Notes - Analysis II HT23, Mean value theorems
Flashcards
Can you state Rolle’s theorem (the one about turning points between two equal values)?
Let $f: [a, b] \to \mathbb R$ be a function. Suppose:
- $f$ is continuous on $[a, b]$.
- $f$ is differentiable on $(a, b)$.
- $f(a) = f(b)$.
Then $\exists \xi \in (a, b)$ such that $f’(\xi) = 0$.
What does Rolle’s theorem say informally?
If $f : [a, b] \to \mathbb R$ is nice and equal at the endpoints, it must turn around somewhere inbetween.
What two ingredients go into proving Rolle’s theorem?
- Boundedness theorem
- Fermat’s theorem on extrema
When proving Rolle’s theorem, the idea is to show that there either exists a local minimum or local maximum of $[a, b]$, which then has zero derivative due to Fermat’s theorem on extrema. What are the three cases?
- $\exists x _ 0 \in [a,b]$ such that $f(x _ 0) \ge f(a)$
- $\exists x _ 0 \in [a, b]$ such that $f(x _ 0) \le f(a)$
- Neither, in which case the function is constant
Can you state the mean value theorem/Lagrange’s theorem (about the derivative between two points of a function)?
Let $f : [a, b] \to \mathbb R$ be a function. Suppose:
- $f$ is continuous on $[a, b]$.
- $f$ is differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that
\[f'(\xi) = \frac{f(b) - f(a)}{b - a}\]What does the mean value theorem/Lagrange’s theorem say informally?
If $f : [a, b] \to \mathbb R$ is nice, then there exists some point where $f$ is tangent to the secant between $f(a)$ and $f(b)$.
Can you state the generalised mean value theorem/Cauchy’s theorem?
Let $f, g : [a, b] \to \mathbb R$ be functions. Suppose:
- $f, g$ are continuous on $[a, b]$.
- $f, g$ are differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that
\[f'(\xi)(g(b) - g(a)) = g'(\xi)(f(b) - f(a))\]The generalised mean value theorem states that
If $f, g : [a, b] \to \mathbb R$ are functions and
- $f, g$ are continuous on $[a, b]$.
- $f, g$ are differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that
\[f'(\xi)(g(b) - g(a)) = g'(\xi)(f(b) - f(a))\]
How can this be written if $g(a) \ne g(b)$ and $g’(x)$ is not zero in $(a, b)$?
How is Rolle’s theorem a special case of the mean value theorem, which itself is a special case of the generalised mean value theorem?
- Rolle’s theorem is the mean value theorem in the case where $f(a) = f(b)$.
- The mean value theorem is the generalised mean value theorem in the case where $g(x) = x$.
When proving the normal mean value theorem, i.e.
Let $f : [a, b] \to \mathbb R$ be a function. Suppose:
- $f$ is continuous on $[a, b]$.
- $f$ is differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that $f’(\xi) = \frac{f(b) - f(a)}{b - a}$.
what function $F$ do you consider so that Rolle’s theorem applies?
Let $f : [a, b] \to \mathbb R$ be a function. Suppose:
- $f$ is continuous on $[a, b]$.
- $f$ is differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that $f’(\xi) = \frac{f(b) - f(a)}{b - a}$.
where $K$ is chosen to be
\[K = \frac{f(b) - f(a)}{b - a}\]so that $F(a) = F(b) = 0$.
When proving the Cauchy/generalised mean value theorem, i.e.
Let $f, g : [a, b] \to \mathbb R$ be functions. Suppose:
- $f, g$ are continuous on $[a, b]$.
- $f, g$ are differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that
\[f'(\xi)(g(b) - g(a)) = g'(\xi)(f(b) - f(a))\]
what function $F$ do you consider so that Rolle’s theorem applies?
Let $f, g : [a, b] \to \mathbb R$ be functions. Suppose:
- $f, g$ are continuous on $[a, b]$.
- $f, g$ are differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that
where $K$ is chosen to be
\[K = \frac{f(b) - f(a)}{g(b) - g(a)}\]so that $F(a) = F(b) = 0$.
Proofs
Prove Rolle’s theorem:
Let $f: [a, b] \to \mathbb R$ be a function. Suppose:
- $f$ is continuous on $[a, b]$.
- $f$ is differentiable on $(a, b)$.
- $f(a) = f(b)$.
Then $\exists \xi \in (a, b)$ such that $f’(\xi) = 0$.
Let $f: [a, b] \to \mathbb R$ be a function. Suppose:
- $f$ is continuous on $[a, b]$.
- $f$ is differentiable on $(a, b)$.
- $f(a) = f(b)$.
Then $\exists \xi \in (a, b)$ such that $f’(\xi) = 0$.
Todo.
Prove the mean value theorem/Lagrange’s theorem:
Let $f : [a, b] \to \mathbb R$ be a function. Suppose:
- $f$ is continuous on $[a, b]$.
- $f$ is differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that $f’(\xi) = \frac{f(b) - f(a)}{b - a}$.
Let $f : [a, b] \to \mathbb R$ be a function. Suppose:
- $f$ is continuous on $[a, b]$.
- $f$ is differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that $f’(\xi) = \frac{f(b) - f(a)}{b - a}$.
Todo.
Prove the generalised mean value theorem/Cauchy’s theorem:
Let $f, g : [a, b] \to \mathbb R$ be functions. Suppose:
- $f, g$ are continuous on $[a, b]$.
- $f, g$ are differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that
\[f'(\xi)(g(b) - g(a)) = g'(\xi)(f(b) - f(a))\]
and furthermore if $g(a) \ne g(b)$ and $g’(x)$ is not zero in $(a, b)$, then this is equivalent to
\[\frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)}\]
Let $f, g : [a, b] \to \mathbb R$ be functions. Suppose:
- $f, g$ are continuous on $[a, b]$.
- $f, g$ are differentiable on $(a, b)$.
Then there exists $\xi \in (a, b)$ such that
and furthermore if $g(a) \ne g(b)$ and $g’(x)$ is not zero in $(a, b)$, then this is equivalent to
Todo.
Prove the relationship between derivatives and monotonicity:
Let $f : I \to \mathbb R$ be a differentiable function defined on an interval. Then
- If $f’(x) \ge 0$ or $f(x) > 0$ for all $x \in I$ then $f$ is increasing or strictly increasing on $I$.
- If $f’(x) \le 0$ or $f(x) < 0$ for all $x \in I$ then $f$ is decreasing or strictly decreasing on $I$.
Todo.