The Inverse Function Theorem and Introduction to Infinite Series Flashcards
What is a strictly increasing function
Let A ⊆ R. A function f : A → R is strictly
increasing, if x1, x2 ∈ A, x1 < x2 implies f(x1) < f(x2)
What is the Inverse Function Theorem
Assume that
the function f is -
* Defined on a closed interval [a, b];
* Strictly increasing;
* Continuous on [a, b].
Then there exists a function g which is -
* Defined on the closed interval [f(a), f(b)];
* Strictly increasing;
* Continuous on [f(a), f(b)],
and which is inverse to f, that is, g(f(x)) = x for all x ∈ [a, b]
Explain the proof of the Existence of the inverse on [f(a), f(b)]
- First change the codomain of f from R to [f(a), f(b)]. We note that for any x ∈ (a, b) we have a < x < b. Since f is strictly increasing, f(a) < f(x) < f(b) and so f(x) ∈
(f(a), f(b)). If x = a or x = b then it is also true that f(x) ∈ [f(a), f(b)]. Therefore, f has domain [a, b] and codomain [f(a), f(b)]. - To prove surjectivity. Take any d ∈ [f(a), f(b)]. Since f
is continuous, by the Intermediate Value Theorem there exists c ∈ [a, b] such that f(c) = d. - To prove injectivity. Take x1, x2 ∈ [a, b] such that x1 ≠ x2. Since f is strictly increasing, if x1 < x2, then f(x1) < f(x2), and if x1 > x2, then f(x1) > f(x2). f(x1) ≠ f(x2), which means that f is injective
- Therefore, f : [a, b] → [f(a), f(b)] is bijective and so By there exists g which is the inverse of f
Explain the proof that g is strictly increasing
- Assume not, then there exist y1, y2 ∈ [f(a), f(b)] such that y1 < y2 but g(y2) ≤ g(y1).
- We cannot have g(y2) = g(y1) because g is injective. So g(y2) < g(y1).
- Since f is strictly increasing, f(g(y2)) < f(g(y1)). This would lead to y2 < y1, contradicting y1 < y2
Explain the proof that g is continuous on [f(a), f(b)] (Won’t ask)
- We take an arbitrary y ∈ [a, b] and show that g is continuous at y. We denote x = g(y).
- Let ε > 0. To verify the definition of “continuous at y”, we need to find δ > 0 such that
whenever z ∈ [f(a), f(b)] and y − δ < z < y + δ, one has x − ε < g(z) < x + ε. - We first find δ1 > 0 such that z ∈ [f(a), f(b)] and z < y + δ1 =⇒ g(z) < x + ε.
- If x+ε ≤ b, then put y+δ1 = f(x+ε). This guarantees δ1 > 0 since f is strictly increasing, and so f(x) = y < f(x + ε). If z < y + δ1, then g(z) < g(y + δ1) = g(f(x + ε)) = x + ε.
- If, however, x + ε > b, we can choose δ1 however we want: e.g., put δ1 = 1000. Indeed, if z ∈ [f(a), f(b)] then g(z) ≤ b and so automatically g(z) < x + ε.
- We now find δ2 > 0 such that z ∈ [f(a), f(b)] and y − δ2 < z =⇒ x − ε < g(z).
- Finding δ2 is completely similar to finding δ1, and we skip the details.
- It remains to take δ = min{δ2, δ1}. Now, if z ∈ [f(a), f(b)] and y − δ < z < y + δ, then z satisfies both (2) and (1), and so x − ε < g(z) and g(z) < x + ε, as required.
What is a strictly decreasing function
Let A ⊆ R. A function f : A → R is strictly
decreasing, if x1, x2 ∈ A, x1 < x2 implies f(x1) > f(x2)
What is a strictly monotonic function
A function that is strictly increasing or decreasing
What is an infinite series
An expression of the form
What is the nth partial sum of an infinite series
What is a geometric series with ratio r
When does a geometric series converge and diverge
- If |r| < 1, the geometric series converges with sum 1/ 1 − r
- If |r| ≥ 1, the geometric series diverges.
What is the formula for sum of geometric series
Multipled by a if the start point is not 1
Describe the proof of geometric convergence
- If r = 1, then sn = n+1, and so sn → +∞ as n → ∞. The geometric series 1+ 1+ 1+. . .diverges to +∞.
What is the Nullity Test
Remember - the Nullity is not sufficient
Describe the proof for the Nullity Test
Describe the proof that the harmonic series diverges
Describe the Algebra Rules of Infinite Sums
What is the Comparison Test
Describe the Proof of the Comparison Test
