Analysis 2 Flashcards
When does a Power Series Converge?
Let Σ0,∞ [anx^n] be a power series with Σant^n convergent. Then
Σ0,∞ [anx^n]
converges absolutely for all x with |x| < |t|.
Proof:
Since Σant^n converges we know that ant^n → 0 as n → ∞ and so the sequence is bounded. There is some M for which |ant^n| < M for all n. Now
Σ0,N [|anx^n|] = Σ0,N [|ant^n||x/t|^n
≤ MΣ0,N [|x/t|^n]
≤ MΣ0,∞ [|x/t|^n]
= M/1 − |x|/|t|
Hence Σ0,∞|anx^n| < ∞ and the series converges absolutely.
What is the Theorem for the possibilities of how a power series converges?
Let Σ0,∞ [anx^n] be a power series. One of the
following holds.
* The series converges only if x = 0.
* The series converges for all real numbers x.
* There is a positive number R with the property that the series converges if |x| < R and diverges if |x| > R.
In the third case the number R is called the radius of convergence. In the first case we say that the radius of convergence is 0 and in the second that it is ∞.
How does the radius of convergence for the absolute power series relate to the original series?
Let Σ0,∞ [anx^n] be a power series with radius of convergence R. Then Σ0,∞|an|x^n also has radius of convergence R.
How are power series continuous?
Let Σ0,∞ [anx^n] with radius of convergence R.
Then the function
x → Σ0,∞ [anx^n] is continuous on the interval (−R, R).
What is the Exponential?
If x ∈ R the series
1 + x + x^2/2 + x^3/6 + · · · + x^n/n! + · · ·
converges. We call the sum exp x.
Use the ratio test of successive terms to show it converges.
What is the Characteristic Property of the Exponential?
If x, y ∈ R then
exp(x + y) = exp(x) exp(y).
Proof For a fixed number z consider the function
x → exp(x) exp(z − x).
We may differentiate this with respect to x using the product rule and the chain rule to get
exp(x) exp(z − x) − exp(x) exp(z − x) = 0.
By the MVT the function is constant. At x = 0 the function is exp z so we know that for all x
exp(x) exp(z − x) = exp(z).
Now if we set z = x + y we get the conclusion we want.
What are the Inequalities for the exponential?
The following estimates hold for the exponential function:
1. 1 + x ≤ e^x for all real x
2. e^x ≤ 1/(1 − x) if x < 1.
Proof If x ≥ 0 then
e^x = 1 + x + x^2/2 + · · · ≥ 1 + x
and if 0 ≤ x < 1
e^x = 1 + x + x^2/2 +x^3/6 + · · · ≤ 1 + x + x^2 + x^3 + · · · = 1/1 − x.
So the inequalities are easy to establish if x ≥ 0. To obtain them for negative x we use the characteristic property of the exponential much as we did to prove positivity.
Suppose x = −u is negative. We know that e^u ≥ 1 + u and hence e^−x ≥ 1 − x. But this implies that
1/1 − x ≥ e^x
so the second inequality is now established for all x < 1.
If x ≤ −1 then 1 + x ≤ 0 whereas e^x > 0 so e^x ≥ 1 + x. It remains to prove the first inequality for −1 < x < 0. If x = −u then 0 < u < 1 and so e^u ≤ 1/(1 − u). This says
that e^−x ≤ 1/(1 + x) and this implies that
1 + x ≤ e^x.
How does the exponential increase?
The exponential function is strictly increasing and its range is (0,∞).
Proof Suppose x < y. Then
e^y = e^(y−x)e^x ≥ (1 + y − x)e^x > e^x.
This shows that the exponential is strictly increasing.
What is the Logarithm?
There is a continuous strictly increasing function x → log x
defined on (0,∞) satisfying
e^log x = x
for all positive x and
log(e^y) = y
for all real y. We have that for all positive u and v,
log(uv) = log u + log v.
Proof We just need to check the last assertion. But
e^log u+log v = e^log u * e^log v = uv.
What are Powers?
If x > 0 and p ∈ R we define
x^p = exp(p log x).
We have the usual rules
1. If n is a positive integer then x^n as defined here is indeed the product x.x. . . . .x of n copies of x.
2. x^p+q = x^px^q for all x > 0 and p, q ∈ R.
3. log(x^) = p log x for x > 0 and p ∈ R.
4. x^pq = (x^p)^q for all x > 0 and p, q ∈ R.
5. exp(p) = e^p for all p ∈ R.
What is the Inequality for the logarithm?
If x > 0 then log x ≤ x − 1.
What is the Limit of a function?
Let I be an open interval, c ∈ I and f a real valued function defined on I except possibly at c. We say that
lim(x→c) f(x) = L
if for every ε > 0 there is a number δ > 0 so that if 0 < |x − c| < δ then |f(x) − L| < ε.
Thus we can guarantee that f(x) is close to L by insisting that x is close to c, but we exclude the possibility that x = c: we don’t care what f does at c itself nor even whether f is defined at c.
How do limits relate to continuity?
If f : I → R is defined on the open interval I and c ∈ I then f is continuous at c if and only if
lim(x→c) f(x) = f(c).
How do limits relate to sequences?
If f : I{c} → R is defined on the interval I except at c ∈ I then
lim(x→c) f(x) = L
if and only if for every sequence (xn) in I/{c} with xn → c we have
f(xn) → L.
What are the Algebra of Limits?
If f, g : I{c} → R are defined on the interval I except at c ∈ I and limx→c f(x) and limx→c g(x) exist then
1. limx→c(f(x) + g(x)) = limx→c f(x) + limx→c g(x)
2. limx→c f(x)g(x) = limx→c f(x) limx→c g(x)
3. if limx→c g(x) /= 0 then
limx→c f(x)/g(x) = (limx→c f(x))/(limx→c g(x)).
What is the Derivative?
Suppose f : I → R is defined on the open interval I and c ∈ I. We say that f is differentiable at c if
lim(h→0) f(c + h) − f(c)/h
exists. If so we call the limit f’(c).
We can rewrite the derivative
f’(c) = limx→c f(x) − f(c)/x − c
since if we put x = c + h and h → 0 we have x → c
How does differentiability imply continuity?
If I is an open interval, f : I → R is differentiable at c ∈ I then f is continuous at c.
Proof We know that
f(x) − f(c)/x − c → f’(c)
as x → c. Hence
f(x) − f(c) = f(x) − f(c)/x − c * (x − c) → f’(c)*0 = 0
as x → c which implies that f(x) → f(c) as required.
What are the sum and product rules of the derivative?
Suppose f, g : I → R are defined on the open interval I and are differentiable at c ∈ I. Then f + g and fg are differentiable at c and (f + g)’(c) = f’(c) + g’(c) and (fg)’(c) = f’(c)g(c) + f(c)g’(c).
Proof of product rule
f(x)g(x) − f(c)g(c)/x − c
=(f(x) − f(c))g(x) + f(c)(g(x) − g(c))/x − c
=f(x) − f(c)/x − c * g(x) + f(c)* g(x) − g(c)/x − c.
The algebra of limits tells us that as x → c this expression approaches
f’(c)g(c) + f(c)g’(c).
What are the derivatives of the monomials?
If n is a positive integer then the derivative of x → x^n is x → nx^(n−1).
Proof We already saw this for n = 1. Assume inductively that we have the result for f(x) = x^n. Then
x^n+1 = xf(x) so by the product rule its derivative is
1.f(x) + xf’(x) = x^n + xnx^(n−1) = (n + 1)x^n
completing the inductive step.
What is the Chain Rule?
Suppose I and J are open intervals, f : I → R and g : J → I,
that g is differentiable at c and f is differentiable at g(c). Then the composition f ◦ g is differentiable at c and
(f ◦ g)’(c) = f’(g(c))* g’(c).
Proof.
f is differentiable at g(c) so for all y
f(y) − f(g(c)) = f’(g(c))(y − g(c)) + ε(y)(y − g(c)).
where ε(g(c)) = 0 and ε is continuous at g(c). Hence
f(g(x)) − f(g(c)) = f’(g(c))(g(x) − g(c)) + ε(g(x))(g(x) − g(c)).
Consequently if x /= c
f(g(x)) − f(g(c))/x − c = f’(g(c)) * g(x) − g(c)/x − c + ε(g(x)) * g(x) − g(c)/x − c.
As x → c,
g(x) − g(c)/x − c → g’(c)
while ε ◦ g is continuous at c so ε(g(x)) → ε(g(c)) = 0. Hence
f(g(x)) − f(g(c))/x − c → f’(g(c)).g’(c) as required.
What is the Mean Value Theorem?
Suppose f : [a, b] → R is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there is a point c ∈ (a, b) where
f’(c) = f(b) − f(a)/b − a.
What is Rolle’s Theorem?
Suppose f : [a, b] → R is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) and that f(a) = f(b). Then there is a point c in the open interval where
f’(c) = 0.
Proof.
If f is constant on the interval then its derivative is zero 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 and so is not positive.
On the other hand 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-negative values and so is not negative. Therefore f’(c) = 0.
What is the proof of the MVT?
Consider the function given by
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.
What is the derivative of the maximum in an interval?
Suppose f : [a, b] → R is continuous and that it is differentiable on the open interval. Then f attains its maximum and minimum either at points in (a, b) where f’ = 0 or at one of the ends a or b.
