Calculus Orthonormal

Question

Use the inner product <f,g>=∫01 f(x)g(x)dx in the vector space C0[0,1] to find the orthogonal projection of f(x)=5x^2+1 onto the subspace V spanned by g(x)=�−1/2 and ℎ(x)=1.

Answer 1

To find the orthogonal projection of f(x)=5x2+1f(x) = 5x^2 + 1 onto the subspace VV spanned by the functions g(x)=−12g(x) = -\frac{1}{2} and h(x)=1h(x) = 1, we will use the concept of orthogonal projection in an inner product space.

Step 1: Set up the orthogonal projection formula

Let f(x)=5x2+1f(x) = 5x^2 + 1 be the function we're projecting, and V=span{g(x),h(x)}V = \text{span}\{ g(x), h(x) \}. The orthogonal projection of f(x)f(x) onto VV is given by the formula:

projVf=⟨f,g⟩⟨g,g⟩g+⟨f,h⟩⟨h,h⟩h\text{proj}_V f = \frac{\langle f, g \rangle}{\langle g, g \rangle} g + \frac{\langle f, h \rangle}{\langle h, h \rangle} hwhere ⟨f,g⟩=∫01f(x)g(x) dx\langle f, g \rangle = \int_0^1 f(x) g(x) \, dx is the inner product between f(x)f(x) and g(x)g(x), and similarly for the other inner products.

Step 2: Compute the inner products

Compute ⟨f,g⟩\langle f, g \rangle

⟨f,g⟩=∫01(5x2+1)(−12)dx=−12∫01(5x2+1) dx\langle f, g \rangle = \int_0^1 (5x^2 + 1) \left( -\frac{1}{2} \right) dx = -\frac{1}{2} \int_0^1 (5x^2 + 1) \, dxNow compute the integral:

∫015x2 dx=[5x33]01=53\int_0^1 5x^2 \, dx = \left[ \frac{5x^3}{3} \right]_0^1 = \frac{5}{3} ∫011 dx=1\int_0^1 1 \, dx = 1So,

⟨f,g⟩=−12(53+1)=−12(83)=−43\langle f, g \rangle = -\frac{1}{2} \left( \frac{5}{3} + 1 \right) = -\frac{1}{2} \left( \frac{8}{3} \right) = -\frac{4}{3}Compute ⟨g,g⟩\langle g, g \rangle

⟨g,g⟩=∫01(−12)2dx=14∫011 dx=14\langle g, g \rangle = \int_0^1 \left( -\frac{1}{2} \right)^2 dx = \frac{1}{4} \int_0^1 1 \, dx = \frac{1}{4}Compute ⟨f,h⟩\langle f, h \rangle

⟨f,h⟩=∫01(5x2+1)(1) dx=∫01(5x2+1) dx\langle f, h \rangle = \int_0^1 (5x^2 + 1)(1) \, dx = \int_0^1 (5x^2 + 1) \, dxWe already computed the individual integrals:

∫015x2 dx=53,∫011 dx=1\int_0^1 5x^2 \, dx = \frac{5}{3}, \quad \int_0^1 1 \, dx = 1So,

⟨f,h⟩=53+1=83\langle f, h \rangle = \frac{5}{3} + 1 = \frac{8}{3}Compute ⟨h,h⟩\langle h, h \rangle

⟨h,h⟩=∫0112 dx=∫011 dx=1\langle h, h \rangle = \int_0^1 1^2 \, dx = \int_0^1 1 \, dx = 1Step 3: Compute the orthogonal projection

Now substitute these inner products into the formula for the projection:

projVf=⟨f,g⟩⟨g,g⟩g+⟨f,h⟩⟨h,h⟩h\text{proj}_V f = \frac{\langle f, g \rangle}{\langle g, g \rangle} g + \frac{\langle f, h \rangle}{\langle h, h \rangle} h projVf=−4314(−12)+831⋅1\text{proj}_V f = \frac{-\frac{4}{3}}{\frac{1}{4}} \left( -\frac{1}{2} \right) + \frac{\frac{8}{3}}{1} \cdot 1Simplifying each term:

projVf=−43⋅4⋅(−12)+83\text{proj}_V f = -\frac{4}{3} \cdot 4 \cdot \left( -\frac{1}{2} \right) + \frac{8}{3} projVf=163⋅12+83\text{proj}_V f = \frac{16}{3} \cdot \frac{1}{2} + \frac{8}{3} projVf=83+83=163\text{proj}_V f = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}Thus, the orthogonal projection of f(x)=5x2+1f(x) = 5x^2 + 1 onto the subspace VV spanned by g(x)=−12g(x) = -\frac{1}{2} and h(x)=1h(x) = 1 is:

projVf=163.\text{proj}_V f = \frac{16}{3}.