Proofs

Question 1

What is the proof of the existence and continuity of the inverse of a continuous and strictly increasing function f?

Answer 1

https://imgur.com/a/V4JuwWE

Question 2

What is the proof of the boundedness of continuous functions with a domain [a, b]?

Answer 2

Suppose f is unbounded. For each n choose x_n ∈ [a, b] where |f(x_n)| >= n. There exists a convergent subsequence x_{n_l} -> x. As the interval [a, b] is closed x ∈ [a, b]. Therefore f(x_{n_l​}) -> f(x) but this is a contradiction as (|f(x_n)| >= n) => (|f(x_{n_l})| >= n). □

Question 3

Prove the following statement: if Σa_nxⁿ is a power series one of the following three properties hold:

(i) The series only converges if x = 0
(ii) The series converges for all real numbers x
(iii) There is a positive number R with the property that the series converges if |x| < R and diverges if |x| > R

Answer 3

Firstly we will prove that if Σa_ntⁿ is convergent then Σa_nxⁿ is convergent for all |x| < |t|. Let M be a constant such that |a_ntⁿ| < M for all n.

Σ|a_nxⁿ| = Σ|a_ntⁿ|*|x/t|ⁿ

=< MΣ|x/t|ⁿ (from 0 to n)

=< MΣ|x/t|ⁿ (from 0 to ∞)

=M/(1-|x|/|t|)

Obviously the series does converge if x = 0. If the first option does not hold than it converges for some other values of x. The set of x for which it converges might be unbounded: it might contain arbitrarily large numbers. In that case the power series will converge absolutely for all real x and defines a function on R. Otherwise the set might contain some numbers other than 0 but be bounded. In that case we let R = sup{|t| : Σa_ntⁿ converges}. Then the series converges absolutely whenever |x| < R. It does not converge if |x| > R by definition of R.

Question 4

Prove that if there is a continuous strictly increasing function defined on the interval (0, ∞) such that:

(i) e^log(x) = x
(ii) log(e^y) = y

We have that log(uv) = log(u) + log(v)

Answer 4

e^{log(u) + log(v)} = e^log(u)e^log(v) = uv

log(e^{log(u) + log(v)}) = log(uv)

log(u) + log(v) = log(uv)

log(uv) = log(u) + log(v)

Question 5

What is the proof of Rolle’s theorem?

Answer 5

If f is constant on the interval then its derivative is 0 everywhere. If not it takes values different from f(a) = f(b).

Assume it is somewhere larger than f(a).

Since f is continuous on the closed interval it attains its maximum value at some point c and this cannot be a or b: so c lies in (a, b).

If x > c then f(x) - f(c) =< 0 while x - c > 0 so the ratio

(f(x) - f(c))/(x-c) =< 0 (*)

So f’(c) is a limit of non-positive values so is not positive.

On the other hand if x < c then f(x) - f(c) =< 0 while x - c < 0 so (*) is greater or equal to 0.

So f’(c) is a limit of non-negative values so is non negative.

Therefore f’(c) = 0. □

Question 6

What is the proof of the Mean Value Theorem?

Answer 6

Consider the function g(x) = f(x) - x(f(b)-f(a)/(b-a).

Then g(b) - g(a) = f(b) - f(a) - (b-a)(f(b)-f(a))/(b-a) = 0.

So by Rolle’s theorem there is a point c where g’(c) = 0.

But this implies f’(c) = (f(b) - f(a))/(b-a). □

Question 7

What are the steps to proving the differentiability of power series?

Answer 7

1. Prove that if Σa_nxⁿ converges absolutely then Σna_nx^n-1 does too using Σ|a_n|(yⁿ-x^{<span>n</span>})/(y - x) with x < y < R
2. Choose |x| < T < R and know that there is a number N where (N+1 to ∞)Σn|a_n|T^n-1 < ε/3 to prove |(N+1 to ∞)Σa_n(yⁿ-x^{<span>n</span>})/(y - x)| < ε/3
3. Choose δ₀ > 0 such that if 0 < |y - x| < δ₀

|(1 to N)Σa_n(yⁿ-x^{<span>n</span>})/(y - x) - (1 to N)Σna_nx^n-1| < ε/3

4. Choose δ to be the smaller of δ₀ and T - |x| so whenever 0 < |y -x| < δ

|(1 to ∞)Σa_n(yⁿ-x^{<span>n</span>})/(y - x) - Σna_nx^n-1| =< |(N+1 to ∞)Σa_n(yⁿ-x^{<span>n</span>})/(y - x)|+ |(1 to N)Σa_n(yⁿ-x^{<span>n</span>})/(y - x) - (1 to N)Σna_nx^n-1| + (N+1 to ∞ |Σna_nx^n-1| < ε □

Question 8

What is the proof that for a power series Σa_nxⁿ of radius of convergence R. then the series Σna_nx^n-1 has the same radius of convergence?

Answer 8

• We know that the absolute series Σ|a_n|xⁿ has the same radius of convergence as Σa_nxⁿ.
• Now if 0 < x < R choose y with x < y < R.
• Then Σ|a_n|xⁿ and Σ|a_n|yⁿ both converge and hence so does (0 to ∞)Σ|a_n|(yⁿ-xⁿ)/(y - x) = (1 to ∞)Σ|a_n|(y^n-1 + y^n-2x + … + x^n-1)

(1 to ∞)Σ|a_n|(y^n-1 + y^n-2x + … + x^n-1) > (1 to ∞)Σ|a_n|nx^n-1 so (1 to ∞)Σ|a_n|nx^n-1 also converges.

This means that Σna_nx^n-1 converges absolutely as required. □

Question 9

Prove that if x and y are real numbers then exp(x + y) = exp(x) exp(y).

Answer 9

• For a fixed number z consider the function x → exp(x)exp(z − x).
• d/dx(exp(x)exp(z − x)) = exp(x)exp(z − x) - exp(x)exp(z − x) = 0
• The function is constant by the MVT
• At x = 0 the function is exp(z) so we know that for all x: exp(x)exp(z − x) = exp(z).
• Let z = x + y and we get the result exp(x)exp(y) = exp(x + y) □

Question 10

What are the proofs for the cos(x+y) identity?

Answer 10

• For a fixed z let f(x) = cos(x)cos(z-x) - sin(x)sin(z-x)
• f’(x) = 0 ∴ f is constant
• f(0) = cos(0)cos(z) - sin(0)sin(z) = cos(z)
• Hence f(x) = cos(z)
• Set z = x + y to get cos(x)cos(y) - sin(x)sin(y) = f(x) = cos(z) = cos(x+y) □