Advanced Calculus Single Variable

8.3.1 The Functions, sin,cos,exp

With this material on power series, it becomes possible to give an understandable treatment of the exponential function exp and the circular functions, sin and cos.

Now that sin

and cos have been defined, the properties of these functions must be considered. First, here is a fundamental lemma.

Proof: Multiply the equation by y^′ and use the chain rule to write

( ) d- 1(y′)2 + 1y2 = 0. dt 2 2

Then by Corollary 7.8.5

² + y² equals a constant. From the initial conditions, y = y^′ = 0, the constant can only be 0. ■

Proof: That sin^′

= cos and cos^′ = −sin follows right away from differentiating the power series term by term using Theorem 8.2.1. It follows from the series that cos = 1 and sin = 0 and sin = −sin while cos = cos because the series for sin only involves odd powers of x while the series for cos only involves even powers.

For x ∈ ℝ, let f

= cos² + sin², it follows from what was just discussed that f = 1. Also from the chain rule,

f′(x) = 2cos(x)(− sin(x)) + 2sin(x)cos(x) = 0

and so by Corollary 7.8.5, f

is constant for all x ∈ ℝ. But f = 1 so the constant can only be 1. Thus

cos2 (x) +sin2(x) = 1

as claimed.

It only remains to verify the identities. Consider 8.7 first. Fixing y and considering both sides as a function of x, it follows from the above that both sides of the identity satisfy the initial value problem

y′′ + y = 0,y (0) = sin(y),y′(0) = cos(y)

Therefore, the difference satisfies the initial value problem of Lemma 8.3.3. Therefore, by this lemma, the difference equals 0. The next identity is handled similarly. This proves the theorem.

Proof: From the definition of sin

given above,

Now whenever < 1. Thus

∞ 2k lim ∑ (− 1)k-(ax)--= 0 x→0 k=1 (2k +1)!

and so

sin(ax) a- lix→m0 bx = b.

The other limit can be handled similarly.

It is possible to verify the functions are periodic.

Proof:To prove this, note that cos

= 1 and so if it is false, it must be the case that cos > 0 for all positive x since otherwise, it would follow from the intermediate value theorem there would exist a point, x where cosx = 0. Assume cos > 0 for all x. Then by Corollary 7.8.6 it would follow that t → sint is a strictly increasing function on . Also note that sin = 0 and so sin > 0 for all x > 0. This is because, by the mean value theorem there exists t ∈ such that

sin (x ) = sin (x) − sin(0) = (cos(t))(x− 0) > 0.

By 8.6,

≤ 1 for f = cos and sin. Let 0 < x < y. Then from the mean value theorem,

− cos (y)− (− cos(x )) = sin(t)(y − x)

for some t ∈

. Since t → sin is increasing, it follows

− cos(y)− (− cos(x)) = sin(t)(y − x) ≥ sin (x)(y − x).

This contradicts the inequality

≤ 1 for all y because the right side is unbounded as y →∞. ■

Proof: Define a number, π such that

π- 2 ≡ inf{x : x > 0 and cos (x) = 0}

Then

> 0 because cos = 1 and cos is continuous. On cos is positive and so it follows sin is increasing on this interval. Therefore, from 8.6, sin = 1. Now from Theorem 8.3.4,

( ) ( ) cos(π) = cos π-+ π = − sin2 π- = − 1, sin(π) = 0 2 2 2

Using Theorem 8.3.4 again,

cos(2π) = cos2(π ) = 1 = cos(0),

and so sin

= 0. From Theorem 8.3.4,

cos(x +2π ) = cos(x )cos(2π)− sin(x)sin (2π) = cos (x)

Thus cos is periodic of period 2π. By Theorem 8.3.4,

sin(x +2π ) = sin(x)cos(2π)+ cos(x)sin (2π) = sin(x)

Using 8.6, it follows sin is also periodic of period 2π. This proves the theorem.

Note that 2π is the smallest period for these functions. This can be seen by observing that the above theorem and proof imply that cos is positive on

( ) (0, π-), 3π,2π 2 2

and negative on

and that similar observations on sin are valid. Also, by considering where these functions are equal to 0, 1, and -1 along with where they are positive and negative, it follows that whenever a² + b² = 1, there exists a unique t ∈ [0,2π) such that cos = a and sin = b. For example, if a and b are both positive, then since cos is continuous and strictly decreases from 1 to 0 on , it follows there exists a unique t ∈ such that cos = a. Since b > 0 and sin is positive on , it follows sin = b. No other value of t in [0,2π) will work since only on are both cos and sin positive. If a > 0 and b < 0 similar reasoning will show there exists a unique t ∈ [0,2π) with cos = a and sin = b and in this case, t ∈. Other cases are similar and are left to the reader. Thus, every point on the unit circle is of the form for a unique t ∈ [0, 2π).

This shows the unit circle is a smooth curve, however this notion will not be considered here.

Proof: Let y

≡ sin − sin. Then from what has been shown above, y^′ = y = 0. It is also clear from the above that y^′′ + y = 0. Therefore, from Lemma 8.3.3 y = 0. Differentiating the identity just obtained yields the second identity. ■

Are these the same as the circular functions you studied very sloppily in calculus and trigonometry? They are.

If sin

defined above and sin studied in a beginning calculus class both satisfy the initial value problem

y′′ +y = 0,y(0) = 0,y′(0) = 1

then they must be the same. However, if you remember anything from calculus you will realize sin

used there does satisfy the above initial value problem. If you don’t remember anything from calculus, then it does not matter about harmonizing the functions. Just use the definition given above in terms of a power series. Similar considerations apply to cos.

Of course all the other trig. functions are defined as earlier. Thus

tan x = sinx-,cotx ≡ cosx ,secx ≡ --1-,cscx ≡ -1--. cosx sinx cosx sinx

Using the techniques of differentiation, you can find the derivatives of all these.

Now it is time to consider the exponential function exp

defined above. To do this, it is convenient to have the following uniqueness theorem.

Proof: The function exp has been defined above in terms of a power series. From this power series and Theorem 8.2.1 it follows that exp solves the above initial value problem. Multiply both sides of the differential equation by exp

. Then using the chain rule and product rule,

-d-(exp(− x)y (x)) = 0 dx

and so exp

y = C, a constant. The constant can only be 0 because of the initial condition. Therefore,

exp (− x)y(x) = 0

for all x.

Now I claim exp

and exp are never equal to 0. This is because by the chain rule, abusing notation slightly,

′ (exp (− x)exp(x)) = − exp (− x)exp(x)+ exp(− x)exp(x) = 0

and so

exp(− x)exp(x) = C

a constant. However, this constant can only be 1 because this is what it is when x = 0, a fact which follows right away from the definition in terms of power series. ■

Proof: To begin with consider 8.10. Fixing y it follows from the chain rule and the definition using power series that

x → exp(x + y)− exp (x)exp(y)

satisfies the initial value problem of Lemma 8.3.9 and so it is 0. This shows 8.10.

8.9 has already been noted. It comes directly from the definition and was proved in Lemma 8.3.9. The claim that exp

> 0 was also established in the proof of this lemma.

Now from the power series, it is obvious that exp

> 0 if x > 0 and by Lemma 8.3.9, exp⁻¹ = exp, so it follows exp is also positive. Since exp > ∑ _k=0², it is clear lim_x→∞exp = ∞ and it follows from this that lim_x→−∞exp = 0.

It only remains to verify 4. Let y ∈

. From the earlier properties, there exist x₁ such that  exp < y and x₂ such that  exp > y. Then by the intermediate value theorem, there exists x ∈ such that exp = y. Thus exp maps onto . It only remains to verify exp is one to one. Suppose then that x₁ < x₂. By the mean value theorem, there exists x ∈ such that

exp(x)(x2 − x1) = exp′(x)(x2 − x1) = exp (x2)− exp (x1).

Since exp

> 0, it follows exp≠exp. ■