- #1
mundane
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I am SO annoyed with this problem. Ready to jump out a window.
Find the first three terms of the Fourier series that approximates f(θ) = tan(θ) from θ = -π/2 to π/2.
So, I know that for an equation on [\frac{-b}{2}, \frac{b}{2}], to define the Fourier series for that equation we use f(x)={a0+a1cosx+a2cos2x+ ... +b1sinx+b2sin2x+ ...}
I only need to find the first three terms, so its just a0+a1cosx+b1sinx.
a0 is defined as \frac{1}{\pi}\intf(x)dx definite integral from -π/2 to π/2.
a1 is defined as \frac{2}{\pi}\int\f(x)cos(2πx/π)dx definite integral from -π/2 to π/2.
b1 is defined as \frac{2}{\pi}\int\f(x)sin(2πx/π)dx definite integral from -π/2 to π/2.For a0, my antiderivative was -log(cos(x)). After substitution, a0=0.
For a1, my antiderivative was log(cos(x))-(1/2)cos(2x). After substitution, a1=0.
I have not done b1 yet. Am I being trolled?
Does this ave something to do with the fact that tan(x) is undefined at those two interval points? Am I going in the wrong direction?
Homework Statement
Find the first three terms of the Fourier series that approximates f(θ) = tan(θ) from θ = -π/2 to π/2.
The Attempt at a Solution
So, I know that for an equation on [\frac{-b}{2}, \frac{b}{2}], to define the Fourier series for that equation we use f(x)={a0+a1cosx+a2cos2x+ ... +b1sinx+b2sin2x+ ...}
I only need to find the first three terms, so its just a0+a1cosx+b1sinx.
a0 is defined as \frac{1}{\pi}\intf(x)dx definite integral from -π/2 to π/2.
a1 is defined as \frac{2}{\pi}\int\f(x)cos(2πx/π)dx definite integral from -π/2 to π/2.
b1 is defined as \frac{2}{\pi}\int\f(x)sin(2πx/π)dx definite integral from -π/2 to π/2.For a0, my antiderivative was -log(cos(x)). After substitution, a0=0.
For a1, my antiderivative was log(cos(x))-(1/2)cos(2x). After substitution, a1=0.
I have not done b1 yet. Am I being trolled?
Does this ave something to do with the fact that tan(x) is undefined at those two interval points? Am I going in the wrong direction?
