Math Real Analysis II

Transcription

1 Math 4 - Real Analysis II Solutions to Homework due May Recall that a function f is called even if f( x) = f(x) and called odd if f( x) = f(x) for all x. We saw that these classes of functions had a particularly nice orthogonality relationship. More precisely we considered the inner product space and its inner product C([ ]) = {f : [ ] R f is continuous} f g = f(t)g(t) dt. Furthermore we considered the susbpace of even functions and odd functions given by In class we proved that E = O. E = {f C([ ]) f( x) = f(x)} O = {f C([ ]) f( x) = f(x)}. (a) Show that if f E O then f is equal to the zero function (i.e. the zero vector in this vector space). (b) Use (a) to show that C([ ]) = E O. To do this show that you can write any f C([ ]) uniquely as f = f e + f o where f e E and f o O. Note that you cannot use the decomposition theorem from class since that requires your subspace to be finite dimensional which this is not. (c) Show that if a function f is odd then f(0) = 0. (d) Show that the derivative of an even function is odd and the derivative of an odd function is even. (e) Let p(x) = a n x n + a n x n + + a x + a 0 be a polynomial. Show p is odd if and only if a k = 0 for all even k. As a hint take a derivative and evaluate the derivative at 0; continue in this pattern. (f) Similar to above show that the polynomial p is even if and only if a k = 0 for all odd k. (g) Consider the polynomial p(x) = 5x 7 7x 6 + x 4 x + x. Decompose p as an even function and an odd function. Use the above to verify that your your components are indeed even and odd. (h) Decompose f(x) = e x as an even plus odd function. The even and odd function components have special names; what are they? Solution. (a) If f E O then f(x) = f( x) = f(x) for all x. Since f(x) = f(x) for all x then f(x) = 0 for all x. Thus f is the zero constant function. (b) In class we learned that we can write any function on [ ] as f = f e + f o where f e (x) = f(x) + f( x) and f o (x) = f(x) f( x). Note that in particular f e E and f o O. Assume that we have another expression of f as an even plus and an odd function. Thus f = g e + g o for g e E and g o O. Then consider f e g e E. Notice that f e g e = (f f o ) (f g o ) = g o f o O. Thus f e g e E O. By above we have that f e g e = 0 the constant 0 function. Thus f e = g e. A similar computation shows that f o = g 0. Thus there is a unique way to decompose f as an even plus an odd function. So C([ ]) = E O.

2 (c) If f is odd then f( x) = f(x). When x = 0 we get that f(0) = f(0). The only way for this to be true is if f(0) = 0. (d) If f is even then f( x) = f(x). Differentiating we get that f ( x) = f (x) and thus f ( x) = f (x). Thus the derivative of an even function is odd. Now assume that g is odd. Then g( x) = g(x). Differentiating we get that g ( x) = g (x) and thus g ( x) = g (x). Thus g is even. (e) If a k = 0 for all even k then computing p( x) we get a constant term that can be factored out. Thus p( x) = p(x). Conversely assume that p is odd. Then plugging in 0 we get that p(0) = a 0. By the above we must have that p(0) = 0 thus a 0 = 0. Taking two derivatives we get that p (x) = n(n )a n x n + (n )(n )a n x n + + a. Since this is two derivatives of an odd function it is itself odd. Thus evaluating at 0 we get that p (0) = a = 0. Thus a = 0. Continuing in this way we get that a k = 0 for all even k. (f) If a k = 0 for all odd k then replacing x by x will not change the polynomial. Thus p( x) = p(x). Conversely assume that p is even. Then its derivative p (x) = na n x n + + a is an odd function. Thus evaluating at 0 should yield 0. Thus we get that 0 = p (0) = a. Taking two more derivatives we get another odd function and after evaluating at 0 we must have that a = 0. Continuing in this way we get that a k = 0 for all odd k. (g) Using the equation for p e and p o we get that p e (x) = ( 5x7 7x 6 + x 4 x + x ) + (5x 7 7x 6 + x 4 x x p o (x) = ( 5x7 7x 6 + x 4 x + x ) (5x 7 7x 6 + x 4 x x ) (h) Using the equation for f e and f o mentioned above we get that f e (x) = ex + e x and f o (x) = ex e x. The function f e is also known as cosh x and f o is known as sinh x. = 7x 6 + x 4 x ; = 5x 7 x. Question. Consider the finite dimensional vector space P n the space of polynomials with real coefficients of degree at most n along with its inner product given by f g = f(t)g(t) dt. (a) A common basis for P is { x}. Use the Gram-Schmidt Process to turn this into an orthonormal basis for P. (b) A common basis for P is { x x }. Use the Gram-Schmidt Process to turn this into an orthonormal basis for P. (c) Let U be the subspace of P of constant functions. Find U. (d) Compute the orthogonal complement of the U you computed in (c). Verify that (U ) = U. Solution.

3 (a) First we normalize the vector. Since = dx = then normalizing the vector we get that e =. To compute the next vector we first compute the dot product of e with x. Doing so we get that Thus To normalize this we first compute Thus e x = x x = x d = x u = x e x e = x. x dx = x e = x. Thus an orthonormal basis for P is { } x. (b) We can use our previous work to find the Gram-Schmidt process for P. First we compute the various components of u : x = x = x dx = x 6 =. ( ) x x x = x dx x = 0. Thus Normalizing we compute that x / x / = 5 9 x + 9 x x 5 = 0. =. u = x x e e x e e = x. (x /) dx = x 4 x + 9 dx = = = = 8. Thus 8 x / =. So normalizing we get that e = 8 (x /). Thus the orthonormal basis obtained via the Gram-Schmidt process is { x 8 (x /) }.

4 Question. In this question we will apply orthogonal projections to the space C([ ]) of continuous functions on [ ]. We will project onto the finite-dimensional subspaces P and P investigated in Question. In what follows using a computer program that can perform numerical integration (like Grapher Sage or Wolfram Alpha) is recommended as well as using decimal expansions. (a) Consider the function f(x) = e x C([ ]). Compute the orthogonal projection of this function onto the subspace P. In other words compute P P (e x ). Your orthonormal basis for P from Question a should be handy in this computation. (b) Compute e x P P (e x ). (c) Compare this to how good of an approximation the first-degree Taylor polynomial T (x) = + x is for e x by comparing the norm computation from (b) to e x ( + x). (d) Repeat steps (a) - (c) but with the subspace P. That is compute P P (e x ) and compare the values of the norms e x P P (e x ) and e x ( + x + x /). Solution. (a) To compute the orthogonal projection of e x we first compute the various inner products of e x with the basis elements: e x e x = dx.66 e x Thus the projection of e x onto P is x = (b) Computing the distance via the norm we get xex dx.90 P P (e x ) = x. x = P P (e x ) e x ) = (P P (e x ) e x ) dx.057 =.95. (c) The distance between e x to + x is given by e x ( + x)) = (e x x) dx. =.48. Thus the projection of e x onto the space of linear functions is better than the first-degree Taylor polynomial for e x. Question 4. For a finite dimensional inner produt space V the Riesz Representation Theorem says that given any linear function ϕ : V F there exists a vector x V such that ϕ(v) = v x. For each of the following ϕ show that it is indeed a linear functional and find the corresponding x V such that ϕ(v) = v x. 4

5 (a) ϕ : P R given by ϕ(p) = p (0). That is ϕ(p) is obtained by taking a derivative and evaluating at 0. (b) ϕ : P R given by ϕ(p) = p(0). (c) ϕ : P R given by ϕ(p) = p (0). Solution 4. { } (a) Consider the orthonormal basis x. for P. Applying ϕ to each of these we get that ϕ(/ ) = ( ) 0 and ϕ x =. Therefore the vector that corresponds to this functional is given by { (b) For the orthonormal basis ( ) x = 0 and ϕ ϕ is ϕ(/ ) / + ϕ (c) For the orthonormal basis ( ) x = ϕ is ϕ(/ ) / + ϕ ( 8 (x /) ( ) x x = x. } x 8 (x /) for P we apply ϕ and get ϕ(/ ) = / ) =. Thus the vector that represents the linear functional 5 8 ( ) ( ) x x + ϕ 8 (x /) 8 (x /) = / 5 8 (x /). { ( and ϕ 8 (x /) ϕ(/ ) / + ϕ } x 8 (x /) for P we apply ϕ and get ϕ(/ ) = 0 ) = 0. Therefore the vector that represents the linear functional ( ) ( ) x x + ϕ 8 (x /) x = x. Question 5. Let V be an n-dimensional inner product space and fix a vector x V. We saw that we can define a linear functional ϕ x V by ϕ x (v) = v x. Let U = span{x}. Notice that U is a -dimensional subspace if x 0. Furthermore recall that the kernel of a linear transformation ϕ is the set of all vectors that are sent to 0: which forms a subspace. (a) Show that ker(ϕ x ) = U. ker(ϕ) = {v V ϕ(v) = 0} (b) Use the Riesz Representation Theorem the Orthogonal Decomposition Theorem and (a) to show that the kernel of a linear functional is always a subspace of dimension n or n. When is it n-dimensional? (c) Let W be any (n )-dimensional subspace of V. Use (a) to show that W is the kernel of some linear functional. Be sure to define what that linear functional is. Solution 5. 5

6 (a) We will show that ker(ϕ x ) = U. First we will show that ker(ϕ x ) U. To do so let v ker(ϕ x ) and let αx U. Then since v ker(ϕ x ) we have that 0 = ϕ x (v) = v x. Thus v αx = α v x = α0 = 0. Thus v U. Next we will show that U ker(ϕ x ). Let v U. Then v is orthogonal to any element of U. In particular since x U v x = 0. Thus Thus v ker(ϕ x ). ϕ x (v) = x v = 0. (b) Let ϕ V be any linear functional of V. By the Riesz Representation Theorem there exists a unique x V such that ϕ(v) = v x. Let U = span{x}. By (a) ker(φ) = U. By the decomposition Theorem we have that V = U U = U ker(ϕ). If x 0 then U is -dimensional and thus ker(ϕ) is n -dimensional. If x = 0 then U is 0-dimensional and ker(ϕ) is n-dimensional. Thus ker(ϕ) has dimension n if and only if x = 0 if and only if ϕ is the zero functional. (c) Let W be an (n ) dimensional vector space and let W be its orthogonal complement. By the Decomposition Theorem V = W W and thus W is -dimensional. Let x be any non-zero vector in W. Then W = span{x}. Consider the linear functional ϕ x given by ϕ x (v) = v x. By (a) we have that ker(ϕ x ) = (W ) = W. 6