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 ddtA2(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.
 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- 1= -A-1AA-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) = 1cv(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 ex since ex is the unique function with the properties in equations 3.1 and 3.2. But then we have ex = u(x) = 1
cv(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
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) = 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
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] can be approximated by a piecewise linear function g : [a,b]


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 > bn. 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 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

|              |
||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 φi : [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] by

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

where δb 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 : 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
  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].

(d)


7


(a)


(b)


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


Define a partition Pn 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(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.

                         ∑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 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) + 3ε
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.