fourier series/ trig calculus problem help

1) In deriving the coefficient of Fourier series (i.e., the related Euler equations), we take advantage of the orthogonality of the trigonometric system which indicated, among other things, that

∫(-pi,pi) cos nx cos mx dx=0 where n≠m

Note: The bounds for the integral -pi to pi.

(integral (-pi,pi) cos(nx) cos(mx) dx=0)

Using the trig identities and depicting relationships graphically, prove the result of the integral above.

2) The Fourier series evaluates respective terms from n=1 to ∞ (infinite), hence at some point it must consider the situation where n=m. how do we handle this case. Show all work and give specific value for the integral in that situation

**Solution** (is this the correct solution please check)

We should be assuming that m, n are positive integers...

Note that ∫(x = -pi to pi) cos(nx) cos(mx) dx

= ∫(x = -pi to pi) (1/2) [cos((n+m)x) + cos((n-m)x)] dx, via trig. identities.

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1) If n ≠ m, then the integral simplifies to

(1/2) [sin((n+m)x)/(n+m) + sin((n-m)x)/(n-m)] {for x = -pi to pi}

= 0, since sin(kpi) = 0 for any integer k.

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2) If n = m, then the integral simplifies to

∫(x = -pi to pi) (1/2) (cos(2nx) + 1) dx

= (1/2)(x + sin(2nx)/(2n)) {for x = -pi to pi}

= pi.

Re: fourier series/ trig calculus problem help

Hey action44.

Your solution looks very good.