Question

Find the half range Fourier cosine series of e

x

in 0 < x < 1​

Answer 1

The Fourier cosine series goes as,

$2\pi [( 1 + e/\pi ^2 + 1) Sin \pi n + (2( 1 - e)/ 4\pi ^2 + 1)Sin2\pi x + (3(1+e)/9\pi ^2+1)Sin3\pi x.......]$

Given:

Using of Fourier cosine series of $e^{x}$

0 < x < 1

To find:

Finding the half-range Fourier cosine series of $e^{x{$

Solution:

We can take $e^{x{$ as a function.

$f(x) = e^{x{$

Since we need to find a solution based on the cosine series we shall use the formula,

$e^{x{$ = ∑ $a_x Sin n \pi x$

$a = \int\limits^2_0 {e^xSinx(\pi x)} \,$

$\int\limits {e^xSin\pi x} \, dx = I$

$= e^xCos\pi x/\pi + 1/\pi \int\limits {e^x cos\pi n} \, dx$

$= -e^x cos\pi x/ \pi + 1/\pi (e^xsin\pi x/\pi - \int\limit {e^xCos\pi x/\pi } \, dx)$

$= I (1 + 1/\pi ^2)$

$= e^xSin\pi x/\pi ^2 - e^xCos\pi x/\pi$

As we can see the limit given is 0 to 2π

$a_1 = 2\pi ( 1 + e ) / \pi ^2 + 1$

$= 2\pi [( 1 + e/\pi ^2 + 1) Sin \pi n + (2( 1 - e)/ 4\pi ^2 + 1)Sin2\pi x + (3(1+e)/9\pi ^2+1)Sin3\pi x.......]$

The series goes continuously and the number difference keeps changing.  

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Answer 2

The function e^x is an odd function, which means its Fourier series only contains sine terms.

However, since we are looking for a cosine series, we need to use the even extension of the function, which is e^x + e^(-x). The half range Fourier cosine series of this function can be found using the formula:

f(x) = a0/2 + Sum(n=1 to infinity)[ancos(npi*x/L)]

where L is the half range of the function, which is 1 in this case.

To find the coefficients an, we can use the formula:

an = (2/L) * Integral(0 to L) [f(x)cos(npi*x/L) dx]

Substituting the even extension of e^x into this formula, we get:

an = (2/L) * Integral(0 to L) [(e^x + e^(-x)) * cos(npix/L) dx]

Evaluating this integral using integration by parts, we get:

an = (2/L) * [(e^L - e^(-L))/((n*pi/L)^2 + 1)]

Thus, the half range Fourier cosine series of e^x in 0 < x < 1 is:

e^x + e^(-x) = a0/2 + Sum(n=1 to infinity)[(2/L) * (e^L - e^(-L))/((npi/L)^2 + 1) * cos(npi*x/L)]

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