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Proving n+1 th differential as zero given lower differentials are 0

If $f(x)=f'(x)+f''(x)$ then show that $f(x)=0$Derivative changes sign for continuous and differentiable functionA problem on Mean Value TheoremCalculus Three time differentiableTrue or false: If $f(x)$ is continuous on $[0, 2]$ and $f(0)=f(2)$, then there exists a number $cin [0, 1]$ such that $f(c) = f(c + 1)$.Rolle's theorem $beta cdot f(x)+f'(x)=0$Let $f(x)=4x^3-3x^2-2x+1,$ use Rolle's theorem to prove that there exist $c,0<c<1$ such that $f(c)=0$If $f(a)=f(b)=0$, then $f'(c)+f(c)g'(c)=0,$ for some $cin(a,b)$Let f be a continuous and differentiable function such that f(a)=f(b)=0Justify differentiability for a parametric function

4

3

$begingroup$

Following is a question I am stuck in.

Let $f : Bbb R to Bbb R$ be an infinitely differentiable function and suppose that for some $n ≥ 1$,
$$f(1) = f(0) = f^(1)(0) = f^(2)(0) = · · · = f^(n)(0) = 0$$
Prove that there exists $x in (0, 1)$ such that $f^(n+1)(x) = 0$.

It is a past question of an entrance exam.

I thought to use Rolle's Theorem, but this requires information about $f^(n) (1)$ that I am unable to get. Only information about behaviour at $1$ I have is $f(1)=0$

real-analysis calculus

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edited 52 mins ago

Ivo Terek

47.2k954147

asked 1 hour ago

A.S. GhoshA.S. Ghosh

232

New contributor

A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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add a comment | 

4

3

$begingroup$

Following is a question I am stuck in.

Let $f : Bbb R to Bbb R$ be an infinitely differentiable function and suppose that for some $n ≥ 1$,
$$f(1) = f(0) = f^(1)(0) = f^(2)(0) = · · · = f^(n)(0) = 0$$
Prove that there exists $x in (0, 1)$ such that $f^(n+1)(x) = 0$.

It is a past question of an entrance exam.

I thought to use Rolle's Theorem, but this requires information about $f^(n) (1)$ that I am unable to get. Only information about behaviour at $1$ I have is $f(1)=0$

real-analysis calculus

share|cite|improve this question

edited 52 mins ago

Ivo Terek

47.2k954147

asked 1 hour ago

A.S. GhoshA.S. Ghosh

232

New contributor

A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Following is a question I am stuck in.

Let $f : Bbb R to Bbb R$ be an infinitely differentiable function and suppose that for some $n ≥ 1$,
$$f(1) = f(0) = f^(1)(0) = f^(2)(0) = · · · = f^(n)(0) = 0$$
Prove that there exists $x in (0, 1)$ such that $f^(n+1)(x) = 0$.

It is a past question of an entrance exam.

I thought to use Rolle's Theorem, but this requires information about $f^(n) (1)$ that I am unable to get. Only information about behaviour at $1$ I have is $f(1)=0$

real-analysis calculus

share|cite|improve this question

edited 52 mins ago

Ivo Terek

47.2k954147

asked 1 hour ago

A.S. GhoshA.S. Ghosh

232

New contributor

A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 52 mins ago

Ivo Terek

47.2k954147

asked 1 hour ago

A.S. GhoshA.S. Ghosh

232

New contributor

A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 52 mins ago

Ivo Terek

47.2k954147

asked 1 hour ago

A.S. GhoshA.S. Ghosh

232

New contributor

A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 52 mins ago

Ivo Terek

47.2k954147

edited 52 mins ago

Ivo Terek

47.2k954147

asked 1 hour ago

A.S. GhoshA.S. Ghosh

232

New contributor

A.S. Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

A.S. GhoshA.S. Ghosh

232

3 Answers
3

active

oldest

votes

3

$begingroup$

Hint: Well, think small. For $n=0$ it is trivial. So try to think about what happens for $n=1$ first. The result would read

If $f(1) = f(0) = f'(0) = 0$, there is $x in (0,1)$ such that $f''(x) = 0$.

How would one go about this? Ok, if $f(1) = f(0)$ there is $y in (0,1)$ such that $f'(y) = 0$. Now there is $x in (0,y)subseteq (0,1)$ such that $f''(x) = 0$, by Rolle's theorem again. Do you see the pattern?

share|cite|improve this answer

answered 54 mins ago

Ivo TerekIvo Terek

47.2k954147

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks...I was fixated on fiding values at 1 that I forgot to consider intermediate values...
  $endgroup$
  – A.S. Ghosh
  48 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Since $f(0)=f(1)=0, exists cin(0,1)$ such that $f'(c)= 0.$ Now since $f'(0) = f'(c)=0, exists d in (0,c)$ such that $f''(d)=0.$ Proceed.

share|cite|improve this answer

answered 52 mins ago

saulspatzsaulspatz

18.1k41636

$endgroup$

add a comment | 

2

$begingroup$

Love the answers. Another method.

Taylor expansion:
$$f(x) = f(0)+ f'(0)x+dots + f^(n)(0)fracx^nn!+fracf^(n+1)(c)(n+1)!x^n+1,$$

$c$ is in the open interval between $0$ and $x$.

share|cite|improve this answer

answered 45 mins ago

FnacoolFnacool

5,506612

$endgroup$

add a comment | 

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3

$begingroup$

Hint: Well, think small. For $n=0$ it is trivial. So try to think about what happens for $n=1$ first. The result would read

If $f(1) = f(0) = f'(0) = 0$, there is $x in (0,1)$ such that $f''(x) = 0$.

How would one go about this? Ok, if $f(1) = f(0)$ there is $y in (0,1)$ such that $f'(y) = 0$. Now there is $x in (0,y)subseteq (0,1)$ such that $f''(x) = 0$, by Rolle's theorem again. Do you see the pattern?

share|cite|improve this answer

answered 54 mins ago

Ivo TerekIvo Terek

47.2k954147

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks...I was fixated on fiding values at 1 that I forgot to consider intermediate values...
  $endgroup$
  – A.S. Ghosh
  48 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Hint: Well, think small. For $n=0$ it is trivial. So try to think about what happens for $n=1$ first. The result would read

If $f(1) = f(0) = f'(0) = 0$, there is $x in (0,1)$ such that $f''(x) = 0$.

How would one go about this? Ok, if $f(1) = f(0)$ there is $y in (0,1)$ such that $f'(y) = 0$. Now there is $x in (0,y)subseteq (0,1)$ such that $f''(x) = 0$, by Rolle's theorem again. Do you see the pattern?

share|cite|improve this answer

answered 54 mins ago

Ivo TerekIvo Terek

47.2k954147

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks...I was fixated on fiding values at 1 that I forgot to consider intermediate values...
  $endgroup$
  – A.S. Ghosh
  48 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Well, think small. For $n=0$ it is trivial. So try to think about what happens for $n=1$ first. The result would read

If $f(1) = f(0) = f'(0) = 0$, there is $x in (0,1)$ such that $f''(x) = 0$.

How would one go about this? Ok, if $f(1) = f(0)$ there is $y in (0,1)$ such that $f'(y) = 0$. Now there is $x in (0,y)subseteq (0,1)$ such that $f''(x) = 0$, by Rolle's theorem again. Do you see the pattern?

share|cite|improve this answer

answered 54 mins ago

Ivo TerekIvo Terek

47.2k954147

$endgroup$

share|cite|improve this answer

answered 54 mins ago

Ivo TerekIvo Terek

47.2k954147

answered 54 mins ago

Ivo TerekIvo Terek

47.2k954147

answered 54 mins ago

Ivo TerekIvo Terek

47.2k954147

2

$begingroup$

Since $f(0)=f(1)=0, exists cin(0,1)$ such that $f'(c)= 0.$ Now since $f'(0) = f'(c)=0, exists d in (0,c)$ such that $f''(d)=0.$ Proceed.

share|cite|improve this answer

answered 52 mins ago

saulspatzsaulspatz

18.1k41636

$endgroup$

add a comment | 

2

$begingroup$

Since $f(0)=f(1)=0, exists cin(0,1)$ such that $f'(c)= 0.$ Now since $f'(0) = f'(c)=0, exists d in (0,c)$ such that $f''(d)=0.$ Proceed.

share|cite|improve this answer

answered 52 mins ago

saulspatzsaulspatz

18.1k41636

$endgroup$

add a comment | 

$begingroup$

Since $f(0)=f(1)=0, exists cin(0,1)$ such that $f'(c)= 0.$ Now since $f'(0) = f'(c)=0, exists d in (0,c)$ such that $f''(d)=0.$ Proceed.

share|cite|improve this answer

answered 52 mins ago

saulspatzsaulspatz

18.1k41636

$endgroup$

share|cite|improve this answer

answered 52 mins ago

saulspatzsaulspatz

18.1k41636

answered 52 mins ago

saulspatzsaulspatz

18.1k41636

answered 52 mins ago

saulspatzsaulspatz

18.1k41636

2

$begingroup$

Love the answers. Another method.

Taylor expansion:
$$f(x) = f(0)+ f'(0)x+dots + f^(n)(0)fracx^nn!+fracf^(n+1)(c)(n+1)!x^n+1,$$

$c$ is in the open interval between $0$ and $x$.

share|cite|improve this answer

answered 45 mins ago

FnacoolFnacool

5,506612

$endgroup$

add a comment | 

2

$begingroup$

Love the answers. Another method.

Taylor expansion:
$$f(x) = f(0)+ f'(0)x+dots + f^(n)(0)fracx^nn!+fracf^(n+1)(c)(n+1)!x^n+1,$$

$c$ is in the open interval between $0$ and $x$.

share|cite|improve this answer

answered 45 mins ago

FnacoolFnacool

5,506612

$endgroup$

add a comment | 

$begingroup$

Love the answers. Another method.

Taylor expansion:
$$f(x) = f(0)+ f'(0)x+dots + f^(n)(0)fracx^nn!+fracf^(n+1)(c)(n+1)!x^n+1,$$

$c$ is in the open interval between $0$ and $x$.

share|cite|improve this answer

answered 45 mins ago

FnacoolFnacool

5,506612

$endgroup$

share|cite|improve this answer

answered 45 mins ago

FnacoolFnacool

5,506612

answered 45 mins ago

FnacoolFnacool

5,506612

answered 45 mins ago

FnacoolFnacool

5,506612