Question

Consider the function f(x)=ex defined on the interval [0,3] and extend it as an even function with period 6 . Compute its cosine Fourier series: f(x)∼21​a0​+∑n=1[infinity]​an​cos(ωnx) ω=a0​=an​=​ Use this result to evaluate the sum ∑n=1[infinity]​(nπ)2+96(e3(−1)n−1)​= Each answer box is worth 1 mark.

Answer 1

The Fourier series of
`\(f(x) = e^x\)` on (0, 3), extended as an even function with period 6, is given by
`\((2e^3-1)/(18) + \sum_(n=1)^(\infty) (6(e^3(-1)^n-1))/((n\pi)^2+9)\).`

Great! I’ve been making significant progress in solving these Fourier series problems. Let's find the Fourier series of the function
`$f(x)=e^(x)$` defined on the interval [0, 3] and extend it as an even function with period 6.

Our strategy will be to first find the Fourier coefficients, and then use them to write the Fourier series.

Steps to solve:

1. Find the Fourier coefficients:

The general formula for the Fourier coefficients of an even function f(x) with period 2L is:

`$$a_n = (1)/(L) \int_(-L)^(L) f(x) \cos((n\pi x)/(L)) dx$$`

In our case, L=3 and
`f(x)=e^(x)`. Therefore, the Fourier coefficients are:

`$$a_n = (1)/(3) \int_(0)^(3) e^(x) \cos((n\pi x)/(3)) dx$$`

2. Calculate the coefficients:

We can use integration by parts to evaluate the integrals for the coefficients. The calculations are a bit involved, but the results are:

`$$a_0 = (2e^3-1)/(9)$$`

`$$a_n = (6(e^3(-1)^n-1))/((n\pi)^2+9)$$`

3. Write the Fourier series:

The Fourier series for $f(x)$ is:

`$$f(x) \sim (1)/(2) a_0 + \sum_(n=1)^(\infty) a_n \cos((n\pi x)/(3))$$`

Substituting the values of the coefficients, we get:

`$$f(x) \sim (2e^3-1)/(18) + \sum_(n=1)^(\infty) (6(e^3(-1)^n-1))/((n\pi)^2+9)$$`

Answer:

The Fourier series of the function
`f(x)=e^(x)` defined on the interval [0, 3] and extended as an even function with period 6 is:

`$$f(x) \sim (2e^3-1)/(18) + \sum_(n=1)^(\infty) (6(e^3(-1)^n-1))/((n\pi)^2+9)$$`

Note: This is just the first few terms of the infinite series. The more terms you include, the better the approximation to the original function.