PHYS 301 HOMEWORK #4

Transcription

1 PHYS HOMEWORK #4 Due : February 7 Do all integrals by hand; you may check your answers via Mathematica but must show all work.. Show for m, n integers :, m n sin (m x) sin (n x) dx = -, m = n - sin (n x) cos (m x) dx = - cos (m x) cos (n x) dx =, m n, m = n, m = n = Solution : We will rewrite each integrand using the sin and cos addition formulae : sin (m ± n) x = sin (m x) cos (n x) ± sin (n x) cos (m x) cos (m ± n) x = cos (m x) cos (n x) sin (m x) sin (n x) a) Adding the sin addition/subtraction formulae gives: sin (m + n) x + sin (m - n) x = sin (m x) cos ( n x) () This allows us to write the second integral above as: - sin (n x) cos (m x) dx = - sin (m + n) x dx + - sin (m - n) x dx () These integrals return cos(p x) where p is an integer evaluated at and -. Since cos is an even function, each integral evaluates to zero when m n. In the case where m = n, the integrals become:. - sin ( m x) dx and dx both of which are easily (or trivially) shown to be zero. b) If we add the cos addition/subtraction formulae, we get: - cos (m x) cos (n x) dx = - cos (m + n) x + cos (m - n) x dx These integrals return sin(p x) (where p is an integer) evaluted at and -; since sin is zero at those values, the integral is zero for all m n. If m = n, the integral becomes

2 phys-7hw4s.nb - cos (m x) dx = - + cos ( x)) dx = (we use the trig identities cos(x) = cos x - sin x and sin x = - cos x) If m = n =, cos (m x) = cos (n x) =, and the integral of on this interval is. c) We subtract the subtraction/addition cos formulae and get : - sin (m x) sin (n x) dx = - (cos (m - n) x - cos (m + n) x) dx Using prior reasining, each integral on the right returns sin(p x) on [-,], so all these terms are zero for m n. If m = n, we have: If m = n =, the integral is zero since sin () =. - sin (m x) dx = For the remaining problems we will use these definitions of the Fourier series: a o = f (x) dx - a n = - f (x) cos (n x) dx b n = - f (x) sin (n x) dx and for the Fourier series:. Consider the function : f (x) = a + Σ an cos (n x) + Σ bn sin (n x) n= n= f (x) =, - < x < x, < x < Find the Fourier coefficients and then write the Fourier series both in closed form and by writing out the first three non zero terms of each series. Solution : We find the Fourier coefficients, using integration by parts where needed : a n = a o = x dx = 4 x cos (n x) dx = - sin (n x) dx = b n = n x sin (n x) n = n (cos (n ) - ) =, n even - n, n odd x sin (n x) = - n x cos (n x) + n cos (n x) dx n cos (n x)

3 phys-7hw4s.nb The last integral goes to zero since it returns sin(n x), so we are left with: The Fourier series is: The first three terms yield: f (x) = 4 - b n = - - ( cos (n ) - ) = n n (-)n f (x) = 4 - cos x + cos ( x) 9 Σ cos (n x) n=odd n cos (5 x) + 5 Plotting two cycles and verifying with Mathematica: In[]:= g = Plot / 4 - SumCos[n x] n, {n,,, } + sin (n x) + Σ (-) n+ n= n sin x - sin x Sum-^n + Sin[n x] / n, {n,, }, {x, -, } sin ( x) Out[]= Just to be sure, we overlay the line y = x in red : In[5]:= g = Plot[x, {x,, }, PlotStyle Red]; Show[g, g] Out[6]= Consider the function :

4 4 phys-7hw4s.nb f (x) =, - < x < x, < x < Find the Fourier coefficients and write the Fourier series both in closed form and by writing out the first three non-zero terms of each series. Solution : We find the Fourier coefficients by making use of integration by parts where needed. a o = x dx = o 6 a n = x cos (n x) dx = n x sin (n x) = - n - n x cos(n x) - n x sin (n x) dx + n cos(n x) dx The final integral above is zero since it returns sin(n x) to be evaluated at and, so the a n coefficients are: or : b n = = - n The Fourier series can be written: or : a n = cos (n ) - n = (-)n n x sin (n x) dx = - n x cos (n x) + n x cos (n x) dx cos (n ) + n - x cos (n x) dx = n (-)n + n n x sin (n x) n - sin (n x) dx = n (-)n - n cos (n x) = - n (-)n + n ( - (-)n ) b n = - / n, n even -4 - n n, n odd f (x) = 6 - Σ (-) n cos (n x) sin (n x) - Σ n= n n=even n f (x) = 6 - cos x - cos ( x) 4 cos ( x) + -,,, sin x sin ( x) n sin (n x) + Σ n=od n sin ( x) sin (4 x) sin (5 x)

5 phys-7hw4s.nb 5 Verifying with Mathematica: Clear[a, a, bodd, beven] a = ^ 6; a[n_] := - n n beven[n_] := - / n bodd[n_] := -4 - n n g = Plot[a + Sum[a[n] Cos[n x], {n,, }] + Sum[beven[n] Sin[n x], {n,,, }] + Sum[bodd[n] Sin[n x], {n,,, }], {x, -, }]; g = Plot[x^, {x,, }, PlotStyle Red]; Show[g, g] Find the Fourier coefficients and write the Fourier series in closed form and also the first three non - zero terms of each series for : f (x) =, - < x < sin ( x), < x < Solution : The danger here is to assume that orthogonality will apply since f (x) = sin ( x). However, since the limits of integration are [, ] and not [-, ], we must do the evaluations explicitly. a o = sin ( x) dx = - cos ( x) We use eqs. ( and ) from problem one to help with the evaluation of a n and b n. sin ( x) cos (n x) dx = = sin (n + ) x - sin (n - ) x dx a n = = cos (n + ) x cos (n - ) x - + = n + n -, n even -4 / n - 4, n odd

6 6 phys-7hw4s.nb b n = = sin (n - ) x sin (n + ) x - n - n + It is common for students to get to this point and conclude that all the b n values are zero since our integrals return sin(n x) where n is an integer on [,]. And this is true for all values of n except n =. In this case, our integral becomes: sin ( x) dx = sin ( x) sin (n x) dx = (cos (n - ) x - cos (n + ) x) dx b = and our Fourier series is: = Verifying with Mathematica: f (x) = sin ( x) Sin ( x) + 4 cos x cos (n x) - 4 Σ n=odd n cos x 5 - cos 5 x -... In[47]:= PlotSin[ x] - (4 / ) SumCos[n x] n^ - 4, {n,,, }, {x, -, }..5 Out[47]= Find the Fourier coefficients and Fourier series for the function : f (x) = Write out the first three non-zero terms of each series. Solution : -, - < x <, < x < This is a relief after all the integration by parts we ve just done. And if we look a litltle more closely, our problem can be simplified even more. Note that f is an odd function, this means that the only non-zero coefficients will be the b n terms. Moreover since the function is odd, we know that:

7 phys-7hw4s.nb 7 In[48]:= In our case, this becomes: b n = and we have simply: b n = - sin (n x) dx = - f (x) = 4 n cos (n x) sin x + sin x Plot[(4 / ) Sum[Sin[n x] / n, {n,,, }], {x, -, }] = n ( - (-)n ) = 4 / n, n odd, n even + sin 5 x f (x) sin (n x) dx = f (x) sin (n x) dx..5 Out[48]= Now, is the problem even simpler than this. Could we have used another Fourier series that we have computed and deduced this one without ever taking an integral? (Hint : compare this series with the first example in class; f (x) = for < x < and f (x) = for - < x < ), see how they compare.)