differentiate and use the chain rule.
     
  9. Dec 31, 2005 #8
    ok.. so f(g(x)) = g(x) + sin(g(x)) = x

    to find g'(x) in terms of g(x) i did

    g'(x) + (g'(x))cos(g(x)) = 1..wait that doesnt seem right
     
  10. Dec 31, 2005 #9

    LeonhardEuler

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    I think that's what they wanted. Now just solve for g'(x).
     
  11. Dec 31, 2005 #10
    ok... so g'(x) = 1/ 1 + cos(g(x))
    now they want g'(pi)...and im stuck again, i cant find any trig identities that can help me in g'(x)
     
  12. Dec 31, 2005 #11
    Remember that if f(x) and g(x) are inverse functions then
    [tex] g'(x) = \frac 1 {f'(g(x))} [/tex]
    and if
    f(x) = y
    then
    g(y) = x
    so to find g'(x) at a point say t, g'(t), just solve f(x) = t for x then that value of x you find is g(t) then to find the derivative at that point just take 1 over the derivative of f evaluated at the x you found.
     
  13. Dec 31, 2005 #12
    well, in this case f(x) = pi so x= pi
    1/ f'(pi) = 1/ 1+ cos( pi) = 1/ 1-1 = 1/0 :confused:
     
  14. Jan 1, 2006 #13

    LeonhardEuler

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    No, f(x)=[itex]\pi[/itex], so
    [tex]\pi=x + \sin{x}[/tex]
    Can you think of the value of x that makes this true? This value is [itex]g(\pi)[/itex] because, by definition of the inverse function, if
    [tex]f(a)=b[/tex]
    then
    [tex]f^{-1}(b)=a[/tex]
    So then what you want to do is find f'(x) and then sub in the value of [itex]g(\pi)[/itex] that you get to find [itex]f'(g(\pi))[/itex]. You know where to go from there.
     
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