Find a and b that minimalize the integral

[rayman123]

Homework Statement

find a and b that the integral is minimaized

Homework Equations

$\int_{0}^{2\pi}|e^x-ae^{ix}-b^{-ix}|^2dx$
$<f,g>=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)\overline{g(x)}dx$
where a $a=<e^x,e^{ix}>= \frac{1}{2\pi}\int_{0}^{2\pi}e^xe^{-ix}dx=\frac{1}{2\pi}\frac{e^{2\pi}-1}{1-i}$

$b=<e^x,e^{-ix}>= \frac{1}{2\pi}\int_{0}^{2\pi}e^xe^{ix}dx=\frac{1}{4\pi}(e^{2\pi}-1)(1-i)$


which is correct but then how to do the same for the following integral?

Homework Equations

$\int_{-1}^{1}|e^x-ax-b|^2dx$
$a=<e^x,>= \frac{1}{2\pi}\int_{-1}^{1}xe^xdx$?
$b=<e^x,-1>= \frac{1}{2\pi}\int_{-1}^{1}-e^xdx$?

the answer to the second integral is
$a= 3e^{-1}, b= (e-e^{-1})/2$
It is a bit strange because i think $$\frac{1}{2\pi}$$ should not be in front of the second integral because in the answer there is not pi term...

 

[mighty2000]

Firstly, let's clarify the question. The first integral is over the interval [0,2π], while the second is over [-1,1]. This means that the second integral is not a Fourier integral, but rather a regular integral. So, the equations for a and b in the second case should be:

a = <e^x,e^x> = \int_{-1}^{1} e^x e^x dx = \frac{1}{2} (e^2 - e^{-2})
b = <e^x,1> = \int_{-1}^{1} e^x dx = e - e^{-1}

To minimize the integral, we use the same method as before by setting the derivative with respect to a and b equal to 0 and solving for a and b. The final result is:

a = e^{-1}, b = \frac{e - e^{-1}}{2}

As for the strange factor of 1/2π in the second integral, it is likely just a typo or a mistake in calculations. The correct answer should not have a factor of 1/2π.