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Prob. 5, Sec. 6.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: How to show this function is strictly decreasing using derivative

Prob. 2, Sec. 3.3, in INTRO. TO REAL ANALYSIS by Bartle & Sherbert: What is the limit of this sequence?Prob. 6, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: How to find this limit?Prob. 7, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent?Prob. 10, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent?Prob. 14, Sec. 3.4, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: The supremum of a bounded sequence not in the rangeProb. 5 (e), Sec. 4.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: How to find $lim_xto 0- fracsqrtx+1x$?Prob. 12, Sec. 5.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Continuity of additive functionsProb. 14, Sec. 5.4, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Every continuous periodic function is bounded and uniformly continuousDefinition 5.6.6 in Bartle & Sherbert's INTRO TO REAL ANALYSIS: How to show that this notion is well-defined?Prob. 17, Sec. 6.1, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Straddle Lemma

3

$begingroup$

Here is Prob. 5, Sec. 6.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

Let $a > b > 0$ and let $n in mathbbN$ satisfy $n geq 2$. Prove that $a^1/n - b^1/n < (a-b)^1/n$. [Hint: Show that $f(x) colon= x^1/n - (x-1)^1/n$ is decreasing for $x geq 1$, and evaluate $f$ at $1$ and $a/b$.]

My Attempt:

We find that for $x > 1$,
$$
beginalign
f^prime(x) &= frac1nx^frac1n - 1 - frac1n(x-1)^frac1n - 1 \
&= frac1n left( fracx^1/nx - frac(x-1)^1/nx-1 right) \
&= frac1nx(x-1) left( x^1/n(x-1) - x(x-1)^1/n right) \
&= fracx^1/n(x-1)^1/nnx(x-1) left( (x-1) - x right) \
&= -frac1nx^1-frac1n (x-1)^1-frac1n .
endalign
$$
Thus $f^prime(x) < 0$ for all $x > 1$. Therefore the function $f$ is strictly decreasing on the interval $[1, +infty)$. That is, for any $x > 1$ we have $f(x) < f(1)$.

Now if $a > b > 0$, then $a/b > 1$, and so we have
$$ f(a/b) < f(1),$$
that is,
$$ left(fracabright)^frac1n - left(fracab - 1right)^frac1n < 1,$$
which amounts to
$$ fraca^1/n - (a-b)^1/n b^1/n < 1, $$
which implies
$$ a^1/n - (a-b)^1/n < b^1/n,$$
and hence
$$ a^1/n - b^1/n < (a-b)^1/n, $$
as required.

Are there any issues with this proof?

real-analysis calculus proof-verification derivatives inequality

share|cite|improve this question

asked 9 hours ago

Saaqib MahmoodSaaqib Mahmood

8,27742583

$endgroup$

add a comment | 

3

$begingroup$

Here is Prob. 5, Sec. 6.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

Let $a > b > 0$ and let $n in mathbbN$ satisfy $n geq 2$. Prove that $a^1/n - b^1/n < (a-b)^1/n$. [Hint: Show that $f(x) colon= x^1/n - (x-1)^1/n$ is decreasing for $x geq 1$, and evaluate $f$ at $1$ and $a/b$.]

My Attempt:

We find that for $x > 1$,
$$
beginalign
f^prime(x) &= frac1nx^frac1n - 1 - frac1n(x-1)^frac1n - 1 \
&= frac1n left( fracx^1/nx - frac(x-1)^1/nx-1 right) \
&= frac1nx(x-1) left( x^1/n(x-1) - x(x-1)^1/n right) \
&= fracx^1/n(x-1)^1/nnx(x-1) left( (x-1) - x right) \
&= -frac1nx^1-frac1n (x-1)^1-frac1n .
endalign
$$
Thus $f^prime(x) < 0$ for all $x > 1$. Therefore the function $f$ is strictly decreasing on the interval $[1, +infty)$. That is, for any $x > 1$ we have $f(x) < f(1)$.

Now if $a > b > 0$, then $a/b > 1$, and so we have
$$ f(a/b) < f(1),$$
that is,
$$ left(fracabright)^frac1n - left(fracab - 1right)^frac1n < 1,$$
which amounts to
$$ fraca^1/n - (a-b)^1/n b^1/n < 1, $$
which implies
$$ a^1/n - (a-b)^1/n < b^1/n,$$
and hence
$$ a^1/n - b^1/n < (a-b)^1/n, $$
as required.

Are there any issues with this proof?

real-analysis calculus proof-verification derivatives inequality

share|cite|improve this question

asked 9 hours ago

Saaqib MahmoodSaaqib Mahmood

8,27742583

$endgroup$

add a comment | 

$begingroup$

Here is Prob. 5, Sec. 6.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

Let $a > b > 0$ and let $n in mathbbN$ satisfy $n geq 2$. Prove that $a^1/n - b^1/n < (a-b)^1/n$. [Hint: Show that $f(x) colon= x^1/n - (x-1)^1/n$ is decreasing for $x geq 1$, and evaluate $f$ at $1$ and $a/b$.]

My Attempt:

We find that for $x > 1$,
$$
beginalign
f^prime(x) &= frac1nx^frac1n - 1 - frac1n(x-1)^frac1n - 1 \
&= frac1n left( fracx^1/nx - frac(x-1)^1/nx-1 right) \
&= frac1nx(x-1) left( x^1/n(x-1) - x(x-1)^1/n right) \
&= fracx^1/n(x-1)^1/nnx(x-1) left( (x-1) - x right) \
&= -frac1nx^1-frac1n (x-1)^1-frac1n .
endalign
$$
Thus $f^prime(x) < 0$ for all $x > 1$. Therefore the function $f$ is strictly decreasing on the interval $[1, +infty)$. That is, for any $x > 1$ we have $f(x) < f(1)$.

Now if $a > b > 0$, then $a/b > 1$, and so we have
$$ f(a/b) < f(1),$$
that is,
$$ left(fracabright)^frac1n - left(fracab - 1right)^frac1n < 1,$$
which amounts to
$$ fraca^1/n - (a-b)^1/n b^1/n < 1, $$
which implies
$$ a^1/n - (a-b)^1/n < b^1/n,$$
and hence
$$ a^1/n - b^1/n < (a-b)^1/n, $$
as required.

Are there any issues with this proof?

real-analysis calculus proof-verification derivatives inequality

share|cite|improve this question

asked 9 hours ago

Saaqib MahmoodSaaqib Mahmood

8,27742583

$endgroup$

share|cite|improve this question

asked 9 hours ago

Saaqib MahmoodSaaqib Mahmood

8,27742583

share|cite|improve this question

asked 9 hours ago

Saaqib MahmoodSaaqib Mahmood

8,27742583

asked 9 hours ago

Saaqib MahmoodSaaqib Mahmood

8,27742583

asked 9 hours ago

Saaqib MahmoodSaaqib Mahmood

8,27742583

1 Answer
1

active

oldest

votes

6

$begingroup$

There is an error; the factorisation $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)-x)$ is incorrect and should be $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)^fracn-1n-x^fracn-1n)$.

To correctly prove this, note that since $frac1n-1$ is negative we have $x^frac1n-1$ is decreasing. This implies $(x-1)^frac1n-1 > x^frac1n-1 $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

auscryptauscrypt

6,321613

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you for the correction. Can you please also suggest as to how to proceed from that point on?
  $endgroup$
  – Saaqib Mahmood
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SaaqibMahmood I've edited my answer to include the corrected proof.
  $endgroup$
  – auscrypt
  8 hours ago
  

  
  
  
  

add a comment | 

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6

$begingroup$

There is an error; the factorisation $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)-x)$ is incorrect and should be $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)^fracn-1n-x^fracn-1n)$.

To correctly prove this, note that since $frac1n-1$ is negative we have $x^frac1n-1$ is decreasing. This implies $(x-1)^frac1n-1 > x^frac1n-1 $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

auscryptauscrypt

6,321613

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you for the correction. Can you please also suggest as to how to proceed from that point on?
  $endgroup$
  – Saaqib Mahmood
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SaaqibMahmood I've edited my answer to include the corrected proof.
  $endgroup$
  – auscrypt
  8 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

There is an error; the factorisation $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)-x)$ is incorrect and should be $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)^fracn-1n-x^fracn-1n)$.

To correctly prove this, note that since $frac1n-1$ is negative we have $x^frac1n-1$ is decreasing. This implies $(x-1)^frac1n-1 > x^frac1n-1 $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

auscryptauscrypt

6,321613

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you for the correction. Can you please also suggest as to how to proceed from that point on?
  $endgroup$
  – Saaqib Mahmood
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SaaqibMahmood I've edited my answer to include the corrected proof.
  $endgroup$
  – auscrypt
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

There is an error; the factorisation $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)-x)$ is incorrect and should be $x^1/n(x-1) - x(x-1)^1/n = x^1/n (x-1)^1/n ((x-1)^fracn-1n-x^fracn-1n)$.

To correctly prove this, note that since $frac1n-1$ is negative we have $x^frac1n-1$ is decreasing. This implies $(x-1)^frac1n-1 > x^frac1n-1 $ since $x-1<x$, and so $f'(x)$ is clearly negative as required.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

auscryptauscrypt

6,321613

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

auscryptauscrypt

6,321613

answered 9 hours ago

auscryptauscrypt

6,321613

answered 9 hours ago

auscryptauscrypt

6,321613