Exam TMA4120 MATHEMATICS 4K. Monday , Time:

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1 Exam TMA4 MATHEMATICS 4K Monday 9.., Time: 9 3 English Hjelpemidler (Kode C): Bestemt kalkulator (HP 3S eller Citizen SR-7X), Rottmann: Matematisk formelsamling Problem. a. Determine the value ( + i) 6 i 3 ( + 4i). ( + i) 6 + i 6 i 3 ( + 4i) = i 3 + 4i. Further + i 6 = 3 = 8; i 3 = ; + 4i = 7. ( + i) 6 i 3 ( + 4i) = 8 7. b. Let ω 3 = and Im ω. Find ω + ω +. ω 3 = ω 3 =. Or ω 3 = (ω )(ω + ω + ) =. Since Iω we have ω. Therefore ω + ω + =. Equation ω 3 = has the following solutions: ω =, ω = e iπ/3, and ω = e 4iπ/3. Since Iω = we have to find only. and ω + ω + and ω + ω + ω = e iπ/3 = cos π 3 + i sin π 3 = + i 3. ω = e 4iπ/3 = cos 4π 3 + i sin 4π 3 = i 3. Respectively ω + ω + =.

2 Relation ω + ω + = can be proved similarly or just mention that ω = ω, hence ω + ω + = ω + ω + =. Problem. a. Find the Laplace transform F (s) = Lf(s) of the function f(t) = te t cos t.. First shifting theorem: L(cos t)(s) = s + 4. L(e t cos t)(s) = Differentiation of transform formula: ( L(te t cos t)(s) = s + (s + ) + 4 s + (s + ) + 4. ) = (s + ) 4 [(s + ) + 4]. Comment: There are many other ways to solve this problem. Each is OK so far it is correct. b. Solve the integral equation y(t) = e t + We rewrite the equation as y(t) = e t + } e τ y(τ)dτ, t. e t τ y(τ)dτ, t, ( ) so the integral in the right-hand side is a convolution of y(t) and e t. Denote Y (s) = Ly(s) and use that L(e t )(s) = (s ). We then have ( ) Y (s) = s + Y (s) Y (s) = s s y(t) = et. We rewrite the equation as or y(t)e t = + z(t) = + z(τ)dτ, e τ y(τ)dτ, t, where z(t) = e t y(t). The integral in the right hand side of (**) is the convolution of the Heaviside function and z(t). Let Z(s) = Lz(s). The Laplace transform of (**) gives ( ) Z(s) = s + Z(s) Z(s) = s s z(t) = et y(t) = e t.

3 3 Problem 3. Let f(x) be the π-periodic function, π < x < ; f(x) =, < x < π. Find its Fourier series. Then determine the sum of the series (n ). Hint: Parseval s formula. Fourier series: c = π Parseval formula: f(x) = n= c n e inx ; c n = f(x)e inx dx. π π dx = ; c n = π e inx dx = π inπ einx π = f(x) = + iπ l= π f(x) dx = π π l ei(l )x. c n. In our case /π π f(x) dx = /, therefore = 4 + π (l ) = 4 + π l= l= (l ) = π 8. l=, n is even; n is odd., inπ (l ). Comment. You can also use Parseval s formula for sin and cos Fourier series. It will lead you to the same result of course. Problem 4. The function u(x, t) satisfies the equation and the boundary conditions u xx = u t u, < x < π, t ( ) u(, t) =, u(π, t) =, t. ( ) a. Find all solutions of this problem having the form u(x, t) = X(x)T (t).

4 4 Let u(x, t) = X(x)T (t). Then ( ) X (x)t (t) = X(x)T (t) + X(x)T (t) here k is an unknown constant. Taking ( ) into account we have and X (x) X(x) = T (t) T (t) = k, X (x) kx(x) =, < x < π, X() =, X(π) =. T (t) (k + )T (t) =, t >. The standard analysis shows that non-trivial solutions exist for k = n, n =,,... and the corresponding functions X n and T n are X n (x) = sin nx and T n (t) = e ( n )t, so u n (x, t) = b n e ( n )t sin nx. b. Find the solution u(x, t) which also satisfies the initial condition u(x, ) = sin x, < x < π. The solution has the form u(x, t) = b n e ( n)t sin nx, n= where b n are the coefficients in the expansion sin x = b n sin nx, < x < π. Finding b n : b n = π sin x sin nxdx = π J,n = π o n= ( ) cos x sin nxdx = π sin nxdx cos x sin nxdx. π o π }}}} J,n sin nxdx = nπ cos nx π = nπ, J,n n is odd;, n is even. In order to find J,n we use the relation sin α cos β = (sin(α + β) + sin(α β))/. Then J,n = cos x sin nxdx = sin(n + )xdx + sin(n )xdx. π π π

5 5 Let n then (similarly to calculation of J,n ): (n+)π J,n = +, n is odd; (n )π, n is even. Direct calculation: b =. π(n ) b n = + +, n is odd πn (n+)π, n is even. Comment You may substitute this expression into ( ), but there is no need for it. Minor punishment will be applied for those who made mistake in calculating the integrals. Problem 5. Find the Fourier transform where ˆf(w) = (/ π) f(x) = f(x)e iwx dx,, if < x < ;, otherwise. Use the formula for the inverse Fourier transform in order to find the value of the integral sin(w) cos w dw. w Fourier transform: Inverse Fourier transform: f(x) = π π ˆf(w) = π u eiux du = π e iwx dx = u sin w π w. cos ux du+i π u x du. The last term in the right-hand side vanishes because the integrand is odd, so we have f(x) = cos ux du. π u Now change variables u = w: f(x) = π sin w w cos wx dw, and set x = / and multiply the both sides by π. We obtain ( ) sin w π = πf = cos w dw. w Problem 6.

6 6 a. Find the singular points of the function f(z) = e 5iz z z +, classify them (poles, essential singularities, removable singularities), and determine the residues. s The singular points of f are zeroes of the denominator i.e. the points z = + i and z = i. They are simple poles. Therefore the residues can be defined by the formula e 5iz Res zk f = (z z + ) z=z k, k =,. In particular Res +i f = i e5( +i), Res i i e5(+i). b. Evaluate the integral Explain your procedure. sin 5x x x + dx = i sin 5x x x + dx. e 5ix x x + dx i } } I e 5ix x x + dx. }} I The function e 5iz decays in the upper half-plane so in order to complement the integral I over the segments of real line to a closed curve one has to use the half-circles located in the upper half-plane. Therefore ( ) e 5iz I = πres +i = π z z + i e5( +i). The function e 5iz decays in the lower half-plane so in order to complement the integral I over the segments of real line to a closed curve one has to use the half-circles located in the lower half-plane. Also orientation should be taken into account. We obtain ( ) e 5iz I = πres i = π z z + i e5( i). sin 5x x x + dx = I I = πe 5 sin 5.

7 Another, and actually simpler way of finding I is to mention that ii = iī. Then you can avoid extra calculations. 7