Math 508: Advanced Analysis

Homework 8

Lawrence Tyler Rush

<me@tylerlogic.com>

November 8, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework08

1 Show that cos x is differentiable at all x.

For h > 0 we have

lim cos(x+-h)--cosx = lim cosx-cos-h--sin-xsin-h--cosx h→0 h h→0 h cos-h--1- sinh = cosx lhim→0 h - sinx lih→m0 h = (cosx)(0)- (sinx)(1) = - sinx

where we make use of the trigonometric identity cos(x + y) = cosxcosy - sinxsiny and the fact that

cosh--1- lhim→0 h = 0

and

sinh lim ----= 1 h→0 h

2 Derivatives of matrices

Let A(t) be an n × n matrix whoose elements depend smoothly on t and that is invertible at t₀.

(a) Compute the derivative of A²(t) in terms of A and A′

The derivative of A²(t) is the derivative with respect to t of each of its elements as functions of t. So

d-( 2 ) d-∑ ∑ ( ′ ′) ′ ′ dt A (t) ij = dt naikakj = n aikakj + aikakj = (A A + AA )ij k=1 k=1

so that we have ddt A²(t) = A′A + AA′. Note the multiplication of A and A′ may not commute.

(b) Show A(t) is invertible for all t near t₀

We know that a matrix is invertible if and only if it has nonzero determinent. Since the determinent of A is simply the sum and product of entries of A and the entries of A depend smoothly on t, then det(A(t)) also depends smoothly on t. Since A(t₀) is invertible, then det(A(t₀))≠0, which, along with the previous sentence, implies that there is a neighborhood of t₀ for which all t in that neighborhood have det(A(t))≠0.

(c) Find a formula for the derivative of A^-1(t) at t₀

For h > 0 we have.

lim A--1(t0 +-h)--A--1(t0) = lim (A-1(t0)A-(t0))A--1(t0-+h-)--A-1(t0)(A(t0 +-h)A-1(t0 +-h)) h→0 h h→0 h -1 A(t0)---A(t0-+-h) -1 = A (t0) lhim→0 h A (t0 + h) -1 ( A(t0 +-h)--A-(t0) -1 ) = A (t0) - hli→m0 h A (t0 + h) ( ) = A -1(t0)(- A ′(t0)) lim A -1(t0 + h) h→0 = - A-1(t0)A ′(t0)A -1(t0)

so that

′ = -A^-1A′A^-1

(d) Find a formula for the derivative of A^-2(t)

By the first part of this problem we have

d-( - 2) ( -1)′ - 1 -1 ( -1)′ dt A = A A + A A

which, through use of the previous part of the problem, leads to

d ( ) ( ) ( ) ( ) -- A -2 = - A-1A ′A -1 A-1 + A-1 - A-1A ′A -1 = - A-1A ′A -2 + A- 2A ′A -1 dt

3

(a) Find the unique solution v(x) such that v′ = v and v(0) = c for constant c.

Let v(x) be such that v′ = v and v(0) = c. Define u(x) =

v(x). Then

u ′(x) = 1 v′(x) = 1 v(x) = u(x) c c

(3.1)

and

1 1 u(0) =-v(0) = -c = 1 c c

(3.2)

Thus, u(x) must be e^x since e^x is the unique function with the properties in equations 3.1 and 3.2. But then we have e^x = u(x) = 1
c v(x) implying v(x) = ce^x.

(b) Prove that e^x+a = e^ae^x for all real a and x

Define u(x) = e^x+a - e^ae^x for real a. Then we have

u(0) = ea - ea = 0

and

′ x+a a x u (x ) = e - e e = u(x)

by the previous part of this problem, u(x) = 0e^x = 0. Hence e^x+a = e^ae^x.

(c) For constant γ show v′- γv ≤ 0 implies v(x) ≤ v(0)e^γx for x ≥ 0

Let v′- γv ≤ 0 and define g(x) = e^-γxv(x). Then we have

g′ = - γe- γxv + v′e-γx = e-γx(v′ - γv) ≤ 0

which implies that g is always decreasing so that in particular g(x) ≤ g(0) for all x ≥ 0. That is to say that

- γxg(x) ≤ g(-0γ)0 e v(x) ≤ e v(0) e- γxv(x) ≤ v(0) v(x) ≤ v(0)eγx

for all x ≥ 0.

4 Show that a continuous f : [a,b] → ℝ can be approximated by a piecewise linear function g : [a,b] → ℝ

Let ε > 0 be given. Because f is continuous and [a,b] is compact, then f is uniformly continuous, and we can thus find a δ > 0 such that

|x- y| < δ ⇒ |f(x)- f(y)| < ε

(4.3)

We divide [a,b] up into chuncks of size less than δ. Let n be an integer such that nδ > b so that δ > . Hence, each of [a,a + δ),[a + δ,a + 2δ),…,[a + (n - 1)δ,b), where δ = b-na- , are intervals of size less than δ. Define M_i = supf(x) and m_i = inf f(x) for each i ∈{0,1, ⋅⋅⋅ ,n - 1} and x ∈ [a + iδ,a + (i + 1)δ) so that in particular

| | ||f(x)- Mi---mi-||< ε | 2 |

(4.4)

for any x ∈ [a + iδ,a + (i + 1)δ) by equation 4.3. Further define a set of functions φ_i : [a,b] →{0,1} by

{ - - φi(x) = 1 x ∈ [a+ iδ,a+ (i+ 1)δ) 0 otherwise

so that we may define g : [a,b] → ℝ by

n-1 g(x) = δb(x)f(b)+ ∑ φi(x)Mi---mi- i=0 2

where δ_b is the Kronecker function. Thus for any x ∈ [a + iδ,a + (i + 1)δ) we have

| | |f(x)- g(x)| = ||f(x) - Mi---mi||< ε | 2 |

by equation 4.4 and for x = b we have |f(x) - g(x)| = |f(b) - f(b)| = 0 < ε as desired.

5

Let f : ℝ → ℝ be a smooth function.

(a) If f′(1) = f′′(1) = f′′′(1) = 0 and f′′′′(1) > 0, show f has a local minimum at x = 1.

Let f′(1) = f′′(1) = f′′′(1) = 0 and f′′′′(1) > 0. Then there is an ε > 0 such that f′′′′(x) > 0 for x ∈ (1 -ε,1 + ε). So if 1 < x < 1 + ε then repeated applications of The Fundamental Theorem of Calculus tells us

∫ x ′ f (x) - f (1) = 1 f (y) dy ∫ x = f ′(y)- f′(1) dy ∫1x∫ y = f′′(z) dzdy ∫1 ∫ 1 = x y f′′(z) - f′′(1) dzdy 1 1 ∫ x∫ y ∫ z ′′′ = 1 1 1 f (w) dwdzdy ∫ x∫ y ∫ z = f′′′(w) - f′′′(1) dwdzdy ∫1x∫ 1y ∫1z∫ w = f′′′′(s) dsdwdzdy 1 1 1 1 > 0

which implies f(x) > f(1). Furthermore, for any x with 1 - ε < x < 1 we have again by repeated applications of The Fundamental theorem of Calculus

∫ 1 f(x )- f(1) = - f ′(y) dy ∫x = - 1f ′(y)- f′(1) dy x ∫ 1 ∫ 1 = f′′(z) dzdy ∫x 1 ∫y 1 = f′′(z) - f′′(1) dzdy x y ∫ 1∫ 1∫ 1 = - f′′′(w) dwdzdy ∫x1∫y1∫ z1 = - f′′′(w)- f′′′(1) dwdzdy x y z ∫ 1 ∫ 1 ∫ 1∫ 1 = f′′′′(s) dsdwdzdy x y z w > 0

again so that f(x) > f(1). Hence if x ∈ (1 - ε,1 + ε) then f(x) ≥ f(1) so that f(1) is a local minimum.

(b) What can be said about f near x = 1 when f′(1) = f′′(1) = 0 and f′′′(1) > 0.

There is nothing that can be generally said about f since it may not be a local maximum or minimum at all.

6

Let u(x) be a smooth solution to the differential equation

u′′ + 3u ′ - (1 +x2)u = 0

(a) Show that u cannot have a positive local maximum

Given the differential equation above, we have

u′′ + 3u′ = (1+ x2)u

for u(x) so that at any local maximum x₀ where u(x₀) = 0, we have

u′′ = (1+ x20)u

(6.5)

So if u(x₀) is positive then u′′(x₀) > 0 , i.e. u is convex at x₀. Thus if u(x₀) is positive then u can only have a local minimum at x₀.

(b) Show that u cannot have a negative local minimum

Let x₀ be as in the previous part of this problem. Equation 6.5 also implies that if u(x₀) is negative, then so is u′′(x₀). Hence u can only have a negative local maximum at x₀.

(c) If u(0) = u(2) = 0, show that u(x) = 0 for x ∈ [0, 2]

If u is zero on [0,2], then we are done. So let x₀ ∈ (0,2) be nonzero. Either

u(x₀) > 0 or
u(x₀) < 0

If the first case, then since u is smooth, it is bounded on [0,2]. Thus there must be a maximum positive value of u on [0,2]. But this contradicts the first part of this problem. This case can thus not happen.

If the second case, then since u is smooth it is bounded on [0,2]. Thus there must be a minimum negative value of u on [0,2]. But this contradicts the second part of this problem. Thus this case can also not happen.

Hence, the only possible scenario is u(x) = 0 for x ∈ [0,2].

(d)

7

(a)

(b)

8 Use the Reimann sum to compute ∫ ₀^b sin x dx

Define a partition P_n of [0,b] by P = {0,θ,2θ,…,(n- 1)θ,b} where θ = b∕n. Then we have

U(Pn,sin x) = θ(sin θ+ ⋅⋅⋅+ sin(n θ))

and

L(Pn,sinx) = θ(sin0+ sinθ+ ⋅⋅⋅+ sin((n- 1)θ))

so that U(P_n,sinx) - L(P_n,sinx) = θ sin(nθ) which approaches zero as n →∞. Thus we know that sinx is indeed integrable.

Now to find the actual value of the integral of sinx we evaluate lim_n→∞U(P_n,sinx). So because θ = b∕n we have the following.

∑n nl→im∞ U(Pn,sin x) = lni→m∞ θ sin(kθ) k(=1 ) = lim θ cos(θ∕2)--cos((n-+-1∕2)θ)- n→∞ 2sin(θ∕2) ( ) ( θ ) = nli→m∞ cos(θ∕2) - cos((n + 1∕2)θ) nli→m∞ 2sin(θ∕2) = (cos(0)- cos(b))(1) = 1- cos(b)

so that ∫ ₀^b sinx dx = 1 - cos(b)

9 Define f(0) = 3 and f(x) = sin(1∕x) for x ∈ (0, 2∕π]. Show f is Reimann integrable.

Let ε > 0 be given. Then on the interval [ε∕8,2π] f is continuous and therefore Reimann integrable on the interval. Hence there exists a partition P such that U(P,f) - L(P,f) < ε∕2. Define a partition P′ = P ∪{0}. Then we have U(P′,f) = U(P,f) + 3 ε
8

and L(P′,f) = L(P,f) - ε∕8. This implies

′ ′ -ε U (P ,f) - L (P ,f) = U(P,f)+ 38 - L(P,f)+ ε∕8 = ε∕2+ (U(P,f)- L(P,f)) < ε∕2 + ε∕2 = ε

Thus f is Riemann integrable.

10 11 Prove the Integral Mean Value Theorem

Let f be real and continuous on [a,b]. Partition [a,b] by P = {a,b}. Then L(P,f) = (b - a)f(y) and U(P,f) = (b - a)f(z) where f(y) = max(f(x)) and f(z) = min(f(x)) for x ∈ [a,b]. Thus because

∫ b L(P,f) ≤ f(x ) dx ≤ U(P,f) a

we have

∫ -1--- b f(y) ≤ b- a a f(x) dx ≤ f(z)

Since f is continuous on [y,z] then The Intermediate Value Theorem tells us that there is a c ∈ [y,x] such that

1 ∫ b ----- f(x) dx = f(c) b- a a

as desired.