$f(x+1)-f(x)=f'(x)$: prove $f(x)$ linear function [closed]

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If I have a differentiable function $f:mathbbRtomathbbR$ satisfies $f(x+1)-f(x)=f'(x)$ and $lim_xtoinftyf'(x)=A$. Can I show $f(x)=ax+b$?

calculus differential-equations

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edited Jul 22 at 12:11

Bernard

110k635103

asked Jul 22 at 11:12

hctb

944310

closed as off-topic by Xander Henderson, user21820, Adrian Keister, choco_addicted, Parcly Taxel Jul 24 at 15:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, user21820, Adrian Keister, choco_addicted, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here is my guess : You can differentiate the given expression $f(x+1)−f(x)=f′(x)$ and then apply limit to the new expression(x tending to infinity) : $f'(x+1)−f'(x)=f′'(x)$ . You will get 0, from which we can conclude that f(x) has to be linear since its 2nd derivate is 0 .
  – Alphanerd
  Jul 22 at 11:22
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I don't see an answer to this question in any of the posts quoted above.
  – Kavi Rama Murthy
  Jul 22 at 12:11
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Note, the TITLE of this is a duplicate, but the actual question in the text is not.
  – GEdgar
  Jul 22 at 12:25
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  @GEdgar why did you vote to close then ?
  – Gabriel Romon
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @hctb What did you try to solve this?
  – Did
  Jul 22 at 13:23
  

  
  
  
  

 | 
show 7 more comments

up vote
8
down vote

favorite

5

If I have a differentiable function $f:mathbbRtomathbbR$ satisfies $f(x+1)-f(x)=f'(x)$ and $lim_xtoinftyf'(x)=A$. Can I show $f(x)=ax+b$?

calculus differential-equations

share|cite|improve this question

edited Jul 22 at 12:11

Bernard

110k635103

asked Jul 22 at 11:12

hctb

944310

closed as off-topic by Xander Henderson, user21820, Adrian Keister, choco_addicted, Parcly Taxel Jul 24 at 15:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, user21820, Adrian Keister, choco_addicted, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here is my guess : You can differentiate the given expression $f(x+1)−f(x)=f′(x)$ and then apply limit to the new expression(x tending to infinity) : $f'(x+1)−f'(x)=f′'(x)$ . You will get 0, from which we can conclude that f(x) has to be linear since its 2nd derivate is 0 .
  – Alphanerd
  Jul 22 at 11:22
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I don't see an answer to this question in any of the posts quoted above.
  – Kavi Rama Murthy
  Jul 22 at 12:11
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Note, the TITLE of this is a duplicate, but the actual question in the text is not.
  – GEdgar
  Jul 22 at 12:25
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  @GEdgar why did you vote to close then ?
  – Gabriel Romon
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @hctb What did you try to solve this?
  – Did
  Jul 22 at 13:23
  

  
  
  
  

 | 
show 7 more comments

up vote
8
down vote

favorite

5

up vote
8
down vote

favorite

5

5

If I have a differentiable function $f:mathbbRtomathbbR$ satisfies $f(x+1)-f(x)=f'(x)$ and $lim_xtoinftyf'(x)=A$. Can I show $f(x)=ax+b$?

calculus differential-equations

share|cite|improve this question

edited Jul 22 at 12:11

Bernard

110k635103

asked Jul 22 at 11:12

hctb

944310

If I have a differentiable function $f:mathbbRtomathbbR$ satisfies $f(x+1)-f(x)=f'(x)$ and $lim_xtoinftyf'(x)=A$. Can I show $f(x)=ax+b$?

calculus differential-equations

share|cite|improve this question

edited Jul 22 at 12:11

Bernard

110k635103

asked Jul 22 at 11:12

hctb

944310

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 12:11

Bernard

110k635103

edited Jul 22 at 12:11

Bernard

110k635103

edited Jul 22 at 12:11

Bernard

110k635103

110k635103

asked Jul 22 at 11:12

hctb

944310

asked Jul 22 at 11:12

hctb

944310

asked Jul 22 at 11:12

hctb

944310

944310

closed as off-topic by Xander Henderson, user21820, Adrian Keister, choco_addicted, Parcly Taxel Jul 24 at 15:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, user21820, Adrian Keister, choco_addicted, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Xander Henderson, user21820, Adrian Keister, choco_addicted, Parcly Taxel Jul 24 at 15:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, user21820, Adrian Keister, choco_addicted, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here is my guess : You can differentiate the given expression $f(x+1)−f(x)=f′(x)$ and then apply limit to the new expression(x tending to infinity) : $f'(x+1)−f'(x)=f′'(x)$ . You will get 0, from which we can conclude that f(x) has to be linear since its 2nd derivate is 0 .
  – Alphanerd
  Jul 22 at 11:22
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I don't see an answer to this question in any of the posts quoted above.
  – Kavi Rama Murthy
  Jul 22 at 12:11
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Note, the TITLE of this is a duplicate, but the actual question in the text is not.
  – GEdgar
  Jul 22 at 12:25
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  @GEdgar why did you vote to close then ?
  – Gabriel Romon
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @hctb What did you try to solve this?
  – Did
  Jul 22 at 13:23
  

  
  
  
  

 | 
show 7 more comments

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here is my guess : You can differentiate the given expression $f(x+1)−f(x)=f′(x)$ and then apply limit to the new expression(x tending to infinity) : $f'(x+1)−f'(x)=f′'(x)$ . You will get 0, from which we can conclude that f(x) has to be linear since its 2nd derivate is 0 .
  – Alphanerd
  Jul 22 at 11:22
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I don't see an answer to this question in any of the posts quoted above.
  – Kavi Rama Murthy
  Jul 22 at 12:11
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  Note, the TITLE of this is a duplicate, but the actual question in the text is not.
  – GEdgar
  Jul 22 at 12:25
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  @GEdgar why did you vote to close then ?
  – Gabriel Romon
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @hctb What did you try to solve this?
  – Did
  Jul 22 at 13:23
  

  
  
  
  

2

2

Here is my guess : You can differentiate the given expression $f(x+1)−f(x)=f′(x)$ and then apply limit to the new expression(x tending to infinity) : $f'(x+1)−f'(x)=f′'(x)$ . You will get 0, from which we can conclude that f(x) has to be linear since its 2nd derivate is 0 .
– Alphanerd
Jul 22 at 11:22

Here is my guess : You can differentiate the given expression $f(x+1)−f(x)=f′(x)$ and then apply limit to the new expression(x tending to infinity) : $f'(x+1)−f'(x)=f′'(x)$ . You will get 0, from which we can conclude that f(x) has to be linear since its 2nd derivate is 0 .
– Alphanerd
Jul 22 at 11:22

4

4

I don't see an answer to this question in any of the posts quoted above.
– Kavi Rama Murthy
Jul 22 at 12:11

I don't see an answer to this question in any of the posts quoted above.
– Kavi Rama Murthy
Jul 22 at 12:11

3

3

Note, the TITLE of this is a duplicate, but the actual question in the text is not.
– GEdgar
Jul 22 at 12:25

Note, the TITLE of this is a duplicate, but the actual question in the text is not.
– GEdgar
Jul 22 at 12:25

5

5

@GEdgar why did you vote to close then ?
– Gabriel Romon
Jul 22 at 12:27

@GEdgar why did you vote to close then ?
– Gabriel Romon
Jul 22 at 12:27

3

3

@hctb What did you try to solve this?
– Did
Jul 22 at 13:23

@hctb What did you try to solve this?
– Did
Jul 22 at 13:23

 | 
show 7 more comments

3 Answers
3

active

oldest

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up vote
14
down vote



accepted

Suppose $f$ is differentiable,
$$f(x+1)-f(x)=f'(x) tag1$$ for all $x$, and $lim_x to +infty f'(x) = A$.

I claim that $f'$ is constant, and therefore that $f$ has the form $ax+b$.

Suppose, for purposes of contradiction, that $f'$ is not constant. Then there is $x_0$ such that $f'(x_0) ne A$. Take the case $f'(x_0) > A$. [The other case $f'(x_0)<A$ is done the same way.]

Differentiate the equation $f(x+1)-f(x)=f'(x)$ to conclude that $f''$ exists and that $f'$ is continuous. Function $f'$ achieves a maximum value $B > A$ on $[x_0,+infty)$. The set where $f'(x)=B$ is nonempty, closed, and bounded above. Let $x_1 in [x_0,+infty)$ be such that $f'(x_1) = B$ and $f'(x) < B$ for all $x in (x_1,+infty)$.

Now note $f'(x) < B$ on $(x_1,x_1+1)$, so $f(x_1+1) - f(x_1) = int_x_1^x_1+1 f'(x);dx < B = f'(x_1)$. This contradicts ($1$).

share|cite|improve this answer

edited Jul 24 at 19:38

amWhy

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answered Jul 22 at 14:19

GEdgar

58.4k264163

add a comment | 

up vote
1
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Define $$g(x)=f(x)+ax+b$$therefore by substitution we get $$g(x+1)=g(x)$$which means that $g(x)$ is periodic with period $1$. Also $g'(x)=f'(x)+a$ and therefore has a limit in $infty$. Since $g'(x)$ is also periodic this is possible only if it is constant over a period or over $Bbb R$ because $$lim_xtoinftyg'(x)=g'(0)\lim_xtoinftyg'(x+a)=g'(a)\g'(a)=g'(0)$$for any $ain [0,1]$. So we have $$g'(x)=c$$concluding that $$g(x)=cx+d$$which means that $$Large f(x)=(c-a)x+d-b$$ or $Large f(x)text is linear$

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answered Jul 24 at 15:24

Mostafa Ayaz

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up vote
0
down vote

A thought

Let $x_0$ be any real number. By the condtion, we have $$f(x_0+1)-f(x_0)=f'(x_0).tag1$$ But by Lagrange mean value theorem, we may also obtain $$f(x_0+1)-f(x_0)=f'(x_1)(x_0+1-x_0)=f'(x_1),$$where $x_0<x_1<x_0+1.tag 2$
Combining $(1)$ and $(2)$, we may claim there exists $x_1 in (x_0,x_0+1)$ such that $$f'(x_0)=f'(x_1).$$

Consider $x_1$. By the condition, we also have $$f(x_1+1)-f(x_1)=f'(x_1).tag3$$
Likewise, we may also claim there exists $x_2 in (x_1,x_1+1)$ such that $$f'(x_1)=f'(x_2).tag4$$

Now, repeat the process above. You may find a sequence of $$x_0<x_1<x_2<cdots<x_n$$ such that $$f'(x_0)=f'(x_1)=f'(x_2)=cdots=f'(x_n).$$
Obviously, if we may prove that $x_n to +infty$ as $n to +infty$, then by the fact that $ limlimits_xtoinftyf'(x)=A,$ we have $limlimits_nto+inftyf'(x_n)=A.$ But $f'(x_n)$ is a constant sequence, thus $$f'(x_0)=f'(x_1)=f'(x_2)=cdots=f'(x_n) =A.$$

Recall the arbitrariness of $x_0$. We may claim that $$f'(x)equiv A,~~~~forall x in mathbbR,$$ which implies that $f(x)=Ax+b$ for all $x in mathbbR.$

But can we prove $x_n to +infty$ as $n to +infty$?

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edited Jul 22 at 15:50

answered Jul 22 at 15:20

mengdie1982

2,897216

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
14
down vote



accepted

Suppose $f$ is differentiable,
$$f(x+1)-f(x)=f'(x) tag1$$ for all $x$, and $lim_x to +infty f'(x) = A$.

I claim that $f'$ is constant, and therefore that $f$ has the form $ax+b$.

Suppose, for purposes of contradiction, that $f'$ is not constant. Then there is $x_0$ such that $f'(x_0) ne A$. Take the case $f'(x_0) > A$. [The other case $f'(x_0)<A$ is done the same way.]

Differentiate the equation $f(x+1)-f(x)=f'(x)$ to conclude that $f''$ exists and that $f'$ is continuous. Function $f'$ achieves a maximum value $B > A$ on $[x_0,+infty)$. The set where $f'(x)=B$ is nonempty, closed, and bounded above. Let $x_1 in [x_0,+infty)$ be such that $f'(x_1) = B$ and $f'(x) < B$ for all $x in (x_1,+infty)$.

Now note $f'(x) < B$ on $(x_1,x_1+1)$, so $f(x_1+1) - f(x_1) = int_x_1^x_1+1 f'(x);dx < B = f'(x_1)$. This contradicts ($1$).

share|cite|improve this answer

edited Jul 24 at 19:38

amWhy

189k25219431

answered Jul 22 at 14:19

GEdgar

58.4k264163

add a comment | 

up vote
14
down vote



accepted

Suppose $f$ is differentiable,
$$f(x+1)-f(x)=f'(x) tag1$$ for all $x$, and $lim_x to +infty f'(x) = A$.

I claim that $f'$ is constant, and therefore that $f$ has the form $ax+b$.

Suppose, for purposes of contradiction, that $f'$ is not constant. Then there is $x_0$ such that $f'(x_0) ne A$. Take the case $f'(x_0) > A$. [The other case $f'(x_0)<A$ is done the same way.]

Differentiate the equation $f(x+1)-f(x)=f'(x)$ to conclude that $f''$ exists and that $f'$ is continuous. Function $f'$ achieves a maximum value $B > A$ on $[x_0,+infty)$. The set where $f'(x)=B$ is nonempty, closed, and bounded above. Let $x_1 in [x_0,+infty)$ be such that $f'(x_1) = B$ and $f'(x) < B$ for all $x in (x_1,+infty)$.

Now note $f'(x) < B$ on $(x_1,x_1+1)$, so $f(x_1+1) - f(x_1) = int_x_1^x_1+1 f'(x);dx < B = f'(x_1)$. This contradicts ($1$).

share|cite|improve this answer

edited Jul 24 at 19:38

amWhy

189k25219431

answered Jul 22 at 14:19

GEdgar

58.4k264163

add a comment | 

up vote
14
down vote



accepted

up vote
14
down vote



accepted

Suppose $f$ is differentiable,
$$f(x+1)-f(x)=f'(x) tag1$$ for all $x$, and $lim_x to +infty f'(x) = A$.

I claim that $f'$ is constant, and therefore that $f$ has the form $ax+b$.

Suppose, for purposes of contradiction, that $f'$ is not constant. Then there is $x_0$ such that $f'(x_0) ne A$. Take the case $f'(x_0) > A$. [The other case $f'(x_0)<A$ is done the same way.]

Differentiate the equation $f(x+1)-f(x)=f'(x)$ to conclude that $f''$ exists and that $f'$ is continuous. Function $f'$ achieves a maximum value $B > A$ on $[x_0,+infty)$. The set where $f'(x)=B$ is nonempty, closed, and bounded above. Let $x_1 in [x_0,+infty)$ be such that $f'(x_1) = B$ and $f'(x) < B$ for all $x in (x_1,+infty)$.

Now note $f'(x) < B$ on $(x_1,x_1+1)$, so $f(x_1+1) - f(x_1) = int_x_1^x_1+1 f'(x);dx < B = f'(x_1)$. This contradicts ($1$).

share|cite|improve this answer

edited Jul 24 at 19:38

amWhy

189k25219431

answered Jul 22 at 14:19

GEdgar

58.4k264163

Suppose $f$ is differentiable,
$$f(x+1)-f(x)=f'(x) tag1$$ for all $x$, and $lim_x to +infty f'(x) = A$.

I claim that $f'$ is constant, and therefore that $f$ has the form $ax+b$.

Suppose, for purposes of contradiction, that $f'$ is not constant. Then there is $x_0$ such that $f'(x_0) ne A$. Take the case $f'(x_0) > A$. [The other case $f'(x_0)<A$ is done the same way.]

Differentiate the equation $f(x+1)-f(x)=f'(x)$ to conclude that $f''$ exists and that $f'$ is continuous. Function $f'$ achieves a maximum value $B > A$ on $[x_0,+infty)$. The set where $f'(x)=B$ is nonempty, closed, and bounded above. Let $x_1 in [x_0,+infty)$ be such that $f'(x_1) = B$ and $f'(x) < B$ for all $x in (x_1,+infty)$.

Now note $f'(x) < B$ on $(x_1,x_1+1)$, so $f(x_1+1) - f(x_1) = int_x_1^x_1+1 f'(x);dx < B = f'(x_1)$. This contradicts ($1$).

share|cite|improve this answer

edited Jul 24 at 19:38

amWhy

189k25219431

answered Jul 22 at 14:19

GEdgar

58.4k264163

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 19:38

amWhy

189k25219431

edited Jul 24 at 19:38

amWhy

189k25219431

edited Jul 24 at 19:38

amWhy

189k25219431

189k25219431

answered Jul 22 at 14:19

GEdgar

58.4k264163

answered Jul 22 at 14:19

GEdgar

58.4k264163

answered Jul 22 at 14:19

GEdgar

58.4k264163

58.4k264163

add a comment | 

add a comment | 

up vote
1
down vote

Define $$g(x)=f(x)+ax+b$$therefore by substitution we get $$g(x+1)=g(x)$$which means that $g(x)$ is periodic with period $1$. Also $g'(x)=f'(x)+a$ and therefore has a limit in $infty$. Since $g'(x)$ is also periodic this is possible only if it is constant over a period or over $Bbb R$ because $$lim_xtoinftyg'(x)=g'(0)\lim_xtoinftyg'(x+a)=g'(a)\g'(a)=g'(0)$$for any $ain [0,1]$. So we have $$g'(x)=c$$concluding that $$g(x)=cx+d$$which means that $$Large f(x)=(c-a)x+d-b$$ or $Large f(x)text is linear$

share|cite|improve this answer

answered Jul 24 at 15:24

Mostafa Ayaz

8,5773630

add a comment | 

up vote
1
down vote

Define $$g(x)=f(x)+ax+b$$therefore by substitution we get $$g(x+1)=g(x)$$which means that $g(x)$ is periodic with period $1$. Also $g'(x)=f'(x)+a$ and therefore has a limit in $infty$. Since $g'(x)$ is also periodic this is possible only if it is constant over a period or over $Bbb R$ because $$lim_xtoinftyg'(x)=g'(0)\lim_xtoinftyg'(x+a)=g'(a)\g'(a)=g'(0)$$for any $ain [0,1]$. So we have $$g'(x)=c$$concluding that $$g(x)=cx+d$$which means that $$Large f(x)=(c-a)x+d-b$$ or $Large f(x)text is linear$

share|cite|improve this answer

answered Jul 24 at 15:24

Mostafa Ayaz

8,5773630

add a comment | 

up vote
1
down vote

up vote
1
down vote

Define $$g(x)=f(x)+ax+b$$therefore by substitution we get $$g(x+1)=g(x)$$which means that $g(x)$ is periodic with period $1$. Also $g'(x)=f'(x)+a$ and therefore has a limit in $infty$. Since $g'(x)$ is also periodic this is possible only if it is constant over a period or over $Bbb R$ because $$lim_xtoinftyg'(x)=g'(0)\lim_xtoinftyg'(x+a)=g'(a)\g'(a)=g'(0)$$for any $ain [0,1]$. So we have $$g'(x)=c$$concluding that $$g(x)=cx+d$$which means that $$Large f(x)=(c-a)x+d-b$$ or $Large f(x)text is linear$

share|cite|improve this answer

answered Jul 24 at 15:24

Mostafa Ayaz

8,5773630

Define $$g(x)=f(x)+ax+b$$therefore by substitution we get $$g(x+1)=g(x)$$which means that $g(x)$ is periodic with period $1$. Also $g'(x)=f'(x)+a$ and therefore has a limit in $infty$. Since $g'(x)$ is also periodic this is possible only if it is constant over a period or over $Bbb R$ because $$lim_xtoinftyg'(x)=g'(0)\lim_xtoinftyg'(x+a)=g'(a)\g'(a)=g'(0)$$for any $ain [0,1]$. So we have $$g'(x)=c$$concluding that $$g(x)=cx+d$$which means that $$Large f(x)=(c-a)x+d-b$$ or $Large f(x)text is linear$

share|cite|improve this answer

answered Jul 24 at 15:24

Mostafa Ayaz

8,5773630

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 15:24

Mostafa Ayaz

8,5773630

answered Jul 24 at 15:24

Mostafa Ayaz

8,5773630

answered Jul 24 at 15:24

Mostafa Ayaz

8,5773630

8,5773630

add a comment | 

add a comment | 

up vote
0
down vote

A thought

Let $x_0$ be any real number. By the condtion, we have $$f(x_0+1)-f(x_0)=f'(x_0).tag1$$ But by Lagrange mean value theorem, we may also obtain $$f(x_0+1)-f(x_0)=f'(x_1)(x_0+1-x_0)=f'(x_1),$$where $x_0<x_1<x_0+1.tag 2$
Combining $(1)$ and $(2)$, we may claim there exists $x_1 in (x_0,x_0+1)$ such that $$f'(x_0)=f'(x_1).$$

Consider $x_1$. By the condition, we also have $$f(x_1+1)-f(x_1)=f'(x_1).tag3$$
Likewise, we may also claim there exists $x_2 in (x_1,x_1+1)$ such that $$f'(x_1)=f'(x_2).tag4$$

Now, repeat the process above. You may find a sequence of $$x_0<x_1<x_2<cdots<x_n$$ such that $$f'(x_0)=f'(x_1)=f'(x_2)=cdots=f'(x_n).$$
Obviously, if we may prove that $x_n to +infty$ as $n to +infty$, then by the fact that $ limlimits_xtoinftyf'(x)=A,$ we have $limlimits_nto+inftyf'(x_n)=A.$ But $f'(x_n)$ is a constant sequence, thus $$f'(x_0)=f'(x_1)=f'(x_2)=cdots=f'(x_n) =A.$$

Recall the arbitrariness of $x_0$. We may claim that $$f'(x)equiv A,~~~~forall x in mathbbR,$$ which implies that $f(x)=Ax+b$ for all $x in mathbbR.$

But can we prove $x_n to +infty$ as $n to +infty$?

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edited Jul 22 at 15:50

answered Jul 22 at 15:20

mengdie1982

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A thought

Let $x_0$ be any real number. By the condtion, we have $$f(x_0+1)-f(x_0)=f'(x_0).tag1$$ But by Lagrange mean value theorem, we may also obtain $$f(x_0+1)-f(x_0)=f'(x_1)(x_0+1-x_0)=f'(x_1),$$where $x_0<x_1<x_0+1.tag 2$
Combining $(1)$ and $(2)$, we may claim there exists $x_1 in (x_0,x_0+1)$ such that $$f'(x_0)=f'(x_1).$$

Consider $x_1$. By the condition, we also have $$f(x_1+1)-f(x_1)=f'(x_1).tag3$$
Likewise, we may also claim there exists $x_2 in (x_1,x_1+1)$ such that $$f'(x_1)=f'(x_2).tag4$$

Now, repeat the process above. You may find a sequence of $$x_0<x_1<x_2<cdots<x_n$$ such that $$f'(x_0)=f'(x_1)=f'(x_2)=cdots=f'(x_n).$$
Obviously, if we may prove that $x_n to +infty$ as $n to +infty$, then by the fact that $ limlimits_xtoinftyf'(x)=A,$ we have $limlimits_nto+inftyf'(x_n)=A.$ But $f'(x_n)$ is a constant sequence, thus $$f'(x_0)=f'(x_1)=f'(x_2)=cdots=f'(x_n) =A.$$

Recall the arbitrariness of $x_0$. We may claim that $$f'(x)equiv A,~~~~forall x in mathbbR,$$ which implies that $f(x)=Ax+b$ for all $x in mathbbR.$

But can we prove $x_n to +infty$ as $n to +infty$?

share|cite|improve this answer

edited Jul 22 at 15:50

answered Jul 22 at 15:20

mengdie1982

2,897216

add a comment | 

up vote
0
down vote

up vote
0
down vote

A thought

Let $x_0$ be any real number. By the condtion, we have $$f(x_0+1)-f(x_0)=f'(x_0).tag1$$ But by Lagrange mean value theorem, we may also obtain $$f(x_0+1)-f(x_0)=f'(x_1)(x_0+1-x_0)=f'(x_1),$$where $x_0<x_1<x_0+1.tag 2$
Combining $(1)$ and $(2)$, we may claim there exists $x_1 in (x_0,x_0+1)$ such that $$f'(x_0)=f'(x_1).$$

Consider $x_1$. By the condition, we also have $$f(x_1+1)-f(x_1)=f'(x_1).tag3$$
Likewise, we may also claim there exists $x_2 in (x_1,x_1+1)$ such that $$f'(x_1)=f'(x_2).tag4$$

Now, repeat the process above. You may find a sequence of $$x_0<x_1<x_2<cdots<x_n$$ such that $$f'(x_0)=f'(x_1)=f'(x_2)=cdots=f'(x_n).$$
Obviously, if we may prove that $x_n to +infty$ as $n to +infty$, then by the fact that $ limlimits_xtoinftyf'(x)=A,$ we have $limlimits_nto+inftyf'(x_n)=A.$ But $f'(x_n)$ is a constant sequence, thus $$f'(x_0)=f'(x_1)=f'(x_2)=cdots=f'(x_n) =A.$$

Recall the arbitrariness of $x_0$. We may claim that $$f'(x)equiv A,~~~~forall x in mathbbR,$$ which implies that $f(x)=Ax+b$ for all $x in mathbbR.$

But can we prove $x_n to +infty$ as $n to +infty$?

share|cite|improve this answer

edited Jul 22 at 15:50

answered Jul 22 at 15:20

mengdie1982

2,897216

A thought

Let $x_0$ be any real number. By the condtion, we have $$f(x_0+1)-f(x_0)=f'(x_0).tag1$$ But by Lagrange mean value theorem, we may also obtain $$f(x_0+1)-f(x_0)=f'(x_1)(x_0+1-x_0)=f'(x_1),$$where $x_0<x_1<x_0+1.tag 2$
Combining $(1)$ and $(2)$, we may claim there exists $x_1 in (x_0,x_0+1)$ such that $$f'(x_0)=f'(x_1).$$

Consider $x_1$. By the condition, we also have $$f(x_1+1)-f(x_1)=f'(x_1).tag3$$
Likewise, we may also claim there exists $x_2 in (x_1,x_1+1)$ such that $$f'(x_1)=f'(x_2).tag4$$

Now, repeat the process above. You may find a sequence of $$x_0<x_1<x_2<cdots<x_n$$ such that $$f'(x_0)=f'(x_1)=f'(x_2)=cdots=f'(x_n).$$
Obviously, if we may prove that $x_n to +infty$ as $n to +infty$, then by the fact that $ limlimits_xtoinftyf'(x)=A,$ we have $limlimits_nto+inftyf'(x_n)=A.$ But $f'(x_n)$ is a constant sequence, thus $$f'(x_0)=f'(x_1)=f'(x_2)=cdots=f'(x_n) =A.$$

Recall the arbitrariness of $x_0$. We may claim that $$f'(x)equiv A,~~~~forall x in mathbbR,$$ which implies that $f(x)=Ax+b$ for all $x in mathbbR.$

But can we prove $x_n to +infty$ as $n to +infty$?

share|cite|improve this answer

edited Jul 22 at 15:50

answered Jul 22 at 15:20

mengdie1982

2,897216

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 15:50

edited Jul 22 at 15:50

edited Jul 22 at 15:50

answered Jul 22 at 15:20

mengdie1982

2,897216

answered Jul 22 at 15:20

mengdie1982

2,897216

answered Jul 22 at 15:20

mengdie1982

2,897216

2,897216

add a comment | 

add a comment |