Problem 1. Possible Solution. Let f be the 2π-periodic functions defined by f(x) = cos ( )

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1 Problem Let f be the π-periodic functions defined by f() = cos ( ) hen [ π, π]. Make a draing of the function f for the interval [ 3π, 3π], and compute the Fourier series of f. Use the result to compute the value of the series n= ( ) n n. First observe that f( ) = cos ( ) ( ) = cos, so f() is an even function. Then the Fourier series of the function f() is of the form Then and π a 0 = π = π π π 0 a 0 + a n cos(n). n= π f() d = f() d π π 0 ( ) cos d = sin ( ) π π a n = f() cos(n) d = f() cos(n) d π π π 0 = π ( ) cos cos(n) d = π ( ( n + cos π 0 π 0 = π ( ) ( ) n + n cos + cos d π 0 = ( ( ) n + π n + sin + ( n n sin = ( ( ) n + π n + sin π + ( n n sin = ( π n + ( )n + ) ( n ( )n+ = ( )n π ( ) ( ) = ( )n = ( )n. π n π n Thus, the Fourier series to f() is =π =0 π + ( ) n n= π n cos(n). = π, ) + cos )] =π =0 ) π 0 0 ( n ) n + n ) )) d

2 Finally to find the value of the series, e evaluate the Fourier series of f() at = 0, and by the Fourier Theorem and since f() is continuous at = 0 e have that ( 0 = cos = f(0) = ) π + ( ) n n= π n cos(0) = π + π hence Problem n= ( ) n n = π ( ) = π. π Find all the non-trivial solutions of the heat equation n= u t = u here 0 π and t 0, ( ) n n, that are of the form u(, t) = F () G(t), and that satisfy the boundary conditions u(0, t) = 0 and u (π, t) = 0 for every t 0. Use this to find a solution satisfying the initial condition ( ) u(, 0) = cos() sin for every 0 π. If u(, t) = F ()G(t) e have that u = F G and u t = F G. Then if e replace this into the heat equation e have that F G = F G, so it follos that F F = G G = K here K R. From this e deduce the to folloing ODEs F KF = 0 and G KG = 0. Additionally, observe that from the boundary conditions it follos that u(0, t) = F (0)G(t) = 0 implies F (0) = 0,

3 u (π, t) = F (π)g(t) = 0 implies F (π) = 0. First e solve the ODE F KF = 0. Suppose that K = 0, then F = 0, from here e have that F () = A + B. But 0 = F (0) = A 0 + B = B and 0 = F (π) = A, thus F = 0. This gives us a trivial solution. No suppose that K > 0. Then the solution of F KF = 0 is of the form F () = A cosh( K) + B sinh( K). But 0 = F (0) = A cosh(0)+b sinh(0) = A and 0 = F (π) = B K cosh(0) = B K = 0, thus F = 0. This gives us also a trivial solution. Finally suppose that K < 0. We rite K = p here p > 0. Then the ODE F + p F = 0 has solution F () = A cos(p) + B sin(p). But 0 = F (0) = A cos 0 + B sin 0 = A and 0 = F (π) = Bp cos(pπ). But this means that pπ = n+ π for n Z, from here it follos that p = n + Then e define the functions ( ) n + F n () = sin ith n Z. for n Z. Observe that since sin is an odd function for every n N e have that ( ) n + F n () = sin = sin ( n ) = F n (). So the negative n s do not give us any ne solutions. Then it is enough to use the functions ( ) n + F n () = sin for n = 0,,,... No e are going to solve the ODE G KG = 0, but since K = p = ( n+ ) for n = 0,,,..., e can rite the ODE ( ) n + G + G = 0, 3

4 n+ that has solution G n (t) = e ( ) t for n = 0,,,.... Then our desired solutions are ( ) n + n+ u n (, t) = F n ()G n (t) = sin e ( ) t for n = 0,,,... No e ant to find a linear combination u(, t) = n=0 B n u n (, t) such that u(, 0) = B n u n (, 0) = B n F n ()G n (0) n=0 n=0 ( ) ( ) n + = B n sin = cos() sin. n=0 But cos() sin ( ) = sin ( 3 ) + sin ( 5 ), so it follos that B = and B = and the rest of B n s are zero. Therefore our desired solution is Problem 3 u(, t) = sin ( 3 ) e ( 3 ) t + sin ( 5 ) e ( 5 ) t. Sho the Fourier transform F( e ) = i π We have that F(e ) = π = π 0 sin ( + ) d. e e i d = π 0 e ( i) d + e ( i) d π = π ( e ( i) ( i) e( i) ( i) = π = π 0 ] =0 =. Use this to compute (+ ) e e i d + e e i d π + ( ] e ( i) = π ( i) e( i) ( i) =0 ( i) + π ( + i) = ( + i) + ( i) π ( + i) ( i) i i (( + i)( i)) = π ( + ), 0

5 as desired. For the last part, e use the inverse of the Fourier transform, that is, e = π i π ( + ) ei d = π No if e set = e have that e = i π ( + ) ei d = π = i π ( + ) cos d + π But then observe that cos d = 0 because (+ ) odd function, so it follos that hence Problem e = π i ( + ) ei d. i (cos + i sin ) d ( + ) sin d ( + ) sin d ( + ) π e = sin d. ( + ) cos is an (+ ) Perfom 3 iterations of the Neton method to find the root of the function f() = e ith 0 = 0. (Use only decimals in your computations). Observe that f () = + e, then Netons metode is the iteration given by Then e have that Problem 5 n+ = n f( n) f ( n ) = n n e n + e n = ne n + e n + e n. 0 = 0, = 0.5, = , 3 = Use the Laplace transform to solve the differential equation ith initial conditions y 3y + y = e 3t, y(0) = 0 and y (0) =. 5

6 We use the Laplace transform in the ODE, and denoting Y the Laplace transform of y, e have that so e have that and and Y = s Y sy(0) y (0) 3(sY y(0)) + Y = s 3, s Y 3sY + Y = s 3, Y (s 3s + ) = s 0 + = s 3 s 3, s 0 (s 3)(s 3s + ) = s 0 (s 3)(s )(s ) = s 3 + s 3 s. So then applying the inverse of the Laplace transform e get y(t) = e 3t + e t 3e t. Problem 6 Find the polynomial of smallest degree that interpolates the points of the function f() i f( i ) 5 Use this polynomial to estimate f(3). Using Neton interpolation, e obtain /3 - / Lagrange interpolation ould be fine as ell 6

7 Then the polynomial is of the form p() = ( )( + )( + ) ( + )( + ) + ( + )( + ) + ( + ) + 3 Then the estimate of f(3) can be done ith p(3) = 6. Problem 7 We ant to numerically evaluate the integral f() d here f() = sin( ), 0 ith the Simpson method such that the approimation error is smaller than What is the largest value of the step size h that this accuracy is guaranteed? Use this h to compute a numerical approimation of the above integral by the Simpson method. (Use only decimals in your computations). (Hint: You can use that ma 0 f () () 30). We use the error estimation for the Simpsons methode, that says that ɛ h b a 80 ma f () () h 0 h 30 = a b Since e ant that ɛ < 0.00 e may choose h such that h < 0.00, so 6 h < Then if e ant to use Simpsons method ith this accuracy e need to pick n such that < So observe that ith n =, e have n that / = 0.5 < Then e compute S = Problem 8 ( sin(0) + sin((0.5) ) + sin((0.5) ) + sin((0.75) ) + sin(() ) ) = Let y() be the function that solves the ODE y = y and y(0) =. Use the Euler method ith h = 0. to approimate the values of y() at the points = 0., = 0. and 3 = 0.3. (Use only decimals in your computations) 7

8 We have the ODE y = f(, y) =. Then the Euler method iteration says y y n+ = y n + hf( n, y n ). Then if e start ith 0 = 0 and y 0 = and ith h = 0., e have that y =, y = 0.99, y 3 = Problem 9 Write the linear system 0 + y z = 8 + y + 0z = 7 + 5y + z = in such a form that you can apply the Gauss-Seidel method and it converges. Then perform 3 iterations ith starting values 0 = y 0 = z 0 = 0. (Use only decimals in your computations). First e can rearrange the linear system in the folloing ay Then e can apply Gauss-Seidel to 0 0 A = = + 0 y 0 z = y + 5 z = y + z = = L+I+U Observe that the method ill converge because A is diagonal dominant. 8

9 We have then the folloing iteration equations (n+) = y(n) + 0 z(n) y (n+) = 5 5 (n+) 5 z(n) z (n+) = (n+) 0 y(n+) Then if ( (0), y (0), z (0) ) = (0, 0, 0) e have that Problem 0 ( (), y (), z () ) = (.8, 0.9, 0.986) ( (), y (), z () ) = (.9906, 0.998, ) ( (3), y (3), z (3) ) = (,, ). Let R be the region defined by the lines L : y = L : y = 0 L 3 : = 0. Let u(, y) be the function defined in R that satisfies the Poisson equation and boundary conditions u + u y = y u(, y) = if (, y) L u(, y) = if (, y) L u(, y) = y if (, y) L 3 y u = y R u = u = 9

10 Let us define the points ( i, y j ) = (i h, j h) ith h =. Use the method of difference equations ith h = in order to set up a linear system for finding approimations of the values u(, ), u(, ) and u(, ). If e define u i,j := u(ih, jh), the difference method gives us the folloing equation h (u i+,j + u i,j + u i,j+ + u i,j u i,j ) = (ih jh) So if h = e have the folloing equations u, + u 0, + u, + u,0 u, = ( ) = u 3, + u, + u, + u,0 u, = ( ) = u, + u 0, + u,3 + u, u, = ( ) = and using the boundary conditions u, + + u, + u, = + u, + + u, = u, u, = Hence the the desired linear system is u, + u, u, = 0 u, u, = 6 u, u, = 6 0