Maths 381 The First Midterm Solutions

Transcription

1 Maths 38 The First Midterm Solutions As Given by Peter Denton October 8, 8 # From class, we have that If we make the substitution y = x 3/, we get # Once again, we use the fact that and substitute into the above to get e y dy = π. dy = 3 xdx y = x = and y = x = 3 xe x3 dx = π xe x 3 dx = π. 3 e y dy = π y = logx dy = x logx dx y = x = and y = x = log x e x log x dx = π x x logx dx = π dx log x = π

2 Maths 38 The First Midterm #3 We have, from class, that for strings fixed at x = and x = with small vibrations, the displacement is u(x, t) = sin (πnx) (a n cosπnt + b n sin πnt) where the coefficients are defined by the initial conditions u(x, ) = a n sin πnx u t (x, ) = b n sin πnx = The second of which implies that all the b n =. Our equation becomes where u(x, t) = a n = a n sin (πnx) cos(πnt) f(x) sin(nπx)dx given that the initial velocity is uniformly, and the initial displacement is u(x, ) = f(x). { x x u(x, ) = x x We calculate the a n. ( ) a n = x sin(nπx)dx a n = a n = / xsin(nπx)dx + x nπ cos(nπx) / + nπ + x nπ cos(nπx) a n = / nπ ( x)sin(nπx)dx / / / ( cos(nπx)dx cos(nπx)dx nπ cos(nπ/) + n π sin(nπx) + nπ cos(nπ) nπ cos(nπ/) a n = n π sin(nπ/) / ( nπ cos(nπx) nπ cos(nπ) n π sin(nπx) / / + ) nπ cos(nπ/) +

3 Maths 38 The First Midterm 3 We obtain # u(x, t) = u(x, t) = u(x, t) = l= l= n π sin(nπ/)sin(nπx)cos(nπat) (l ) sin((l )π/)sin((l )πx)cos((l )πat) π ( ) l (l ) sin((l )πx)cos((l )πat) π Since f(x) = x is an even function, all the terms will be cosine terms. The Fourier series thus becomes f(x) a + a k coskx where a k = π In this case, we find the coefficients to be a k = π x coskxdx a k = xcoskxdx + π π a k = x π k sinkx + k a k = sinkxdx kπ a k = k π a k = k π We separately calculate a. Thus coskx + coskx π ( ( ) k ) a = π a = π f(x) π k= f(x)coskxdx. xcos kxdx sin kxdx + x k sinkx π sin kxdx k= f(x) π l= x dx xdx = π. k ( ( ) k ) k coskx; π cos((l )x) π(l ) sin kxdx

4 Maths 38 The First Midterm Remark. Note that the Fourier coefficients decay quadratically, and therefore the Fourier series we have obtained converges uniformly on, π but then, as we showed in class, it must converge to f. Thus in fact we have #5 f(x) = π k= f(x) = π l= ( ( ) k ) k coskx π cos((l )x) π(l ) Since f(x) = x is an even function, all the terms will be cosine terms. The Fourier series thus becomes f(x) = a + a k coskx where a k = π In this case, we find the coefficients to be k= a k = x coskxdx π a k = x π sin kx π k k a k = x kπ k coskx π a k = kπ a k = kπ We separately calculate a. Thus we have a k = ( )k k f(x)coskxdx. + k xsin kxdx cos kxdx π k cosπk π k cosπk + ( k sin kx π πk ( )k a = π a = a = 3π π x dx 3π x3 f(x) π 3 + π 3 + π 3 = π 3 k= ( ) k coskx k

5 Maths 38 The First Midterm 5 Remark. Note that the Fourier coefficients decay quadratically, and therefore the Fourier series we have obtained converges uniformly on, π but then, as we showed in class, it must converge to f. Thus in fact we have #6 We showed in # above that f(x) = π 3 + x = f(x) = π k= ( ) k coskx k cos((n )x). π(n ) (As remarked above, the Fourier coefficients decay quadratically, and therefore the Fourier series we have obtained converges uniformly on, π but then, as we showed in class, it must converge to f.) Setting x = and rearranging gives = π π(n ) cos() We showed in #5 above that π = π (n ) π 8 = (n ) x = f(x) = π 3 + ( ) n cosnx. (As remarked above, the Fourier coefficients decay quadratically, and therefore the Fourier series we have obtained converges uniformly on, π but then, as we showed in class, it must converge to f.) Setting x = and rearranging gives π 3 = π = n ( ) n n ( ) n+ Setting x = π into the equation from #5 and rearranging gives π = π 3 + n ( ) n cos(πn) n

6 Maths 38 The First Midterm 6 #7 J n(x) = π 3 = J n (x) = k= ( ) n ( ) n n π 6 = n k= ( ) ( ) k x k+n k!(k + n)! (k + ( n)( )k x k!(k + n)! ) k+n Every term in J n() = except that for when k + n = or k = ( n)/. Thus for n even, J n () =. We look at n odd. J n() = ( )( n)/ k!( k)! We note that the numerator is bounded, and look at the denominator k!( k)! = Γ(k + )Γ( k). The first of which goes to ± for k and the second goes to ± for k k =, n = ± (note: all the k are integers). Thus J n () = (since the denominator goes to ± ) except for n = ±. We look at J () (n = and k = ). For J (), n = and k =. J () = J () = ( )( )/ =!( )!. ( )(+)/ =!( )! Thus, in summary, J n () = unless n = ± in which case J ± () = ±/. #8 We showed in class the following relations d dx (xp J p (x)) = x p J p (x) d dx (x p J p (x)) = x p J p+ (x). Expanding the LHS of each equation and rearranging, we get px p J p (x) + x p d dx J p(x) = x p J p (x) d dx J p(x) = J p (x) p x J p(x)

7 Maths 38 The First Midterm 7 px p J p (x) + x p d dx J p(x) = x p J p+ (x) d dx J p(x) = J p+ (x) + p x J p(x). We then add these two equations to get #9 d dx J p(x) = J p (x) J p+ (x) J p(x) = J p (x) J p+ (x) Since J n (x) = ( ) n J n (x) (J n (x)) = (J n (x)) we can write the RHS as =(J (x)) + = (J n (x)) (J n (x)) + We take the derivative of both sides of the equation to get = Using the identity from #8, this becomes = = = (J n (x)) J n (x)j n(x). J n (x)j n (x) J n (x)j n+ (x) J n (x)j n (x) J n (x)j n (x) J n (x)j n+ (x) J n n(x)j n (x) the desired identity. Thus we have proved that the RHS is constant and need to evaluate the constant, which is independent of x. We choose x =, and we have, by definition of the Bessel functions, that J () =, J n () = for n, which concludes the proof of the identity. Rearranging, we get that (J (x)) = (J n (x)). Since we get that (J n (x)) (J (x))

8 Maths 38 The First Midterm 8 Similarly, we look at J (x) Rearranging, we get that n = (J (x)) + (J k (x)) + (J n (x)) + k= n (J n (x)) = (J (x)) (J k (x)) k= l=n+ l=n+ (J l (x)). (J l (x)). Since we have that n (J (x)) + (J k (x)) + k= (J n (x)) l=n+ (J l (x)) J n (x).