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How to prove by mathematical induction that $(y-x)x^n leq frac{y^{n+1}-x^{n+1}}{n+1} leq (y-x)y^n$? [closed]

0

1

$begingroup$

Prove by induction that:
$$(y-x)x^n leq frac{y^{n+1}-x^{n+1}}{n+1} leq (y-x)y^n .
$$

---

As a hint, the professor told us to use the following expression that we had previously proven:
$$sum_{i=0}^n{x^i}= frac{1-x^{n+1}}{1-x}
$$

I already tried several things, but I can’t manage to get to the solution.

inequality induction

share|cite|improve this question

edited Mar 23 at 14:04

Jack

27.7k1784204

asked Mar 22 at 13:30

Facu50196Facu50196

396

$endgroup$

closed as off-topic by Saad, John Omielan, user21820, RRL, Alexander Gruber♦ Mar 24 at 2:54

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, John Omielan, user21820, RRL, Alexander Gruber

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I suppose you assume that $y>x$?
  $endgroup$
  – Yanko
  Mar 22 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It definitely holds for $n=0$.
  $endgroup$
  – Wuestenfux
  Mar 22 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes sorry. It’s for 0<x<y
  $endgroup$
  – Facu50196
  Mar 22 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And what exactly have you tried?
  $endgroup$
  – Saad
  Mar 22 at 13:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Were you really told to use induction and to use the result you had previously proved (presumably by using induction)? I can see no need to use both. You could use induction, but this would probably duplicate the proof of the previous result. You can use the previous result directly (i.e. without using induction again) by observing that if $x < 1$, then the left hand side is less than $n+1$ and greater than $(n+1)x^n$.
  $endgroup$
  – Calum Gilhooley
  Mar 23 at 15:37
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Prove by induction that:
$$(y-x)x^n leq frac{y^{n+1}-x^{n+1}}{n+1} leq (y-x)y^n .
$$

---

As a hint, the professor told us to use the following expression that we had previously proven:
$$sum_{i=0}^n{x^i}= frac{1-x^{n+1}}{1-x}
$$

I already tried several things, but I can’t manage to get to the solution.

inequality induction

share|cite|improve this question

edited Mar 23 at 14:04

Jack

27.7k1784204

asked Mar 22 at 13:30

Facu50196Facu50196

396

$endgroup$

closed as off-topic by Saad, John Omielan, user21820, RRL, Alexander Gruber♦ Mar 24 at 2:54

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, John Omielan, user21820, RRL, Alexander Gruber

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I suppose you assume that $y>x$?
  $endgroup$
  – Yanko
  Mar 22 at 13:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It definitely holds for $n=0$.
  $endgroup$
  – Wuestenfux
  Mar 22 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes sorry. It’s for 0<x<y
  $endgroup$
  – Facu50196
  Mar 22 at 13:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And what exactly have you tried?
  $endgroup$
  – Saad
  Mar 22 at 13:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Were you really told to use induction and to use the result you had previously proved (presumably by using induction)? I can see no need to use both. You could use induction, but this would probably duplicate the proof of the previous result. You can use the previous result directly (i.e. without using induction again) by observing that if $x < 1$, then the left hand side is less than $n+1$ and greater than $(n+1)x^n$.
  $endgroup$
  – Calum Gilhooley
  Mar 23 at 15:37
  

  
  
  
  

add a comment | 

$begingroup$

Prove by induction that:
$$(y-x)x^n leq frac{y^{n+1}-x^{n+1}}{n+1} leq (y-x)y^n .
$$

---

As a hint, the professor told us to use the following expression that we had previously proven:
$$sum_{i=0}^n{x^i}= frac{1-x^{n+1}}{1-x}
$$

I already tried several things, but I can’t manage to get to the solution.

inequality induction

share|cite|improve this question

edited Mar 23 at 14:04

Jack

27.7k1784204

asked Mar 22 at 13:30

Facu50196Facu50196

396

$endgroup$

share|cite|improve this question

edited Mar 23 at 14:04

Jack

27.7k1784204

asked Mar 22 at 13:30

Facu50196Facu50196

396

share|cite|improve this question

edited Mar 23 at 14:04

Jack

27.7k1784204

asked Mar 22 at 13:30

Facu50196Facu50196

396

edited Mar 23 at 14:04

Jack

27.7k1784204

edited Mar 23 at 14:04

Jack

27.7k1784204

asked Mar 22 at 13:30

Facu50196Facu50196

396

asked Mar 22 at 13:30

Facu50196Facu50196

396

1 Answer
1

active

oldest

votes

1

$begingroup$

I'm not quit sure how to use the hint given by the professor. But here's one way to do this:

Recall that $$(y^{n+1}-x^{n+1}) = (y-x)(y^n+y^{n-1}x+....+yx^{n-1}+x^{n})$$

Now assuming that $x<y$ you can replace all the $y$'s with $x$'s in the second multiple¹ and get that $$y^{n+1}-x^{n+1} geq (y-x) (n+1)x^n$$ Divide by $n+1$ and you get the first inequality. For the other one replace the $x$'s with $y$'s.

---

^{1.Edit It's possible that you need to prove this part by induction. That is, prove by induction that $$(n+1)x^n leq y^n+y^{n-1}x+...+yx^{n-1}+x^n$$}

share|cite|improve this answer

edited Mar 23 at 12:34

answered Mar 22 at 13:35

YankoYanko

8,4692830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks!! I’ll try that
  $endgroup$
  – Facu50196
  Mar 22 at 13:58
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm not quit sure how to use the hint given by the professor. But here's one way to do this:

Recall that $$(y^{n+1}-x^{n+1}) = (y-x)(y^n+y^{n-1}x+....+yx^{n-1}+x^{n})$$

Now assuming that $x<y$ you can replace all the $y$'s with $x$'s in the second multiple¹ and get that $$y^{n+1}-x^{n+1} geq (y-x) (n+1)x^n$$ Divide by $n+1$ and you get the first inequality. For the other one replace the $x$'s with $y$'s.

---

^{1.Edit It's possible that you need to prove this part by induction. That is, prove by induction that $$(n+1)x^n leq y^n+y^{n-1}x+...+yx^{n-1}+x^n$$}

share|cite|improve this answer

edited Mar 23 at 12:34

answered Mar 22 at 13:35

YankoYanko

8,4692830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks!! I’ll try that
  $endgroup$
  – Facu50196
  Mar 22 at 13:58
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm not quit sure how to use the hint given by the professor. But here's one way to do this:

Recall that $$(y^{n+1}-x^{n+1}) = (y-x)(y^n+y^{n-1}x+....+yx^{n-1}+x^{n})$$

Now assuming that $x<y$ you can replace all the $y$'s with $x$'s in the second multiple¹ and get that $$y^{n+1}-x^{n+1} geq (y-x) (n+1)x^n$$ Divide by $n+1$ and you get the first inequality. For the other one replace the $x$'s with $y$'s.

---

^{1.Edit It's possible that you need to prove this part by induction. That is, prove by induction that $$(n+1)x^n leq y^n+y^{n-1}x+...+yx^{n-1}+x^n$$}

share|cite|improve this answer

edited Mar 23 at 12:34

answered Mar 22 at 13:35

YankoYanko

8,4692830

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks!! I’ll try that
  $endgroup$
  – Facu50196
  Mar 22 at 13:58
  

  
  
  
  

add a comment | 

$begingroup$

I'm not quit sure how to use the hint given by the professor. But here's one way to do this:

Recall that $$(y^{n+1}-x^{n+1}) = (y-x)(y^n+y^{n-1}x+....+yx^{n-1}+x^{n})$$

Now assuming that $x<y$ you can replace all the $y$'s with $x$'s in the second multiple¹ and get that $$y^{n+1}-x^{n+1} geq (y-x) (n+1)x^n$$ Divide by $n+1$ and you get the first inequality. For the other one replace the $x$'s with $y$'s.

---

^{1.Edit It's possible that you need to prove this part by induction. That is, prove by induction that $$(n+1)x^n leq y^n+y^{n-1}x+...+yx^{n-1}+x^n$$}

share|cite|improve this answer

edited Mar 23 at 12:34

answered Mar 22 at 13:35

YankoYanko

8,4692830

$endgroup$

share|cite|improve this answer

edited Mar 23 at 12:34

answered Mar 22 at 13:35

YankoYanko

8,4692830

answered Mar 22 at 13:35

YankoYanko

8,4692830

answered Mar 22 at 13:35

YankoYanko

8,4692830