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

How to make the letter "K" that denote Krylov space

2019 gold coins to share

Possible runaway argument using circuitikz

Are polynomials with the same roots identical?

The usage of kelvin in formulas

If I leave the US through an airport, do I have to return through the same airport?

Does putting salt first make it easier for attacker to bruteforce the hash?

Why we don’t make use of the t-distribution for constructing a confidence interval for a proportion?

Is Lambda Calculus purely syntactic?

Is it safe to change the harddrive power feature so that it never turns off?

Did Apple bundle a specific monitor with the Apple II+ for schools?

Please figure out this Pan digital Prince

Should I put programming books I wrote a few years ago on my resume?

The origin of the Russian proverb about two hares

Is there a set of positive integers of density 1 which contains no infinite arithmetic progression?

Was Self-modifying-code possible just using BASIC?

Is it possible to fly backward if you have really strong headwind?

Why is long-term living in Almost-Earth causing severe health problems?

tabular: caption and align problem

Why did Intel abandon unified CPU cache?

Can the removal of a duty-free sales trolley result in a measurable reduction in emissions?

Can I utilise a baking stone to make crepes?

Write a function that checks if a string starts with or contains something

Getting UPS Power from One Room to Another



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?










share|cite|improve this question









$endgroup$
















    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?










    share|cite|improve this question









    $endgroup$














      3












      3








      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?










      share|cite|improve this question









      $endgroup$




      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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 9 hours ago









      Saaqib MahmoodSaaqib Mahmood

      8,27742583




      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











          $endgroup$












          • $begingroup$
            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











          Your Answer








          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "69"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3255422%2fprob-5-sec-6-2-in-bartle-sherberts-intro-to-real-analysis-4th-ed-how-to%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          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











          $endgroup$












          • $begingroup$
            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















          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











          $endgroup$












          • $begingroup$
            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













          6












          6








          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











          $endgroup$



          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














          share|cite|improve this answer



          share|cite|improve this answer








          edited 8 hours ago

























          answered 9 hours ago









          auscryptauscrypt

          6,321613




          6,321613











          • $begingroup$
            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
















          • $begingroup$
            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















          $begingroup$
          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





          $begingroup$
          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




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

















          draft saved

          draft discarded
















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3255422%2fprob-5-sec-6-2-in-bartle-sherberts-intro-to-real-analysis-4th-ed-how-to%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Invision Community Contents History See also References External links Navigation menuProprietaryinvisioncommunity.comIPS Community ForumsIPS Community Forumsthis blog entry"License Changes, IP.Board 3.4, and the Future""Interview -- Matt Mecham of Ibforums""CEO Invision Power Board, Matt Mecham Is a Liar, Thief!"IPB License Explanation 1.3, 1.3.1, 2.0, and 2.1ArchivedSecurity Fixes, Updates And Enhancements For IPB 1.3.1Archived"New Demo Accounts - Invision Power Services"the original"New Default Skin"the original"Invision Power Board 3.0.0 and Applications Released"the original"Archived copy"the original"Perpetual licenses being done away with""Release Notes - Invision Power Services""Introducing: IPS Community Suite 4!"Invision Community Release Notes

          Canceling a color specificationRandomly assigning color to Graphics3D objects?Default color for Filling in Mathematica 9Coloring specific elements of sets with a prime modified order in an array plotHow to pick a color differing significantly from the colors already in a given color list?Detection of the text colorColor numbers based on their valueCan color schemes for use with ColorData include opacity specification?My dynamic color schemes

          Ласкавець круглолистий Зміст Опис | Поширення | Галерея | Примітки | Посилання | Навігаційне меню58171138361-22960890446Bupleurum rotundifoliumEuro+Med PlantbasePlants of the World Online — Kew ScienceGermplasm Resources Information Network (GRIN)Ласкавецькн. VI : Літери Ком — Левиправивши або дописавши її