PHYS 301 HOMEWORK #5

Transcription

1 PHY 3 HOMEWORK #5 olutions On this homework assignment, you may use Mathematica to compute integrals, but you must submit your Mathematica output with your assignment.. For f x = -, - < x <, < x < Find the Fourier coefficients and write out the first three non - zero terms of the series expansion. olution : This is an odd function on (-, ), so we know that the a coefficients are zero. ince the function is L = periodic, L =. Computing the b coefficients : b n = - f x sin x dx = ÿ sin x dx = - cos x = Therefore, our Fourier series is : - cos - = - - n =, odd, even f x = sin 3 p x si x + + sin 5 p x +... p 3 5 Verifying via Mathematica : Plot um in n x n, n,, 3,, x, 3, For f x = x, < x < - x, < x <

2 phys3-hw5s.nb extend f to construct a) an odd function on (-, ) and b) an even function on (-, ). Compute the Fourier coefficients for each series and write out the first three non - zero terms of each expansion. ( pts for this problem). olution : a) If we construct an odd function on (-, ), we know that only the b n coefficients will be non - zero; we compute these via : b n = f x dx = x sin x dx + - x sin x dx In[]:= We obtain : Integrate x in n x, x,, Integrate x^ in n x, x,, Out[]= In[3]:= n Cos n in n n 8 n Cos n n Cos n n in n in n implifying this output using the fact that n is an integer : implify, Assumptions n Integers Out[3]= n n Cos n n 3n in If we examine this expression, we see that the argument of the trig functions (/) indicates we will need to consider separate cases (i.e., separate values) of n. In[]:= ince we have terms involving sin and cos of /, we know that we need to consider cases where n = {,, 3, }. We can use the Mod command successively in conjuction with Assumptions : implify, Assumptions Mod n, Out[]= In[5]:= 3n n (These are the coefficients when n =, 5, 9,...) implify, Assumptions Mod n, Out[5]= In[6]:= 8 n (These are the coefficients when n =, 6,,...) implify, Assumptions Mod n, 3 Out[6]= 3n n (These are the coefficients for n = 3, 7,,...)

3 phys3-hw5s.nb 3 In[7]:= implify, Assumptions Mod n, Out[7]= n (Finally, these are the coefficients for n =, 8,,...) Using these data, the first few coefficients are : b = b = b 3 = p-p p p 8 p p-36 p 7 p 3 b = p And we can write the first few non - zero terms of the expansion, using these coefficients, as : f x = b sin p x + b sin p x + b 3 sin 3 p x + b sin p x You can verify these results by writing the following short code : In[67]:= Clear bn, f bn Integrate x in n x, x,, Integrate x^ in n x, x,, ; f Which x, x, x, x^ ; Plot f, um bn in n x, n,, x,, Out[7]= b) For the even solution, we extend f as an even function on (-, ). We can employ symmetry to argue that the b n coefficients are zero, and that the a n coefficients can be determined from : a = f x dx = xdx + - x dx = 6

4 phys3-hw5s.nb In[]:= We use Mathematica to compute the a n coefficients and plot the function : Clear a, an, f f Which x, x, x, x^ ; a Integrate x, x,, Integrate x^, x,, ; an Integrate f Cos n x, x,, ; Print "a ",a,"; ","an ", an Print "The first five terms of the Fourier expansion are: ", a um an Cos n x, n, 5 Plot a um an Cos n x, n,, x,, a n 3n Cos n n n Cos n in in n 6 ; an n in n The first five terms of the Fourier expansion are: 3 Cos x 8 Cos x 9 Cos 3 x Cos x And the plot of this Fourier expansion :. 5 Cos 5 x Out[5]= Problem, p. 37 of the text. olution : The first thing we need to do to solve this problem is write the equation of the string on (, L). Using simple geometry, we have : f x = hx L, < x < L h-hx L, L < x < L, L < x < L Knowing we are constructing a sin series, we know we need only worry about the b coefficients, and we can calculate them via : b n = L L f x sin x L dx = L -L L f x sin x L dx =

5 phys3-hw5s.nb 5 In[6]:= Integrate hx L in n x L, x,, L L Integrate h hx L in n x L, x, L, L and when we do this integration we get : hl -npcos + in hl npcos + n p L + in - in n p Which by inspection becomes (well, with a little help from) : In[6]:= implify, Assumptions n Integers Out[6]= In[67]:= 6 h Cos in n p The nth term of the Fourier expansion is the nth b coefficient times in[ x/l]. We can verify that these coefficients yield the proper curve by setting L and h to arbitrary but reasonable values (Mathematica won' t print symbolic functions) : Clear f,bn,x,h,l h.; L.; f Which x L,hx L, L x L, h hx L, L x L, ; bn L Integrate hxin n x L L, x,, L Integrate h hx L in n x L, x, L, L ; Plot um bn in n x L, n,, x,, L Consider the following graph of one complete cycle of voltage vs. time :

6 6 phys3-hw5s.nb corresponding to V t = 5, < t < -5, < t < 5 where V is measured in volts and t in seconds. Write the Fourier series representing this pattern (assume this part of the graph is repeated 5 times/second). olution : ince the entire cycle lasts a duration of /5 s, the cycle is L = /5 periodic, implying that L = / in this calculation. We use the familiar equations to find coefficients : 5 a = 5 dt - 5 dt = a n = b n = 5 sin t 5 5 cos t dt - 5 cos t = - sin r 5 5 sin - sin np - sin = 5 = 5 sin t dt - 5 sin t dt = 5 - cos t + cos t cos - + cos np -cos = - - n = =, n odd, n even V t = sin t p odd n Verifying : = sin 3 ÿ p t sin 5 ÿ p t sin p t p 3 5

7 phys3-hw5s.nb 7 Plot um in n t n, n,, 3,, t,, Use the function iroblem of this assignment to find the value of odd n n olution : We will use Parseval' s theorem : average value of f x on -, = a + an + n= ince the length of this interval is, we write the average of f as: bn dx = - The Fourier series for this problem shows that all the a n coefficients are zero, and that b n = for odd values of n Therefore, odd n = 8 p odd n n = fl odd n n We could have figured this out from the result derived in class for the infinite sum of all the reciprocal squares, namely n= n = p 6 We can find the inifinite sum for the even numbers by noting that all even numbers are divisible by, so that we can represent them as n. Thus, the sum of all even terms is simply : even n = p 8 = n= n = n= n = ÿ p 6 = p The sum of the odd terms is just the difference between the sum over all numbers and the sum over n-

8 8 phys3-hw5s.nb the evens, or : n= n - even n = odd n = p 6 - p = p as we found before. 8 Finally, direct verification with Mathematica : um n, n,,, 8 6. You are familiar with Fibonacci numbers (fib[n]). Let' s define a similar set of numbers, the so called Loyola U numbers which have the properties : lu[] = lu[] = lu[n] = lu[n - ] + lu[n - ] (Note that the lu numbers start at n = ). Write a short Mathematica program to test the conjecture : lu[n] = fib[n - ] + fib[n + ] for n 3. If the conjecture is true for a given value of n, your program should output the ratio of lu[n]/fib[n]. If the conjecture is false for a given value of n, your program should print "The conjecture is false." Your printout should include both your program and all results. ( pts for this question) olution : We will define two functions, lu[n] and f[n] to describe the Loyola and Fibonacci numbers respectively. We will initialize values and describe functions : Clear lu, f f ; f ; lu ; lu ; f n_ : f n f n f n lu n_ : lu n lu n lu n These definitions allow us to store previously computed values of lu n and f n to minimize computing time. Do If lu n f n f n, Print "For n ", n, " The value of lu n f n ", lu n f n N, Print "The conjecture is false.", n,, 3 We nest an If statement inside a Do loop; the If statement tests to see if the conjecture is true make sure you notice that this requires the use of a double equal sign. Note also the limits of the Do loop; what would happen and why would it happen if you started with n? Now execute the program:

9 phys3-hw5s.nb 9 For n The value of lu n f n 3. For n 3 The value of lu n f n. For n The value of lu n f n For n 5 The value of lu n f n. For n 6 The value of lu n f n.5 For n 7 The value of lu n f n.377 For n 8 The value of lu n f n.38 For n 9 The value of lu n f n.359 For n The value of lu n f n.3636 For n The value of lu n f n.3596 For n The value of lu n f n.36 For n 3 The value of lu n f n.365 For n The value of lu n f n.367 For n 5 The value of lu n f n.367 For n 6 The value of lu n f n.367 For n 7 The value of lu n f n.367 For n 8 The value of lu n f n.367 For n 9 The value of lu n f n.367 For n The value of lu n f n.367 For n The value of lu n f n.367 For n The value of lu n f n.367 For n 3 The value of lu n f n.367 For n The value of lu n f n.367 For n 5 The value of lu n f n.367 For n 6 The value of lu n f n.367 For n 7 The value of lu n f n.367 For n 8 The value of lu n f n.367 For n 9 The value of lu n f n.367 For n 3 The value of lu n f n.367 Pretty clearly the conjecture seems reasonable. In fact lu numbers really exist and are known as Lucas Numbers.