November 8, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework08
For h > 0 we have

and

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

The derivative of A

so that we have A^{2}(t) = A′A + AA′. Note the multiplication of A and A′ may not commute.

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(t

For h > 0 we have.

By the first part of this problem we have

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

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 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) = v(x) implying v(x) = ce^{x}.

Define u(x) = e

and

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

Let v′- γv ≤ 0 and define g(x) = e

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

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 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

| (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

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

where δ_{b} 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.

Let f : ℝ → ℝ be a smooth function.

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

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

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

Given the differential equation above, we have

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

| (6.5) |

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

Let x

If u is zero on [0,2], then we are done. So let x

- u(x
_{0}) > 0 or - u(x
_{0}) < 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].

Define a partition P

and

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.

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.

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.