2 cos x+y 2 cosx+ cosy= 2cos x+y 2 cos x y 2 cosx cosy= 2sin x+y 2 sin x y 2 The Law of Sines sinA a = sinB b = sinC c Suppose you are given two sides, a;band the
sin(x) = 0, lim x→0− (1 − cos(x)) = 0 The left and the right limits are equal, thus lim x→0 sin(x) = 0, lim x→0 (1 − cos(x)) = 0 or, lim x→0 sin(x) = 0, lim x→0 cos(x) = 1 – Typeset by FoilTEX – 8
satisfying x2 + y2 = 1, we have cos2 + sin2 = 1 Other trignometric identities re ect a much less obvious property of the cosine and sine functions, their behavior under addition of angles This is given by the following two formulas, which are not at all obvious cos( 1 + 2) =cos 1 cos 2 sin 1 sin 2 sin( 1 + 2) =sin 1 cos 2 + cos 1 sin 2 (1)
cos2 = cos2 sin2 = 2cos2 1 = 1 2sin2 (10) sin2 = 2sin cos (11) tan2 = 2tan 1 tan2 (12) Note that you can get (5) from (4) by replacing B with B, and using the fact
2 cos(A B) + 1 2 cos(A+ B) sinAcosB = 1 2 sin(A B) + 1 2 sin(A+ B) sinAsinB = 1 2 cos(A B) 1 2 cos(A+ B): These identities allow us to transform any product of trigonometric functions into a sum By applying them repeatedly, we can remove all of the multiplications from a trigonometric polynomial, resulting in a Fourier sum
2 4 ˇ X1 n=0 cos(0) (2n + 1)2 = ˇ 2 4 ˇ X1 n=0 1 (2n + 1)2 Solving for the series gives the result Remark: In Calculus II you learned that this series converges, but were unable to obtain its exact value Daileda Fourier Coe cients
Multiply both sides of (2) by 2/π: Sine coefficients S(−x)=−S(x) b k = 2 π π 0 S(x)sinkxdx= 1 π π −π S(x)sinkxdx (6) Notice that S(x)sinkx is even (equal integrals from −π to 0 and from 0 to π) I will go immediately to the most important example of a Fourier sine series S(x) is an odd square wavewith SW(x)=1for0
1 x = 1 + x + x2 + x3 + x4 + ::: note this is the geometric series just think of x as r = X1 n=0 xn x 2( 1;1) ex = 1 + x + x2 2 + x3 3 + x4 4 + ::: so: e = 1 + 1 + 1 2 + 3 + 1 4 + ::: e(17x) = P 1 n=0 (17 x)n = X1 n=0 17n n n = X1 n=0 xn n x 2R cosx = 1 x2 2 + x4 4 x6 6 + x8 8::: note y = cosx is an even function (i e , cos( x
Example Suppose that a curve is given by the equation x2 + y3 = 2x2y Verify that the point (x;y) = (1;1) lies on the curve Assume that the curve is given by a function y= y(x) for xnear 1 and approximate y(1:2) Solution To verify that (x;y) = (1;1) lies on the curve, we need to know that 13 + 12 = 2 12 1 which is true
[PDF] cos x 2
[PDF] sin x pi 2
[PDF] cos x 1
[PDF] cos(a+b) démonstration exponentielle
[PDF] cos a+b démonstration géométrique
[PDF] tan 2a démonstration
[PDF] cos(a-b) démonstration
[PDF] cos(a+b)=cos(a) cos(b)-sin(a) sin(b) démonstration
[PDF] cos a + cos b
[PDF] cos(2x)
[PDF] les misérables tome 3 pdf
[PDF] portrait thénardier
[PDF] cosette chez les thénardier
[PDF] résumé cosette une enfance malheureuse
[PDF] les misérables cosette résumé
CHAPTER 4
FOURIER SERIES AND INTEGRALS
4.1 FOURIER SERIES FOR PERIODIC FUNCTIONS
This section explains three Fourier series:sines, cosines, and exponentialse ikx Square waves (1 or 0 or-1) are great examples, with delta functions in the derivative. We look at a spike, a step function, and a ramp"and smoother functions too. Start with sinx.Ithasperiod2since sin(x+2)=sinx. It is an odd function since sin(-x)=-sinx, and it vanishes atx=0andx=. Every function sinnx
has those three properties, and Fourier looked atinnite combinations of the sines:Fourier sine seriesS(x)=b
1 sinx+b 2 sin2x+b 3 sin3x+···= n=1 b n sinnx(1)
If the numbersb1
,b 2 ,...drop offi quickly enough (we are foreshadowing the im- portance of the decay rate) then the sumS(x) will inherit all three properties:
PeriodicS(x+2)=S(x)OddS(-x)=-S(x)S(0) =S()=0
200 years ago, Fourier startled the mathematicians in France by suggesting thatany
functionS(x) with those properties could be expressed as an infinite series of sines. This idea started an enormous development of Fourier series. Our first step is to compute fromS(x)thenumberbk that multiplies sinkx.
SupposeS
(x)=bn sinnx.Multiply both sides bysinkx.Integrate from0to: 0
S(x)sinkxdx=
0 b 1 sinxsinkx dx+···+ 0 b k sinkxsinkxdx+···(2) On the right side, all integrals are zero except the highlighted one withn=k. This property of "orthogonality" will dominate the whole chapter. The sines make 90
angles in function space, when their inner products are integrals from 0 to:Orthogonality 0 sinnxsinkxdx=0 ifn?=k.(3) 317
318Chapter 4 Fourier Series and Integrals
Zero comes quickly if we integrate
cosmxdx= sinmx m 0 =00. So we use this:
Product of sinessinnxsinkx=1
2cos(nk)x12cos(n+k)x.(4)
Integrating cosmxwithm=nkandm=n+kproves orthogonality of the sines. The exception is whenn=k. Then we are integrating (sinkx) cos2kx: sinkxsinkxdx= 1
2dx
1
2cos2kxdx=π2.(5)
The highlighted term in equation (2) isb
k
Γ/2. Multiply both sides of (2) by 2/π:
Sine coeffcients
S(x)=S(x)
b k =2
S(x)sinkxdx=1
S(x)sinkxdx.(6)
Notice thatS(x)sinkxiseven(equal integrals fromπto 0 and from 0 toπ). I will go immediately to the most important example of a Fourier sine series.S(x) is anodd square wavewithSW(x)=1for0
π0π2π Figure 4.1: The odd square wave withSW(x+2π)=SW(x)={1or0or1}. Example 1Find the Fourier sine coefficientsb
k of the square waveSW(x). SolutionFork=1,2,...use the first formula(6)withS(x)=1between0andπ: b k =2 sinkxdx=2 coskxkffl
0 =2 21,02,23,04,25,06,..."
(7) The even-numbered coecientsb
?k are all zero because cos2kπ= cos0 = 1. The odd-numbered coecientsb k =4/πkdecrease at the rate 1/k. We will see that same 1/kdecay rate for all functions formed fromsmooth pieces and jumps.
Put those coecients 4/πkand zero into the Fourier sine series forSW(x): Square waveSW(x)=4
(8) Figure 4.2 graphs this sum after one term, then two terms, and then five terms. You can see the all-importantGibbs phenomenonappearing as these "partial sums" 4.1 Fourier Series for Periodic Functions319
include more terms. Away from the jumps, we safely approachSW(x)=1or1. Atx=Γ/2, the series gives a beautiful alternating formula for the numberΓ: 1= 4 1113+1517+···?
so that≈=4?1113+1517+···? .(9) The Gibbs phenomenon is the overshoot that moves closer and closer to the jumps. Its height approaches 1.18...and it does not decrease with more terms of the series! Overshoot is the one greatest obstacle to calculation of all discontinuous functions (like shock waves in fluid flow). We try hard to avoid Gibbs but sometimes we can"t. xxΓΓDashed 4 Γsinx1Solid curve
4 sinx1+sin3x3? 5terms:4Γ?
sinx1+···+sin9x9? overshoot?SW=1 2 Figure 4.2:Gibbs phenomenon: Partial sums?
N 1 b n sinnxovershoot near jumps. Fourier Coefficients are Best
Let me look again at the first termb
1 sinx=(4/Γ)sinx.Thisistheclosest possible approximationto the square waveSW, by any multiple of sinx(closest in the least squares sense). To see this optimal property of the Fourier coecients, minimize the error over allb 1 The error is
0 (SWb 1 sinx) 2 dxTheb 1 derivative is2? 0 (SWb 1 sinx)sinxdx. The integral of sin
2 xisΓ/2. So the derivative is zero whenb 1 =(2/Γ)? 0 S(x)sinxdx.
This is exactly equation (6) for the Fourier coecient. Eachb k sinkxis as close as possible toSW(x). We can find the coecientsb k one at a time,because the sines are orthogonal. The square wave hasb 2 = 0 because all other multiples of sin2xincrease the error. Term by term, we are "projecting the function onto each axis sinkx." Fourier Cosine Series
The cosine series applies toeven functionswithC(x)=C(x): Cosine seriesC(x)=a
0 +a 1 cosx+a 2 cos2x+···=a 0 n=1 a n cosnx.(10) 320Chapter 4 Fourier Series and Integrals
Every cosine has period 2Γ. Figure 4.3 shows two even functions, therepeating rampRR(x)andtheup-down trainUD(x) of delta functions. That sawtooth rampRRis the integral of the square wave. The delta functions inUDgive the derivative of the square wave. (For sines, the integral and derivative are cosines.) RRandUDwill be valuable examples, one smoother thanSW, one less smooth. First we find formulas for the cosine coecientsa
anda k . The constant terma is theaverage valueof the functionC(x): a 0 =Averagea =1 ffi C(x)dx=1
2Γ ffi ffi
C(x)dx.(11)
I just integrated every term in the cosine series (10) from 0 toΓ.Ontherightside, the integral ofa isa Γ(divide both sides byΓ). All other integrals are zero: ffi cosnxdx=sinnx n ffi =00=0.(12) In words, the constant function 1 is orthogonal to cosnxover the interval [0,Γ]. The other cosine coecientsa
k come from theorthogonality of cosines.Aswith sines, we multiply both sides of (10) by coskxand integrate from 0 toΓ: ffi C(x)coskxdx=
ffi a coskxdx+ ffi a cosxcoskxdx+··+ 0 a k (coskx) 2 dx+·· You know what is coming. On the right side, only the highlighted term can be nonzero. Problem 4.1.1 proves this by an identity for cosnxcoskx"now (4) has a plus sign. The bold nonzero term isa k σ/2and we multiply both sides by 2/Γ:
Cosine coe?cients
C(x)=C(x)
a k =2 ffi C(x)coskxdx=1
ffi ffi
C(x)coskxdx.(13)
Again the integral over a full period fromΓtoΓ(also 0 to 2Γ) is just doubled. xΓ0Γ2ΓRR(x)=|x| Repeating RampRR(x)
Integral of Square Wave
xΓ0Γ2Γ 2Σ(x+Γ)2Σ(x)
2Σ(xΓ)2Σ(x2Γ)
Up-downUD(x)
Figure 4.3: The repeating rampRRand the up-downUD(periodic spikes) are even. The derivative ofRRis the odd square waveSW.The derivative ofSWisUD. 4.1 Fourier Series for Periodic Functions321
Example 2Find the cosine coefficients of the rampRR(x)and the up-downUD(x). SolutionThe simplest way is to start with the sine series for the square wave: SW(x)=4
sinx1+sin3x3+sin5x5+sin7x7+···? Take the derivative of every term to produce cosines in the up-down delta function: Up-down seriesUD(x)=4
Those coefficients don"t decay at all. The terms in the series don"t approach zero, so officially the series cannot converge. Nevertheless it is somehow correct and important. Unofficially this sum of cosines has all1"s atx=0and all-1"s atx=Γ.Then+∞ and-∞are consistent with2Σ(x)and-2Σ(x-Γ). The true way to recognizeΣ(x)isquotesdbs_dbs32.pdfusesText_38
Trigonometric Identities - Miami