Chris

Show that cos(x)=x has a root on R.

Let f,g be continuous from R to R, and suppose that f(r)=g(r) for all rational numbers r. Is it true that f(x)=g(x) for all x in R?

Using the alternating harmonic series, show that the series sum(((-1)^n)/sqrt(n), n=1..infinity) converges.

a) use the definition of limit to show that lim(x_n)=0.

b) Find a specific value of K(epsilon) as required in the definition of limit for each of i) epsilon=1/2, and ii) epsilon=1/10

If I_n is a nested sequence of intervals and if I_n=[an,bn], show that a1 <=a2<=...<=an<=... and b1>=b2>=...>=bn>=...

Let S be a nonempty subset of R that is bounded below. Prove that inf(S)= -sup{-s: s in S}

Also, if you can, Determine and sketch the set of pairs (x,y) in RxR that satisfy abs(x*y)=2.

Let K := {s + t√2:s,t ∈ Q}. Show that K satisfies the following:

a) If x1,x2 ∈ K, then x1+x2 ∈ K and x1x2 ∈ K.

b) If x not=0 and x in K, then 1/x in K

Specify the field axiom at each step

By induction of course

Prove that 1^2-2^2+3^2+...+(-1)^(n+1)*n^2=(-1)^(n+1)*n*(n+1)/2 for all n in N. (By induction)

Briefly explain why.

a) f(x)=x^2 on [-3,3), period 6.

b) f(x)=cos(x) on [-Pi/3, Pi/3), period 2*Pi/3

c) f(x)=x^3 on [-1,1), period 2.

d) f(x)=abs(x)-x on [-2,2), on period 4.

e) f(x)=sin(e^(abs(x)) on [-1,1), period 2.

prove the following orthogonality theorem for the set of functions {sin(n*Pi*x/a)} n=1..infinity ov erthe interval [0,a]: For m not equal to n, int(sin(n*Pi*x/a)sin(m*Pi*x/a), x=0..a)=0 and int(sin^2(n*Pi*x/a), x=0..a)=a/2

1. Find a recurrence relation and initial conditions for the sequence {S_n}.

2. Show that S_n=f_n+1, n=1, 2,... where f denotes the Fibonacci sequence.

The order of the subsets is not taken into account.

a) Show that S_n,k=0 if k>n

b) Show that S_n,n=1 for all n>=1

c) Show that S_n,1=1 for all n>=1

d)Show that S_3,2=3

e) Show that S_4,2=7

f) Show that S_4,3=6

g) Show that S_n,2=2^(n-1)-1 for all n>=2

h) Show that S_n,n-1=C(n,2) for all n>=2

i) Find a formula for S_n,n-2, n>=3, and prove it.

I know this is extremely long. Best answer will be awarded.