Proof that two definitions of a point of inflection are equivalent
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I have seen that many online sources (including other math stack exchange questions) say that the following are equivalent definitions of a point of inflection:
If $f$ is differentiable on $I$ we say that $f$ has a point of inflection at $a$ if $f'$ has an isolated local extremum at $a$.
We say that $f$ has a point of inflection at $a$ if $f$ changes concavity at $a$. This definition is somewhat ambiguous to me, so I decided that it would mean that $exists delta>0$ such that $f$ is strictly convex on $[a-delta,a]$ and strictly concave $[a,a+delta]$ or $f$ is strictly concave on $[a-delta,a]$ and strictly convex $[a,a+delta]$.
I want to prove that these definitions are equivalent (I know that this might require clarifying definition 2. in a different way than I have).
I was able to prove that 2. implies 1.
This is my proof:
- Suppose that $exists delta>0$ such that $f$ is strictly convex on $[a-delta,a]$ and strictly concave $[a,a+delta]$.
Then $f'$ is strictly increasing on $[a-delta,a]$ and $f'$ is strictly decreasing on $[a,a+delta]$, so $f'$ has an isolated local maximum at $a$.
The case when $f$ is strictly concave on $[a-delta,a]$ and strictly convex $[a,a+delta]$ is similar.
Now I want to show that 1. implies 2.
- Suppose that $f'$ has a isolated relative extremum. Without loss of generality, suppose that $f'$ has a isolated relative minimum. I need to show that $f'$ is strictly increasing on $[a-delta,a]$ and $f'$ is strictly decreasing on $[a,a+delta]$ (which is equivalent to showing that $f$ is convex on $[a-delta,a]$ and concave on $[a,a+delta]$).
I know that this is not true for a general function $g$ that if $g$ has an isolated extremum at $g$, $g$ is strictly monotone on either side of $a$. (for example, if the g has a jump discontinuity, which I know is not possible for $f'$). Why is true for $f'$?
real-analysis derivatives convex-analysis
asked Nov 17 at 0:03
Andrew Murdza
254
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up vote
2
down vote
favorite
I have seen that many online sources (including other math stack exchange questions) say that the following are equivalent definitions of a point of inflection:
If $f$ is differentiable on $I$ we say that $f$ has a point of inflection at $a$ if $f'$ has an isolated local extremum at $a$.
We say that $f$ has a point of inflection at $a$ if $f$ changes concavity at $a$. This definition is somewhat ambiguous to me, so I decided that it would mean that $exists delta>0$ such that $f$ is strictly convex on $[a-delta,a]$ and strictly concave $[a,a+delta]$ or $f$ is strictly concave on $[a-delta,a]$ and strictly convex $[a,a+delta]$.
I want to prove that these definitions are equivalent (I know that this might require clarifying definition 2. in a different way than I have).
I was able to prove that 2. implies 1.
This is my proof:
- Suppose that $exists delta>0$ such that $f$ is strictly convex on $[a-delta,a]$ and strictly concave $[a,a+delta]$.
Then $f'$ is strictly increasing on $[a-delta,a]$ and $f'$ is strictly decreasing on $[a,a+delta]$, so $f'$ has an isolated local maximum at $a$.
The case when $f$ is strictly concave on $[a-delta,a]$ and strictly convex $[a,a+delta]$ is similar.
Now I want to show that 1. implies 2.
- Suppose that $f'$ has a isolated relative extremum. Without loss of generality, suppose that $f'$ has a isolated relative minimum. I need to show that $f'$ is strictly increasing on $[a-delta,a]$ and $f'$ is strictly decreasing on $[a,a+delta]$ (which is equivalent to showing that $f$ is convex on $[a-delta,a]$ and concave on $[a,a+delta]$).
I know that this is not true for a general function $g$ that if $g$ has an isolated extremum at $g$, $g$ is strictly monotone on either side of $a$. (for example, if the g has a jump discontinuity, which I know is not possible for $f'$). Why is true for $f'$?
real-analysis derivatives convex-analysis
asked Nov 17 at 0:03
Andrew Murdza
254
Here is a differentiable $f$ whose derivative is not continuous: calculus.subwiki.org/wiki/…
– helper
Nov 17 at 1:22
I know that it is possible for a derivative to be discontinuous, but a derivative cannot have a jump discontinuity. I was able to prove that if both of the one sided limits of the derivative exist they must be equal and the derivative must be continuous at that point. If the derivative is continuous, it is because of an essential discontinuity, such as the one-sided limit tending to infinity or oscillations as in that example.
– Andrew Murdza
Nov 17 at 1:44
Note that 2 cannot imply 1 in general (without assuming $f$ everywhere differentiable), as the function may fail to be differentiable at $a$. Consider $f(x) = begin{cases} sqrt{-x}, & x<0 \ x^2, & xge 0end{cases}$.
– Ted Shifrin
Nov 17 at 17:42
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have seen that many online sources (including other math stack exchange questions) say that the following are equivalent definitions of a point of inflection:
If $f$ is differentiable on $I$ we say that $f$ has a point of inflection at $a$ if $f'$ has an isolated local extremum at $a$.
We say that $f$ has a point of inflection at $a$ if $f$ changes concavity at $a$. This definition is somewhat ambiguous to me, so I decided that it would mean that $exists delta>0$ such that $f$ is strictly convex on $[a-delta,a]$ and strictly concave $[a,a+delta]$ or $f$ is strictly concave on $[a-delta,a]$ and strictly convex $[a,a+delta]$.
I want to prove that these definitions are equivalent (I know that this might require clarifying definition 2. in a different way than I have).
I was able to prove that 2. implies 1.
This is my proof:
- Suppose that $exists delta>0$ such that $f$ is strictly convex on $[a-delta,a]$ and strictly concave $[a,a+delta]$.
Then $f'$ is strictly increasing on $[a-delta,a]$ and $f'$ is strictly decreasing on $[a,a+delta]$, so $f'$ has an isolated local maximum at $a$.
The case when $f$ is strictly concave on $[a-delta,a]$ and strictly convex $[a,a+delta]$ is similar.
Now I want to show that 1. implies 2.
- Suppose that $f'$ has a isolated relative extremum. Without loss of generality, suppose that $f'$ has a isolated relative minimum. I need to show that $f'$ is strictly increasing on $[a-delta,a]$ and $f'$ is strictly decreasing on $[a,a+delta]$ (which is equivalent to showing that $f$ is convex on $[a-delta,a]$ and concave on $[a,a+delta]$).
I know that this is not true for a general function $g$ that if $g$ has an isolated extremum at $g$, $g$ is strictly monotone on either side of $a$. (for example, if the g has a jump discontinuity, which I know is not possible for $f'$). Why is true for $f'$?
real-analysis derivatives convex-analysis
asked Nov 17 at 0:03
Andrew Murdza
254
I have seen that many online sources (including other math stack exchange questions) say that the following are equivalent definitions of a point of inflection:
If $f$ is differentiable on $I$ we say that $f$ has a point of inflection at $a$ if $f'$ has an isolated local extremum at $a$.
We say that $f$ has a point of inflection at $a$ if $f$ changes concavity at $a$. This definition is somewhat ambiguous to me, so I decided that it would mean that $exists delta>0$ such that $f$ is strictly convex on $[a-delta,a]$ and strictly concave $[a,a+delta]$ or $f$ is strictly concave on $[a-delta,a]$ and strictly convex $[a,a+delta]$.
I want to prove that these definitions are equivalent (I know that this might require clarifying definition 2. in a different way than I have).
I was able to prove that 2. implies 1.
This is my proof:
- Suppose that $exists delta>0$ such that $f$ is strictly convex on $[a-delta,a]$ and strictly concave $[a,a+delta]$.
Then $f'$ is strictly increasing on $[a-delta,a]$ and $f'$ is strictly decreasing on $[a,a+delta]$, so $f'$ has an isolated local maximum at $a$.
The case when $f$ is strictly concave on $[a-delta,a]$ and strictly convex $[a,a+delta]$ is similar.
Now I want to show that 1. implies 2.
- Suppose that $f'$ has a isolated relative extremum. Without loss of generality, suppose that $f'$ has a isolated relative minimum. I need to show that $f'$ is strictly increasing on $[a-delta,a]$ and $f'$ is strictly decreasing on $[a,a+delta]$ (which is equivalent to showing that $f$ is convex on $[a-delta,a]$ and concave on $[a,a+delta]$).
I know that this is not true for a general function $g$ that if $g$ has an isolated extremum at $g$, $g$ is strictly monotone on either side of $a$. (for example, if the g has a jump discontinuity, which I know is not possible for $f'$). Why is true for $f'$?
real-analysis derivatives convex-analysis
real-analysis derivatives convex-analysis
asked Nov 17 at 0:03
Andrew Murdza
254
asked Nov 17 at 0:03
Andrew Murdza
254
asked Nov 17 at 0:03
Andrew Murdza
254
asked Nov 17 at 0:03
Andrew Murdza
254
asked Nov 17 at 0:03
Andrew Murdza
254
254
Here is a differentiable $f$ whose derivative is not continuous: calculus.subwiki.org/wiki/…
– helper
Nov 17 at 1:22
I know that it is possible for a derivative to be discontinuous, but a derivative cannot have a jump discontinuity. I was able to prove that if both of the one sided limits of the derivative exist they must be equal and the derivative must be continuous at that point. If the derivative is continuous, it is because of an essential discontinuity, such as the one-sided limit tending to infinity or oscillations as in that example.
– Andrew Murdza
Nov 17 at 1:44
Note that 2 cannot imply 1 in general (without assuming $f$ everywhere differentiable), as the function may fail to be differentiable at $a$. Consider $f(x) = begin{cases} sqrt{-x}, & x<0 \ x^2, & xge 0end{cases}$.
– Ted Shifrin
Nov 17 at 17:42
add a comment |
Here is a differentiable $f$ whose derivative is not continuous: calculus.subwiki.org/wiki/…
– helper
Nov 17 at 1:22
I know that it is possible for a derivative to be discontinuous, but a derivative cannot have a jump discontinuity. I was able to prove that if both of the one sided limits of the derivative exist they must be equal and the derivative must be continuous at that point. If the derivative is continuous, it is because of an essential discontinuity, such as the one-sided limit tending to infinity or oscillations as in that example.
– Andrew Murdza
Nov 17 at 1:44
Note that 2 cannot imply 1 in general (without assuming $f$ everywhere differentiable), as the function may fail to be differentiable at $a$. Consider $f(x) = begin{cases} sqrt{-x}, & x<0 \ x^2, & xge 0end{cases}$.
– Ted Shifrin
Nov 17 at 17:42
Here is a differentiable $f$ whose derivative is not continuous: calculus.subwiki.org/wiki/…
– helper
Nov 17 at 1:22
Here is a differentiable $f$ whose derivative is not continuous: calculus.subwiki.org/wiki/…
– helper
Nov 17 at 1:22
I know that it is possible for a derivative to be discontinuous, but a derivative cannot have a jump discontinuity. I was able to prove that if both of the one sided limits of the derivative exist they must be equal and the derivative must be continuous at that point. If the derivative is continuous, it is because of an essential discontinuity, such as the one-sided limit tending to infinity or oscillations as in that example.
– Andrew Murdza
Nov 17 at 1:44
I know that it is possible for a derivative to be discontinuous, but a derivative cannot have a jump discontinuity. I was able to prove that if both of the one sided limits of the derivative exist they must be equal and the derivative must be continuous at that point. If the derivative is continuous, it is because of an essential discontinuity, such as the one-sided limit tending to infinity or oscillations as in that example.
– Andrew Murdza
Nov 17 at 1:44
Note that 2 cannot imply 1 in general (without assuming $f$ everywhere differentiable), as the function may fail to be differentiable at $a$. Consider $f(x) = begin{cases} sqrt{-x}, & x<0 \ x^2, & xge 0end{cases}$.
– Ted Shifrin
Nov 17 at 17:42
Note that 2 cannot imply 1 in general (without assuming $f$ everywhere differentiable), as the function may fail to be differentiable at $a$. Consider $f(x) = begin{cases} sqrt{-x}, & x<0 \ x^2, & xge 0end{cases}$.
– Ted Shifrin
Nov 17 at 17:42
add a comment |
1 Answer
1
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up vote
1
down vote
$1$ does not imply $2$ !
Take $f(x) = int_{0}^{x} g(t) dt$, where $g(t)= 2|
t | +t sin frac{1}{t} $ , with $g(0) = 0 $. Note that $g $ is continuous, so $f $ is differentiable and $ f' = g $.
it is easy to check that $$ |x |
leq g(x) leq 3 |x|
$$ for all $x in Bbb{R} $.
This implies $ g $ gets an isolated minimum at $ x = 0 $ but clearly in not monotone function around $ 0 $.
answered Nov 17 at 17:31
Red shoes
4,556621
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$1$ does not imply $2$ !
Take $f(x) = int_{0}^{x} g(t) dt$, where $g(t)= 2|
t | +t sin frac{1}{t} $ , with $g(0) = 0 $. Note that $g $ is continuous, so $f $ is differentiable and $ f' = g $.
it is easy to check that $$ |x |
leq g(x) leq 3 |x|
$$ for all $x in Bbb{R} $.
This implies $ g $ gets an isolated minimum at $ x = 0 $ but clearly in not monotone function around $ 0 $.
answered Nov 17 at 17:31
Red shoes
4,556621
add a comment |
up vote
1
down vote
$1$ does not imply $2$ !
Take $f(x) = int_{0}^{x} g(t) dt$, where $g(t)= 2|
t | +t sin frac{1}{t} $ , with $g(0) = 0 $. Note that $g $ is continuous, so $f $ is differentiable and $ f' = g $.
it is easy to check that $$ |x |
leq g(x) leq 3 |x|
$$ for all $x in Bbb{R} $.
This implies $ g $ gets an isolated minimum at $ x = 0 $ but clearly in not monotone function around $ 0 $.
answered Nov 17 at 17:31
Red shoes
4,556621
add a comment |
up vote
1
down vote
up vote
1
down vote
$1$ does not imply $2$ !
Take $f(x) = int_{0}^{x} g(t) dt$, where $g(t)= 2|
t | +t sin frac{1}{t} $ , with $g(0) = 0 $. Note that $g $ is continuous, so $f $ is differentiable and $ f' = g $.
it is easy to check that $$ |x |
leq g(x) leq 3 |x|
$$ for all $x in Bbb{R} $.
This implies $ g $ gets an isolated minimum at $ x = 0 $ but clearly in not monotone function around $ 0 $.
answered Nov 17 at 17:31
Red shoes
4,556621
$1$ does not imply $2$ !
Take $f(x) = int_{0}^{x} g(t) dt$, where $g(t)= 2|
t | +t sin frac{1}{t} $ , with $g(0) = 0 $. Note that $g $ is continuous, so $f $ is differentiable and $ f' = g $.
it is easy to check that $$ |x |
leq g(x) leq 3 |x|
$$ for all $x in Bbb{R} $.
This implies $ g $ gets an isolated minimum at $ x = 0 $ but clearly in not monotone function around $ 0 $.
answered Nov 17 at 17:31
Red shoes
4,556621
edited Nov 17 at 17:37
answered Nov 17 at 17:31
Red shoes
4,556621
answered Nov 17 at 17:31
Red shoes
4,556621
answered Nov 17 at 17:31
Red shoes
4,556621
4,556621
add a comment |
add a comment |
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Here is a differentiable $f$ whose derivative is not continuous: calculus.subwiki.org/wiki/…
– helper
Nov 17 at 1:22
I know that it is possible for a derivative to be discontinuous, but a derivative cannot have a jump discontinuity. I was able to prove that if both of the one sided limits of the derivative exist they must be equal and the derivative must be continuous at that point. If the derivative is continuous, it is because of an essential discontinuity, such as the one-sided limit tending to infinity or oscillations as in that example.
– Andrew Murdza
Nov 17 at 1:44
Note that 2 cannot imply 1 in general (without assuming $f$ everywhere differentiable), as the function may fail to be differentiable at $a$. Consider $f(x) = begin{cases} sqrt{-x}, & x<0 \ x^2, & xge 0end{cases}$.
– Ted Shifrin
Nov 17 at 17:42