What can you do if the higher-order derivative test fails?

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The second-derivative test states that if $x$ is a real number such that $f'(x)=0$, then:

1. If $f''(x)>0$, then $f$ has a local minimum at $x$.


2. If $f''(x)<0$, then $f$ has a local maximum at $x$.


3. If $f''(x)=0$, then the text is inconclusive.

But there's no need to despair if the second-derivative test is inconclusive, because there is the higher-order derivative test. It states that if $x$ is a real number such that $f'(x)=0$, and $n$ is the smallest natural number such that $f^(n)(x)neq 0$, then:

1. If $n$ is even and $f^(n)>0$, then $f$ has a local minimum at $x$.


2. If $n$ is even and $f^(n)<0$, then $f$ has a local manimum at $x$.


3. If $n$ is odd, then $f$ has an inflection point at $x$.

But the higher-order derivative test can also be inconclusive, if $f^(n)(x)=0$ for all $n$. My question, what can you do if the higher-order derivative test is inconclusive?

Is the first-derivative test the only option at that point, or are there other options?

EDIT: I’m interested in finding a method that depends only on the germ of $f$.

calculus real-analysis derivatives optimization maxima-minima

share|cite|improve this question

edited yesterday

asked 2 days ago

Keshav Srinivasan

1,88811338

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The case you are talking about is for the polynomials of degree $m<n$. However, if you see that $f^n$$(x)$ $=$ $0$ $forall nin N$, then then the function must be a constant one, right ?? @Keshav Srinivasan
  – Anik Bhowmick
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AnikBhowmick I'm not talking about polynomials, I'm talking about smooth functions in general. If you have an infinitely differentiable function $f$ whose derivatives of all orders evaluated at a given point $x$ is zero, then the higher-order derivative test fails. But that doesn't mean that $f$ is a constant function. For instance consider the example function given here: en.wikipedia.org/wiki/Non-analytic_smooth_function
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I assume you meant to ask what can you do if the higher-order derivative test is inconclusive?
  – N. F. Taussig
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @N.F.Taussig Yes
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

2

The second-derivative test states that if $x$ is a real number such that $f'(x)=0$, then:

1. If $f''(x)>0$, then $f$ has a local minimum at $x$.


2. If $f''(x)<0$, then $f$ has a local maximum at $x$.


3. If $f''(x)=0$, then the text is inconclusive.

But there's no need to despair if the second-derivative test is inconclusive, because there is the higher-order derivative test. It states that if $x$ is a real number such that $f'(x)=0$, and $n$ is the smallest natural number such that $f^(n)(x)neq 0$, then:

1. If $n$ is even and $f^(n)>0$, then $f$ has a local minimum at $x$.


2. If $n$ is even and $f^(n)<0$, then $f$ has a local manimum at $x$.


3. If $n$ is odd, then $f$ has an inflection point at $x$.

But the higher-order derivative test can also be inconclusive, if $f^(n)(x)=0$ for all $n$. My question, what can you do if the higher-order derivative test is inconclusive?

Is the first-derivative test the only option at that point, or are there other options?

EDIT: I’m interested in finding a method that depends only on the germ of $f$.

calculus real-analysis derivatives optimization maxima-minima

share|cite|improve this question

edited yesterday

asked 2 days ago

Keshav Srinivasan

1,88811338

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The case you are talking about is for the polynomials of degree $m<n$. However, if you see that $f^n$$(x)$ $=$ $0$ $forall nin N$, then then the function must be a constant one, right ?? @Keshav Srinivasan
  – Anik Bhowmick
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AnikBhowmick I'm not talking about polynomials, I'm talking about smooth functions in general. If you have an infinitely differentiable function $f$ whose derivatives of all orders evaluated at a given point $x$ is zero, then the higher-order derivative test fails. But that doesn't mean that $f$ is a constant function. For instance consider the example function given here: en.wikipedia.org/wiki/Non-analytic_smooth_function
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I assume you meant to ask what can you do if the higher-order derivative test is inconclusive?
  – N. F. Taussig
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @N.F.Taussig Yes
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

2

up vote
3
down vote

favorite

2

2

The second-derivative test states that if $x$ is a real number such that $f'(x)=0$, then:

1. If $f''(x)>0$, then $f$ has a local minimum at $x$.


2. If $f''(x)<0$, then $f$ has a local maximum at $x$.


3. If $f''(x)=0$, then the text is inconclusive.

But there's no need to despair if the second-derivative test is inconclusive, because there is the higher-order derivative test. It states that if $x$ is a real number such that $f'(x)=0$, and $n$ is the smallest natural number such that $f^(n)(x)neq 0$, then:

1. If $n$ is even and $f^(n)>0$, then $f$ has a local minimum at $x$.


2. If $n$ is even and $f^(n)<0$, then $f$ has a local manimum at $x$.


3. If $n$ is odd, then $f$ has an inflection point at $x$.

But the higher-order derivative test can also be inconclusive, if $f^(n)(x)=0$ for all $n$. My question, what can you do if the higher-order derivative test is inconclusive?

Is the first-derivative test the only option at that point, or are there other options?

EDIT: I’m interested in finding a method that depends only on the germ of $f$.

calculus real-analysis derivatives optimization maxima-minima

share|cite|improve this question

edited yesterday

asked 2 days ago

Keshav Srinivasan

1,88811338

The second-derivative test states that if $x$ is a real number such that $f'(x)=0$, then:

1. If $f''(x)>0$, then $f$ has a local minimum at $x$.


2. If $f''(x)<0$, then $f$ has a local maximum at $x$.


3. If $f''(x)=0$, then the text is inconclusive.

But there's no need to despair if the second-derivative test is inconclusive, because there is the higher-order derivative test. It states that if $x$ is a real number such that $f'(x)=0$, and $n$ is the smallest natural number such that $f^(n)(x)neq 0$, then:

1. If $n$ is even and $f^(n)>0$, then $f$ has a local minimum at $x$.


2. If $n$ is even and $f^(n)<0$, then $f$ has a local manimum at $x$.


3. If $n$ is odd, then $f$ has an inflection point at $x$.

But the higher-order derivative test can also be inconclusive, if $f^(n)(x)=0$ for all $n$. My question, what can you do if the higher-order derivative test is inconclusive?

Is the first-derivative test the only option at that point, or are there other options?

EDIT: I’m interested in finding a method that depends only on the germ of $f$.

calculus real-analysis derivatives optimization maxima-minima

share|cite|improve this question

edited yesterday

asked 2 days ago

Keshav Srinivasan

1,88811338

share|cite|improve this question

share|cite|improve this question

edited yesterday

edited yesterday

edited yesterday

asked 2 days ago

Keshav Srinivasan

1,88811338

asked 2 days ago

Keshav Srinivasan

1,88811338

asked 2 days ago

Keshav Srinivasan

1,88811338

1,88811338

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The case you are talking about is for the polynomials of degree $m<n$. However, if you see that $f^n$$(x)$ $=$ $0$ $forall nin N$, then then the function must be a constant one, right ?? @Keshav Srinivasan
  – Anik Bhowmick
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AnikBhowmick I'm not talking about polynomials, I'm talking about smooth functions in general. If you have an infinitely differentiable function $f$ whose derivatives of all orders evaluated at a given point $x$ is zero, then the higher-order derivative test fails. But that doesn't mean that $f$ is a constant function. For instance consider the example function given here: en.wikipedia.org/wiki/Non-analytic_smooth_function
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I assume you meant to ask what can you do if the higher-order derivative test is inconclusive?
  – N. F. Taussig
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @N.F.Taussig Yes
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The case you are talking about is for the polynomials of degree $m<n$. However, if you see that $f^n$$(x)$ $=$ $0$ $forall nin N$, then then the function must be a constant one, right ?? @Keshav Srinivasan
  – Anik Bhowmick
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AnikBhowmick I'm not talking about polynomials, I'm talking about smooth functions in general. If you have an infinitely differentiable function $f$ whose derivatives of all orders evaluated at a given point $x$ is zero, then the higher-order derivative test fails. But that doesn't mean that $f$ is a constant function. For instance consider the example function given here: en.wikipedia.org/wiki/Non-analytic_smooth_function
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I assume you meant to ask what can you do if the higher-order derivative test is inconclusive?
  – N. F. Taussig
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @N.F.Taussig Yes
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  

The case you are talking about is for the polynomials of degree $m<n$. However, if you see that $f^n$$(x)$ $=$ $0$ $forall nin N$, then then the function must be a constant one, right ?? @Keshav Srinivasan
– Anik Bhowmick
2 days ago

The case you are talking about is for the polynomials of degree $m<n$. However, if you see that $f^n$$(x)$ $=$ $0$ $forall nin N$, then then the function must be a constant one, right ?? @Keshav Srinivasan
– Anik Bhowmick
2 days ago

@AnikBhowmick I'm not talking about polynomials, I'm talking about smooth functions in general. If you have an infinitely differentiable function $f$ whose derivatives of all orders evaluated at a given point $x$ is zero, then the higher-order derivative test fails. But that doesn't mean that $f$ is a constant function. For instance consider the example function given here: en.wikipedia.org/wiki/Non-analytic_smooth_function
– Keshav Srinivasan
2 days ago

@AnikBhowmick I'm not talking about polynomials, I'm talking about smooth functions in general. If you have an infinitely differentiable function $f$ whose derivatives of all orders evaluated at a given point $x$ is zero, then the higher-order derivative test fails. But that doesn't mean that $f$ is a constant function. For instance consider the example function given here: en.wikipedia.org/wiki/Non-analytic_smooth_function
– Keshav Srinivasan
2 days ago

1

1

I assume you meant to ask what can you do if the higher-order derivative test is inconclusive?
– N. F. Taussig
2 days ago

I assume you meant to ask what can you do if the higher-order derivative test is inconclusive?
– N. F. Taussig
2 days ago

@N.F.Taussig Yes
– Keshav Srinivasan
2 days ago

@N.F.Taussig Yes
– Keshav Srinivasan
2 days ago

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote

The proofs of derivative tests for optimality are based on Taylor's theorem. The Taylor series of bump functions do not provide useful information, so the tests are not helpful. You can study the function values of a sequence that converges to $x$ to rule out maximality or minimality. If $f(x_i)<f(x)$ (or vice versa) for all $i$ and $x_irightarrow x$ as $irightarrowinfty$, then $x$ cannot be a minimum (or maximum).

share|cite|improve this answer

answered 2 days ago

Asdf

547

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there any way to do it other than actually looking at the values of the function outside the point? In principle there should be some way to do it that depends only on the germ of $f$.
  – Keshav Srinivasan
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  On a related note, I posted a question here about what information characterizes the germ of a smooth function: math.stackexchange.com/q/2870699/71829
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you have to look outside of the point.
  – Asdf
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But what about my point about the germ?
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't see how it makes a difference. If you only look at the values of $f^(n)$ at $x$, then $f$ could be any function in the germ. This includes functions which have a max, a min, or a saddle point at $x$. If the germ consists of analytic functions, then equal derivatives of all order implies that there is only one function (by Taylor's theorem).
  – Asdf
  2 days ago
  

  
  
  
  

 | 
show 6 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

The proofs of derivative tests for optimality are based on Taylor's theorem. The Taylor series of bump functions do not provide useful information, so the tests are not helpful. You can study the function values of a sequence that converges to $x$ to rule out maximality or minimality. If $f(x_i)<f(x)$ (or vice versa) for all $i$ and $x_irightarrow x$ as $irightarrowinfty$, then $x$ cannot be a minimum (or maximum).

share|cite|improve this answer

answered 2 days ago

Asdf

547

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there any way to do it other than actually looking at the values of the function outside the point? In principle there should be some way to do it that depends only on the germ of $f$.
  – Keshav Srinivasan
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  On a related note, I posted a question here about what information characterizes the germ of a smooth function: math.stackexchange.com/q/2870699/71829
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you have to look outside of the point.
  – Asdf
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But what about my point about the germ?
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't see how it makes a difference. If you only look at the values of $f^(n)$ at $x$, then $f$ could be any function in the germ. This includes functions which have a max, a min, or a saddle point at $x$. If the germ consists of analytic functions, then equal derivatives of all order implies that there is only one function (by Taylor's theorem).
  – Asdf
  2 days ago
  

  
  
  
  

 | 
show 6 more comments

up vote
1
down vote

The proofs of derivative tests for optimality are based on Taylor's theorem. The Taylor series of bump functions do not provide useful information, so the tests are not helpful. You can study the function values of a sequence that converges to $x$ to rule out maximality or minimality. If $f(x_i)<f(x)$ (or vice versa) for all $i$ and $x_irightarrow x$ as $irightarrowinfty$, then $x$ cannot be a minimum (or maximum).

share|cite|improve this answer

answered 2 days ago

Asdf

547

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there any way to do it other than actually looking at the values of the function outside the point? In principle there should be some way to do it that depends only on the germ of $f$.
  – Keshav Srinivasan
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  On a related note, I posted a question here about what information characterizes the germ of a smooth function: math.stackexchange.com/q/2870699/71829
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you have to look outside of the point.
  – Asdf
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But what about my point about the germ?
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't see how it makes a difference. If you only look at the values of $f^(n)$ at $x$, then $f$ could be any function in the germ. This includes functions which have a max, a min, or a saddle point at $x$. If the germ consists of analytic functions, then equal derivatives of all order implies that there is only one function (by Taylor's theorem).
  – Asdf
  2 days ago
  

  
  
  
  

 | 
show 6 more comments

up vote
1
down vote

up vote
1
down vote

The proofs of derivative tests for optimality are based on Taylor's theorem. The Taylor series of bump functions do not provide useful information, so the tests are not helpful. You can study the function values of a sequence that converges to $x$ to rule out maximality or minimality. If $f(x_i)<f(x)$ (or vice versa) for all $i$ and $x_irightarrow x$ as $irightarrowinfty$, then $x$ cannot be a minimum (or maximum).

share|cite|improve this answer

answered 2 days ago

Asdf

547

The proofs of derivative tests for optimality are based on Taylor's theorem. The Taylor series of bump functions do not provide useful information, so the tests are not helpful. You can study the function values of a sequence that converges to $x$ to rule out maximality or minimality. If $f(x_i)<f(x)$ (or vice versa) for all $i$ and $x_irightarrow x$ as $irightarrowinfty$, then $x$ cannot be a minimum (or maximum).

share|cite|improve this answer

answered 2 days ago

Asdf

547

share|cite|improve this answer

share|cite|improve this answer

answered 2 days ago

Asdf

547

answered 2 days ago

Asdf

547

answered 2 days ago

Asdf

547

547

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there any way to do it other than actually looking at the values of the function outside the point? In principle there should be some way to do it that depends only on the germ of $f$.
  – Keshav Srinivasan
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  On a related note, I posted a question here about what information characterizes the germ of a smooth function: math.stackexchange.com/q/2870699/71829
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you have to look outside of the point.
  – Asdf
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But what about my point about the germ?
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't see how it makes a difference. If you only look at the values of $f^(n)$ at $x$, then $f$ could be any function in the germ. This includes functions which have a max, a min, or a saddle point at $x$. If the germ consists of analytic functions, then equal derivatives of all order implies that there is only one function (by Taylor's theorem).
  – Asdf
  2 days ago
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there any way to do it other than actually looking at the values of the function outside the point? In principle there should be some way to do it that depends only on the germ of $f$.
  – Keshav Srinivasan
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  On a related note, I posted a question here about what information characterizes the germ of a smooth function: math.stackexchange.com/q/2870699/71829
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you have to look outside of the point.
  – Asdf
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  But what about my point about the germ?
  – Keshav Srinivasan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't see how it makes a difference. If you only look at the values of $f^(n)$ at $x$, then $f$ could be any function in the germ. This includes functions which have a max, a min, or a saddle point at $x$. If the germ consists of analytic functions, then equal derivatives of all order implies that there is only one function (by Taylor's theorem).
  – Asdf
  2 days ago
  

  
  
  
  

Is there any way to do it other than actually looking at the values of the function outside the point? In principle there should be some way to do it that depends only on the germ of $f$.
– Keshav Srinivasan
2 days ago

Is there any way to do it other than actually looking at the values of the function outside the point? In principle there should be some way to do it that depends only on the germ of $f$.
– Keshav Srinivasan
2 days ago

On a related note, I posted a question here about what information characterizes the germ of a smooth function: math.stackexchange.com/q/2870699/71829
– Keshav Srinivasan
2 days ago

On a related note, I posted a question here about what information characterizes the germ of a smooth function: math.stackexchange.com/q/2870699/71829
– Keshav Srinivasan
2 days ago

I think you have to look outside of the point.
– Asdf
2 days ago

I think you have to look outside of the point.
– Asdf
2 days ago

But what about my point about the germ?
– Keshav Srinivasan
2 days ago

But what about my point about the germ?
– Keshav Srinivasan
2 days ago

I don't see how it makes a difference. If you only look at the values of $f^(n)$ at $x$, then $f$ could be any function in the germ. This includes functions which have a max, a min, or a saddle point at $x$. If the germ consists of analytic functions, then equal derivatives of all order implies that there is only one function (by Taylor's theorem).
– Asdf
2 days ago

I don't see how it makes a difference. If you only look at the values of $f^(n)$ at $x$, then $f$ could be any function in the germ. This includes functions which have a max, a min, or a saddle point at $x$. If the germ consists of analytic functions, then equal derivatives of all order implies that there is only one function (by Taylor's theorem).
– Asdf
2 days ago

 | 
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