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Orthogonal Functions Calculator

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Example
Created on 2024-06-20Asked by Levi Anderson (Solvelet student)
Show that the functions sin⁡(nπx) \sin(n \pi x) and cos⁡(mπx) \cos(m \pi x) are orthogonal on the interval [0,1][0, 1].

Solution

To show that sin⁡(nπx) \sin(n \pi x) and cos⁡(mπx) \cos(m \pi x) are orthogonal on the interval [0,1][0, 1]: 1. Compute the inner product: ∫01sin⁡(nπx)cos⁡(mπx) dx. \int_0^1 \sin(n \pi x) \cos(m \pi x) \, dx. 2. Use the product-to-sum identities: sin⁡(nπx)cos⁡(mπx)=12[sin⁡((n+m)πx)+sin⁡((n−m)πx)]. \sin(n \pi x) \cos(m \pi x) = \frac{1}{2} [\sin((n+m) \pi x) + \sin((n-m) \pi x)]. 3. Integrate: ∫0112[sin⁡((n+m)πx)+sin⁡((n−m)πx)] dx=12[∫01sin⁡((n+m)πx) dx+∫01sin⁡((n−m)πx) dx]. \int_0^1 \frac{1}{2} [\sin((n+m) \pi x) + \sin((n-m) \pi x)] \, dx = \frac{1}{2} \left[ \int_0^1 \sin((n+m) \pi x) \, dx + \int_0^1 \sin((n-m) \pi x) \, dx \right]. 4. Evaluate the integrals: ∫01sin⁡((n+m)πx) dx=[−cos⁡((n+m)πx)(n+m)π]01=0, \int_0^1 \sin((n+m) \pi x) \, dx = \left[ -\frac{\cos((n+m) \pi x)}{(n+m) \pi} \right]_0^1 = 0, ∫01sin⁡((n−m)πx) dx=[−cos⁡((n−m)πx)(n−m)π]01=0. \int_0^1 \sin((n-m) \pi x) \, dx = \left[ -\frac{\cos((n-m) \pi x)}{(n-m) \pi} \right]_0^1 = 0. Therefore, the inner product is zero, proving that sin⁡(nπx) \sin(n \pi x) and cos⁡(mπx) \cos(m \pi x) are orthogonal on the interval [0,1][0, 1]. Solved on Solvelet with Basic AI Model
Some of the related questions asked by James Lopez on Solvelet
1. Determine whether the functions sin⁡(x) \sin(x) and cos⁡(x) \cos(x) are orthogonal on the interval [0,π] [0, \pi] .2. Find an orthogonal basis for the subspace spanned by the functions exe^x and e−xe^{-x} on the interval [−∞,∞][-\infty, \infty].
DefinitionOrthogonal functions are functions that over the given interval, if we multiply them together and integrate on the interval, the result will be zero. They are the foundation of function spaces and appear in Fourier series, among related expansions. Case: The sine and cosine functions are orthogonal on [0,2π].
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