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A Fourier analysis of the instantaneous value of a waveform can be represented by: $$y=(t+\dfrac{π}{4})+\sin(t)+\dfrac{1}{8}\sin(3t)$$ Use Newton's method to determine the value of $t$ near to $0.04$, correct to 4 decimal places, when the amplitude, $y$, is $0.880$

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To make things easier I generated a graph with a few points including the root and I derived the above function:

$$\dfrac{d}{dt}((t+\dfrac{π}{4})+\sin(t)+\dfrac{1}{8}\sin(3t)) = \cos(t)+\dfrac{3}{8}\cos(3t)+1$$

I've been staring at the question and graph, reading them both over and over but still don't understand how the Newton Raphson method can be used to determine the value...I thought it was only used for calculating roots? for difficult equations? you can see on graph that $x$ is $0.04$ at $0.880$ already, this confuses me...how do you use NR method to determine the value near $0.04$? I know how to use it to find roots..that's easy, just this question confuses me. If anyone can give me a hint on how to proceed I'd be grateful. Apologies for the lack of leg work.

Modrisco
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The general formula is: $$x_{n+1}=x_{n}-\frac{f(x_n)}{f'(x_n)}$$ You got it right, but you have to use an initial guess (say $x_0=1$), and iterate.

For a basic understanding of Newton-Raphson method, it assumes that the system is basically linear, and predicts the zero for a linear approximation: $$y=\frac{\Delta y}{\Delta x}x+b=0 \rightarrow x=-b*\frac{1}{\frac{\Delta y}{\Delta x}}$$ Which why you get the derivative in the denominator. In other words: $$\frac{(y-0)}{1}*\frac{\Delta x}{\Delta y}=\Delta x_{new}$$

FundThmCalculus
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  • I understand the method and how it works...I just don't understand how it works in this problem, I'm not finding a root this time. – Modrisco Jun 24 '15 at 21:42
  • Okay, have you tried changing your starting guess? Because Newton's method can fail, it can end up oscillating between two non-solutions, and never getting any closer. – FundThmCalculus Jun 24 '15 at 21:51
  • I'll try another guess and I'll get back to you, thank you. – Modrisco Jun 24 '15 at 21:55
  • I'm sorry but I think I'm lost on this one...I'm fine with figuring out the starting guess when aprroxitmating the root of the function using the NR method but this time the question is different? why does it say "When y is 0.880" when it should be zero? if you could explain what the question is really asking I'd be grateful. – Modrisco Jun 24 '15 at 22:12
  • Then recast your problem as $f(x)=a \rightarrow g(x)=f(x)-a=0$. Then it will find the solution. Sorry. Missed that bit of the problem. – FundThmCalculus Jun 25 '15 at 00:40