Kenneth Kuttler

166 CHAPTER 8. POWER SERIES

Proof: blogb(x) ≡ exp(ln(b) logb (x)) = exp(

ln(b) ln(x)ln(b)

)= exp(ln(x)) = x and also

logb (bx) = ln(bx)

ln(b) = x ln(b)ln(b) = x. This proves 1.

Now consider 2.

logb (xy) =ln(xy)ln(b)

=ln(x)ln(b)

+ln(y)ln(b)

= logb (x)+ logb (y) .

Finally, logb (xα) = ln(xα )

ln(b) = αln(x)ln(b) = α logb (x) .

8.5 The Complex ExponentialWhat does eix mean? Here i2 =−1. Recall the complex numbers are of the form a+ ib andare identified as points in the plane. For f (x) = eix, you would want

f ′′ (x) = i2 f (x) =− f (x)

so f ′′ (x)+ f (x) = 0. Also, you would want f (0) = e0 = 1, f ′ (0) = ie0 = i. One solutionto these conditions is f (x) = cos(x)+ isin(x) . Is it the only solution? Suppose g(x) isanother solution. Consider u(x) = f (x)−g(x) . Then it follows

u′′ (x)+u(x) = 0, u(0) = 0 = u′ (0) .

Thus both Reu and Imu solve the differential equation and 0 initial condition. By Lemma8.3.3 both Reu and Imu are equal to 0. Thus the above is the only solution. Recall byDe’Moivre’s theorem

(cosx+ isinx)n = cos(nx)+ isin(nx)

for any integer n and so(eix)n

= einx.If you have a complex number x+ iy, you can write it as

√x2 + y2

(x√

x2 + y2+ i

y√x2 + y2

)

and then note that(

x√x2+y2

, y√x2+y2

)is a point on the unit circle. Thus there is θ such that

this ordered pair is (cosθ ,sinθ) and so if you let r =√

x2 + y2, the distance to the origin,the complex number can be written in the form r (cosθ + isinθ) . From the above, this isof the form reiθ . This is called the polar form of a complex number. You should verify thatwith this convention, reiθ r̂eiθ̂ = rr̂ei(θ+θ̂). This reduces to using the trig identities for thecosine and sine of the sum of two angles.

In particular, this shows how to parametrize a circle in C centered at 0 which has radiusr. It is just γ (t) = reit where t ∈ [0,2π]. By this is meant that as t moves from 0 to 2π, thepoint γ (t) is on the circle of radius r and moves in the counter clockwise direction over thecircle.

8.6 The Binomial TheoremThe following is a very important example known as the binomial series. It was discoveredby Newton.

166 CHAPTER 8. POWER SERIESProof: b'°2“) = exp (In(b) log, (x)) = exp (in (b) it) = exp(In(x)) =x and alsoIn(b)log, (b*) = ay = a = x. This proves 1.Now consider 2.loss (9) = Taps = mrBy t in(py — BHO) + 10g (9).Finally, log, (x) =8.5 The Complex ExponentialWhat does e”* mean? Here i” = —1. Recall the complex numbers are of the form a+ ib andare identified as points in the plane. For f (x) = e™, you would wantf" (x) =P f (*~) =f (x)so f” (x) + f (x) =0. Also, you would want f (0) = e° = 1, f’ (0) = ie® =i. One solutionto these conditions is f (x) = cos (x) +isin(x). Is it the only solution? Suppose g(x) isanother solution. Consider u (x) = f (x) — g(x). Then it followsu" (x) +u(x) =0, u(0) =0=u' (0).Thus both Reu and Imu solve the differential equation and 0 initial condition. By Lemma8.3.3 both Reu and Imu are equal to 0. Thus the above is the only solution. Recall byDe’ Moivre’s theorem(cosx +isinx)” = cos (nx) +isin (nx)for any integer n and so (e” \" = ink,If you have a complex number x + iy, you can write it asatestVery Veryand then note that ( Jane as) is a point on the unit circle. Thus there is @ such thatthis ordered pair is (cos @,sin@) and so if you let r = \/x? +”, the distance to the origin,the complex number can be written in the form r(cos @ +isin@). From the above, this isof the form re’®. This is called the polar form of a complex number. You should verify thatwith this convention, re‘? a = reei(9+8)cosine and sine of the sum of two angles.In particular, this shows how to parametrize a circle in C centered at 0 which has radiusr. It is just y(t) = re where t € [0,27]. By this is meant that as f moves from 0 to 27, thepoint y(t) is on the circle of radius r and moves in the counter clockwise direction over thecircle.. This reduces to using the trig identities for the8.6 The Binomial TheoremThe following is a very important example known as the binomial series. It was discoveredby Newton.