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

Derivatives and continuity of one variable functionsDerivative changes sign for continuous and differentiable functionCalculus 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 $f(c) = f(c + 1)$.A question on an inequality relating a function and its derivativeLet $f : Bbb R rightarrow Bbb R$ be a func such that $p>0$, that $f(x+p) = f(x)$ for all $x in Bbb R$ . Show that $f$ has an absolute max and minLet $f(x)=4x^3-3x^2-2x+1,$ 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 $f(x)$ be differentiable over $left[0,1right]$ with $f(0) = f(1) = 0$. Prove that $f'(x) - 2f(x)$ has a zero in $(0,1)$Justify 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 58 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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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 58 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 58 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 58 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 58 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 58 mins ago

Ivo Terek

47.2k954147

edited 58 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 1 hour 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
  54 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 59 mins ago

saulspatzsaulspatz

18.1k41636

$endgroup$

add a comment | 

2

$begingroup$

Love the answers. Another method.

Taylor expansion:
$$f(x) = f(0)+ f'(0)x+dots + f^{(n)}(0)frac{x^n}{n!}+frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1},$$

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

share|cite|improve this answer

answered 51 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 1 hour 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
  54 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 1 hour 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
  54 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 1 hour ago

Ivo TerekIvo Terek

47.2k954147

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Ivo TerekIvo Terek

47.2k954147

answered 1 hour ago

Ivo TerekIvo Terek

47.2k954147

answered 1 hour 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 59 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 59 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 59 mins ago

saulspatzsaulspatz

18.1k41636

$endgroup$

share|cite|improve this answer

answered 59 mins ago

saulspatzsaulspatz

18.1k41636

answered 59 mins ago

saulspatzsaulspatz

18.1k41636

answered 59 mins ago

saulspatzsaulspatz

18.1k41636

2

$begingroup$

Love the answers. Another method.

Taylor expansion:
$$f(x) = f(0)+ f'(0)x+dots + f^{(n)}(0)frac{x^n}{n!}+frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1},$$

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

share|cite|improve this answer

answered 51 mins ago

FnacoolFnacool

5,506612

$endgroup$

add a comment | 

2

$begingroup$

Love the answers. Another method.

Taylor expansion:
$$f(x) = f(0)+ f'(0)x+dots + f^{(n)}(0)frac{x^n}{n!}+frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1},$$

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

share|cite|improve this answer

answered 51 mins ago

FnacoolFnacool

5,506612

$endgroup$

add a comment | 

$begingroup$

Love the answers. Another method.

Taylor expansion:
$$f(x) = f(0)+ f'(0)x+dots + f^{(n)}(0)frac{x^n}{n!}+frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1},$$

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

share|cite|improve this answer

answered 51 mins ago

FnacoolFnacool

5,506612

$endgroup$

share|cite|improve this answer

answered 51 mins ago

FnacoolFnacool

5,506612

answered 51 mins ago

FnacoolFnacool

5,506612

answered 51 mins ago

FnacoolFnacool

5,506612