Fourier cosine series correspondence for f(x)= x, o < x < pie given by x ~ pie / 2 - 4/n, E infinity on top and n=1 on bottom. cos (an-1)/x / (2n-1)squared, (0 < x < Pie).
Explain why this correspondence is actually an equality for 0 is less than or equal to x and x is less than or equal to Pie. Then explain how we can write
{x} = Pie /2 - 4 / n E infinity on top and n=1 on bottom
cos (2n - 1)x/ (2n-1) squared, (-n is less than or equal to x and x is less than or equal to Pie)
The Fourier cosine series is a way to represent a periodic function as an infinite sum of cosine functions with different frequencies and amplitudes. In this case, we are considering the function f(x) = x defined on the interval (0, π).
To prove that the Fourier cosine series correspondence is actually an equality for 0 ≤ x ≤ π, we need to show that the given formula for the Fourier coefficients converges to the function f(x) within this interval.
The formula for the nth Fourier cosine coefficient is:
c_n = (2 / π) ∫[0, π] f(x) cos((nπx) / π) dx
Substituting f(x) = x and simplifying, we have:
c_n = (2 / π) ∫[0, π] x cos(n x) dx
To evaluate this integral, we can use integration by parts. Let's set u = x and dv = cos(n x) dx. Differentiating u, we get du = dx, and integrating dv, we get v = (1 / n) sin(n x).
Applying the integration by parts formula:
∫ u dv = u v - ∫ v du
We get:
c_n = (2 / π) [(x / n) sin(n x)] [0, π] - (2 / π) ∫[0, π] (1 / n) sin(n x) dx
Evaluating the first term, we obtain:
c_n = (2 / π) [(π / n) sin(n π) - (0 / n) sin(0)]
Since sin(0) = 0, the second term disappears, and we are left with:
c_n = (2 / π) (π / n) sin(n π)
Now, based on the properties of sine function, we know that sin(nπ) equals zero when n is an integer. Therefore, c_n = 0 for all integer values of n except n = 0.
For n = 0, the coefficient c_0 is given by:
c_0 = (2 / π) ∫[0, π] f(x) dx = (2 / π) ∫[0, π] x dx = (2 / π) [(x^2 / 2)] [0, π] = π / 2
Thus, the Fourier cosine series for f(x) = x on (0, π) becomes:
f(x) ≈ (π / 2) + ∑[(2 / π) (π / n) sin(nπ)] cos(nx)
Since sin(nπ) is zero for all integers except when n = 0, the above series simplifies to:
f(x) ≈ (π / 2) + ∑[2 cos(nx)] [n = 1 to infinity]
Now, to express {x} (the fractional part of x) as a series, we need to determine its behavior for different intervals.
For -1 ≤ x ≤ 0, the function {x} varies from 0 to 1 as x moves from -1 to 0. Hence, {x} = x + 1.
For 0 ≤ x ≤ 1, the function {x} varies from 0 to x. Hence, {x} = x.
For π - 1 ≤ x ≤ π, the function {x} varies from 0 to 1 as x moves from π - 1 to π. Hence, {x} = x - (π - 1).
Combining these cases, we find that {x} = x for x ∈ [0, π].
Thus, the Fourier cosine series representation of {x} for -π ≤ x ≤ π becomes:
{x} ≈ (π / 2) + ∑[(2 / π) (π / n) sin(nπ)] cos(nx) [n = 1 to infinity]
Since sin(nπ) is zero for all integers except when n = 0, the above series simplifies to:
{x} ≈ (π / 2) + ∑[2 cos(nx)] [n = 1 to infinity]
Conclusion:
The given Fourier cosine series correspondence is actually an equality for 0 ≤ x ≤ π because we have derived the series representation for the function f(x) = x and the function {x} using the appropriate Fourier coefficients and their corresponding trigonometric functions.