Math Assignment 13

Transcription

1 Math 8 - Assignment 3 Dylan Zwick Spring 4 Section 9. -, 8,, 3, Section 9. -, 9, 5, 7, Section 9.3 -, 5, 8, 3,

2 Section 9. - Periodic Series Functions and Trigonometric Sketch the graph of the function f defined for all t by the given formula, and determine whether it is periodic. If so, find its smallest period. f(t) = sin3t. Solution - Periodic with period fl.

3 Sketch the graph of the function f defined for all t by the given formula, and determine whether it is periodic. If so, find its smallest period. f(t) = sinhiit. Solution Not periodic. 3

4 J - + (cos -- cos The value of a period ir function f(t) in one full period is given below. Sketch several periods of its graph and find its Fourier series. f(t) =, t Solution - Just the constant function. The Fourier series for this will be just, but let s formally calculate it anyways. a= f ir dt= (n ( r))=. For n : = J f?r cos(nt)dt= sin(nt) = =. = 7T pir sin (nt)dt = cos (nt) = fl7. (nir) (nir)) = So, the Fourier series for is: a. 9 a cos (nt) + b sin (ni) = =. Wow! 4

5 J The value of a period ir function f(t) in one full period is given below. Sketch several periods of its graph and find its Fourier series. f(t)={ 7<t<O Solution - The Fourier coefficients are: -Fr?J7 ao = ff(t)dt =f dt= =. For n : a = it = ff(t)cos(nt)dt = it ir cos(nt)clt = =. sin (nt) = r7 J f(t) sin (nt)dt f sin (nt)dt = cos(nt) = if o flit ( (cos(nit) ) J noaa neven So, the Fourier series is: flit- -+- it (sill(t) + sin(3t) + sin(5t) +...). 5

6 I I The value of a period ii- function f(t) in one full period is given below. Sketch several periods of its graph and find its Fourier series., <t<n f(t)=t The Fourier coefficients are: a = fir tdt = t dt t 3 n = in o 3 For in > : a = t cos (nt)dt ( t sin (nt) t cos (nt) sin (nt) \ + ( in cos (nm) 3 4(_)Th } \ b = t sin (nt)dt = (Odd function.) So, the Fourier series is: / cos(ri) cos(3t) cos(4t) 4icos(t) \\

7 (sin Section 9. - gence General Fourier Series and Conver The values of a periodic function f(t) in one full period are given below; at each discontinuity the value of f(t) is that given by the average value condition. Sketch the graph of f and find its Fourier series. f(t)={ 3<t<O O<t<3 Solution The function f(t) is odd, so all the cosine terms will be. The sine terms are: bfl=-i 4f3 3 3 nt = sin dt n7rt \ n7rt f(t)sin()dt=j sin)dt \\3) nir The Fourier series for f(t) is: ( )) = { 8 nodd even 8/. /t I n. + sin /3irt \ )+sin()+). 7

8 cos The values of a periodic function f(t) in one full period are given below; at each discontinuity the value of f(t) is that given by the average value condition. Sketch the graph of f and find its Fourier series. f(t)t, <t< Solution - The function f(t) is even, so all the sine terms in its Fourier series are zero. As for the cosine terms we have: t fin a= Itdt= = I I=. i i 3 3 3J 3 a = f t cos (nt)dt = f t cos (nt)dt ft sin (riirt) t cos (nlft) sin (nirt) N 4 = ( + fir fit Jo fit 4(_)Th - 3 J = cos (nn) f ff So, the Fourier series for the function f(t) is: 4 / cos (irt) cos (3irt) (irt) + 9 8

9 (a) - Suppose that f is a function of period Ti with f(t) = t for o <t < ir. Show that cos nt +4 47r: n= n= sin nt f(t)rrrand sketch the graph of f, indicating the value at each disconti nuity. (b) - Deduce the series summations Ti = 6 m= and n+ 7= p from the Fourier series in part (a). Solution - (a) - Sketch of the graph: I -ZTI The Fourier coefficients are: Yir 9

10 π a = π π t dt = π π ( ) t 3 π a n = t cos (nt)dt = π ( t sin (nt) t cos (nt) + n n 9π cos (nπ) = = 4 πn n. π 3 = 8π 3. sin (nt) n 3 ) π b n = t sin (nt)dt = π ( ) t cos (nt) t sin (nt) sin (nt) π + π n n n 3 4π cos (nπ) = = 4π n n. So, the Fourier series for the function f(t) is: a + a n cos (nt) + n= = 4π n= cos (nt) n b n sin (nt) n= 4π n= sin (nt). n (b) - The averaged value of our function at t = is 4π + So, = π. π = 4π n n= π 6 = n. n= At t = π we get:

11 π = 4π n= π 3 = 4 n= π = π = n= n= ( ) n n ( ) n n ( ) n n ( ) n+ n.

12 _i (tcos (o (a) - Supose that f is a funciton of period with f(t) = t for <t <. Show that f(t) = sin n_irt and sketch the graph of f, indicating the value at each disconti nuity. (b) - Substitute an appropriate value of t to deduce Leibniz s series it 4 Solution - Sketch of the graph: (a) - -L[ The Fourier coefficients will be: ao=i t tdt= =. t sin (nirt) cos (flirt) aflj tcos(nirt)= flit 7_iit (sin(nir) cos(nir) \ + _i_i) = =. ititiit ) L G = It5ffi = (nirt) sin (nt)n nit + )L nit

13 So, the Fourier series for f(t) is: a + a n cos (nt) + n= = π n= b n sin (nt) n= sin (nπt). n (b) - If we plug in t = we get: = sin ( ) nπ π n n= sin ( ) nπ = π n n= π 4 = sin ( ) nπ = n n= 3

14 j 6irt 7r 3irt 67t Derive the Fourier series given below, and graph the period ir function to which the series converges. cosnt 3t 6irt+7r ( < t < ir) Solution - If we take the even extension of f(t) = must have b,, = for all n, and 3t + 7 we = f (3t + 7) dt = : ( 87 ± 4ir (t3 )= r tn As for the a terms, these are (after a nasty integral, which you can do using Mathematica) = f 3t 6irt+ir cos (nt)dt = 7J \ j fl \ So, the Fourier series for the even extension is: Graph: : n= cos (nt) 3

15 Section Fourier Sine and Cosine Series For the given function f(t) defined on the given interval find the Fourier cosine and sine series of f and sketch the graphs of the two extensions of f to which these two series converge. f(t) =, O<t<ir. Solution - Even extension Cosine series: a= / 7t J dt=, a7,. (See problem 9..) So, the cosine series is: a Wow!

16 (sin(t) Odd extension: - The sine series has coefficients: J 7 f sin(nt)dt 4 n7r = cos(nt) fl7t n odd n even So, the sine series is: 4/ + Sill (3t) + sin (5t) + ). 5

17 cos For the given function f(t) defined on the given interval find the Fourier cosine and sine series of f and sketch the graphs of the two extensions of f to which these two series converge. f(t)rr <t< <t< <t<3 Solution - Even extension: (9 The cosine series will have the coefficients: f = f(t)dt= nirt an=f f(t)cos()dtf / \ 9 in irt = Shill I Sin fl7 flhi i\ ( rurt 3 /nir 3) Sfl -- ( nir /3 TT So, the cosine series is: n 6k.6k+,6k+3.6k+5 n 6k + n = 6k +4 /( (irt) (4irt) cos 3 ir 3 4 /8rrt +COS\

18 I sin - cos The odd extension is: C) The sine series will have coefficients I 7 = - cos fl7 flq J Jo The sine series is: ) ={ nir = - f(t)sin ()dt= f /n7rtn sin dt 3J 7ritI / nn fir = = Ti 6k, 6k +, 6k + 4 n 6k+,6k+5 TL7 n = 6k +3 ()) /. /irt sin I Ti /37rt I 3 \\3 /5irt 7

19 For the given function f(t) defined on the given interval find the Fourier cosine and sine series of f and sketch the graphs of the two extensions of f to which these two series converge. f(t)=t t, O<t< Solution - The even extension is: z z j The cosine series has the coefficients: = f(t -t)dt = an=f(t_t)cos(lt)dt= (-) { = Veven The integral above can be evaluated with a good graphing calculator, a computer, Wolfram alpha, integrals.com, or by hand with a cou pie integrations by parts. I just cut to the chase. The corresponding Fourier cosine series will be: 4 /cos (t) cos (47rt)

20 The odd extension is: - - The sine series has the coefficients: n odd b [(t t) sin (nnt)dt = J neveri The corresponding Fourier sine series will be: 8 / sin (3nt) sill (5nt) sin(t)

21 The odd extension is: The sine series has the coefficients: b n = (t t ) sin (nπt)dt = { 8 n 3 π 3 n odd n even The corresponding Fourier sine series will be: ( ) 8 sin (3πt) sin (5πt) sin (πt) π 3 7 5

22 Find a formal Fourier series solution to the endpoint value problem x + x = t x() = x() =. Solution - We note that sin (nπt) has value at t = and t = for all n, so we ll want to use the sine series. x(t) = a n sin (nπt) n= x (t) = a n ( n π ) sin (nπt). n= The Fourier coefficients for the sine series of t are: b n = ( t cos (nπt) t sin (nπt)dt = + nπ ) sin (nπt) n π = ( )n+. nπ Plugging these into our differential equation we get: a n ( n π ) = ( )n+ nπ a n = ( )n+ nπ( n π ). So, x(t) = π n= ( ) n sin (nπt) n(n π ).

23 Substitute t = π/ and t = π in the series 4 t4 = π t n= ( ) n n 4 cosnt + ( ) n n= n 4, π < t < π, to obtain the summations n= n 4 = π4 9, ( ) n+ n 4 n= = 7π4 7, and = π4 96. For t = π we get: 4 ( ) ( ) π 4 = π π ( ) n nπ cos( 6 4 n 4 n= ) + n= ( ) n n 4. Now, ( nπ cos ) = n = 4k n = 4k + n = 4k +, 4k + 3 So,