Homework 8 <me@tylerlogic.com>
November 8, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework08

### 1 Show that cosx is differentiable at all x.

For h > 0 we have
where we make use of the trigonometric identity cos(x + y) = cosxcosy - sinxsiny and the fact that

and

### 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

so that we have A2(t) = AA + AA. Note the multiplication of A and Amay 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.
so that = -A-1AA-1

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

By the first part of this problem we have

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

### 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
 (3.1)

and

 (3.2)

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

#### (b) Prove that ex+a= eaex for all real a and x

Define u(x) = ex+a - eaex for real a. Then we have

and

by the previous part of this problem, u(x) = 0ex = 0. Hence ex+a = eaex.

#### (c) For constant γ show v′- γv ≤ 0 implies v(x) ≤ v(0)eγx for x ≥ 0

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

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

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
 (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 = , are intervals of size less than δ. Define Mi = supf(x) and mi = inf f(x) for each i ∈{0,1,,n - 1} and x [a + iδ,a + (i + 1)δ) so that in particular

 (4.4)

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

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

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

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 : 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
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
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

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

Given the differential equation above, we have

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

 (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
1. u(x0) > 0 or
2. 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].

### 8 Use the Reimann sum to compute ∫0bsinx dx

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

and

so that U(Pn,sinx) - L(Pn,sinx) = θ sin() 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 limn→∞U(Pn,sinx). So because θ = b∕n we have the following.

so that 0b 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 and L(P,f) = L(P,f) - ε∕8. This implies

Thus f is Riemann integrable.

### 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

we have

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

as desired.