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Taylor polynomial of degree 1 - solve for theta

  1. Feb 24, 2012 #1
    1. The problem statement, all variables and given/known data

    I was given the following problem, but I am having a hard time interpreting what some parts mean.

    We're given the equation

    sinθ+b(1+cos^2(θ)+cos(θ))=0

    Assume that this equation defines θ as a function, θ(b), of b near (0,0). Computer the Taylor polynomial of degree one for θ(b) near b=0


    2. Relevant equations

    sinθ+b(1+cos^2(θ)+cos(θ))=0

    3. The attempt at a solution

    Not sure exactly what they mean by θ(b).
     
  2. jcsd
  3. Feb 24, 2012 #2

    LCKurtz

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    Replace b by x for a moment so you are thinking of ##\theta## as a function of x, say f(x). Can you write the Taylor series for a function f(x) near x = 0? Do that first. Then replace ##f## by ##\theta## and ##x## by ##b##. There will be a couple of things you will need to calculate from your equation.
     
  4. Feb 24, 2012 #3
    So I took the derivative with respect to b and got

    dθ/db= (1+cos^2(θ)+cos(θ))/(2bcos(θ)sin(θ)+sin(θ)-cos(θ))

    knowing the general form of the taylor series, I used the value of dθ/db at b=0 with θ=0 to get a value of -3 and plugged it into the tayor series general form

    θ(b)=θ(0)+(-3)(b) to get θ(b)=-3b, the same value I got when I took the taylor polynomial of this function and solved for θ in another step.

    Is this right?
     
  5. Feb 24, 2012 #4

    SammyS

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    That looks right to me, but there is a small error in your θ'(b) .

    θ'(b) = (1+cos^2(θ)+cos(θ))/(2bcos(θ)sin(θ)+bsin(θ)-cos(θ))
     
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