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A Real Analysis textbook says the identity $$b^n-a^n = (b-a)(b^{n-1}+\cdots+a^{n-1})$$ yields the inequality $$b^n-a^n < (b-a)nb^{n-1} \text{ when } 0 < a< b.$$ (Note that $n$ is a positive integer)

No matter how I look at it, the inequality seems to be wrong. Take for instance, the inequality does not hold for $n=1$ when one tries mathematical induction. It does not hold for other values of $n$ too. I guess there is something I am missing here and I will appreciate help.

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    $\begingroup$ How can it be wrong? When $n=1$, the inequality is trivially true since $b-a=b-a$. $\endgroup$ – Clayton Apr 5 '18 at 0:15
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    $\begingroup$ @Clayton : Since it was stated as a strict inequality, it is trivially false when $n=1. \qquad$ $\endgroup$ – Michael Hardy Apr 5 '18 at 0:19
  • $\begingroup$ From what I can see, when n=1, b - a < b - a . $\endgroup$ – Mr Prof Apr 5 '18 at 0:21
  • $\begingroup$ Sorry @MichaelHardy: the text would have been easier to read if the OP had used $\LaTeX$. I mistook the written $<$ for $\leq$. $\endgroup$ – Clayton Apr 5 '18 at 0:24
  • $\begingroup$ @Clayton : True. But you shouldn't call it LaTeX. It's MathJax. LaTeX is immensely more elaborate than MathJax. $\endgroup$ – Michael Hardy Apr 5 '18 at 0:33
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\begin{align} b^n-a^n & = (b-a)(b^{n-1}+ b^{n-2}a + b^{n-3}a^2 + b^{n-4}a^3 + b^{n-5} a^4 +\cdots+a^{n-1}) \\[10pt] & < (b-a)(b^{n-1} + b^{n-2} b + b^{n-3}b^2 + b^{n-4}b^3+ b^{n-5}b^4 + \cdots + b^{n-1}) \\[10pt] & = (b-a)(b^{n-1} + b^{n-1} + b^{n-1} + b^{n-1} + b^{n-1} + \cdots + b^{n-1}) \\[10pt] & = (b-a) n b^{n-1}. \end{align} The only positive integer $n$ for which this does not work is $n=1,$ where the second factor has only one term, which is $1.$ And in that case it works if you say $\text{“}\le\text{''}$ instead of $\text{“}<\text{''}.$ \begin{align} b^2-a^2 & = (b-a)(b+a) < (b-a)(b+b) & & = (b-a)2b. \\[10pt] b^3-a^3 & = (b-a)(b^2 + ba + a^2) < (b-a)(b^2+b^2+b^2) & & = (b-a)3b^2. \\[10pt] b^4 - a^4 & = (b-a)(b^3+b^2a+ba^2+a^3) \\ & < (b-a)(b^3+b^3+b^3+b^3) & & = (b-a)4b^3. \\[10pt] b^5-a^5 & = (b-a)(b^4 + b^3a + b^2 a^2 + ba^3 + a^4) \\ & < (b-a)(b^4+b^4+b^4+b^4+b^4) & & = (b-a)5b^4. \\[10pt] & \qquad\qquad\text{and so on.} \end{align}

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    $\begingroup$ +$1$: a superb answer per your usual style. I like that you gave additional examples, too, to better illustrate the method. $\endgroup$ – Clayton Apr 5 '18 at 0:55
  • $\begingroup$ @Clayton : Thank you. $\endgroup$ – Michael Hardy Apr 5 '18 at 10:55
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Observe that $n > 1$ for the assertion to be valid. Thus: $b^{n-1-k}a^k< b^{n-1-k}b^k=b^{n-1}$. Letting $k$ runs from $0$ to $n-1$ and add them up: $b^{n-1}+b^{n-2}a+\cdots+a^{n-1} < nb^{n-1}$ which implies the inequality in question.

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  • $\begingroup$ Ok, I get the argument. What I am confused about is the sign "strictly less than". I think it makes b - a < b - a when n=1. If the sign had been "less than or equal to", I would have easily seen that when n=1, b - a = b - a. I will like to know if what I am thinking does not matter or if it is wrong. $\endgroup$ – Mr Prof Apr 5 '18 at 0:32
  • $\begingroup$ I think it is a mistake in the text, it should hold for $n>1$. When $n=1$ then the $\ge$ sign should be used. $\endgroup$ – Btzzzz Apr 5 '18 at 0:33
  • $\begingroup$ I think my source of confusion has been addressed. In addition, I understand the proof better now. I appreciate you guys for your contributions. I particularly appreciate Micheal Hardy's effort. Thanks $\endgroup$ – Mr Prof Apr 5 '18 at 0:44
  • $\begingroup$ @ Btzzzz. Ok I see $\endgroup$ – Mr Prof Apr 5 '18 at 0:48
  • $\begingroup$ Perhaps something in the context of that section of the textbook implies $n>1$? $\endgroup$ – Michael Hardy Apr 5 '18 at 10:56
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We basically need to show $$ b^{n-1}+\ldots+a^{n-1}<nb^{n-1}$$ Since $a<b$, then $a^{n-1}<b^{n-1}$. There are $n$ terms, and so the inequality holds only for when $n>1$.

Note that there is the same question here which uses a $\le$ sign, so i think it is a misprint in the text.

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    $\begingroup$ I have to say, that's one of the most vexing usages of "$\dots$" I've ever seen. $\endgroup$ – Joker_vD Apr 5 '18 at 8:02
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As $0 < a < b$ then $a^k < b^k$ and....

\begin{align} b^n-a^n & = (b-a)(b^{n-1}+b^{n-2}a+\cdots+a^{n-2}b + a^{n-1}) \\ & =(b-a)\sum_{k=0}^{n-1} b^{n-k-}a^k \\ & < (b-a)\sum_{k= 0}^{n-1}b^{n-k-} b^k \\ & = (b-a)\sum_{k=0}^{n-1} b^{n-1} \\ & =(b-a)nb^{n-1}. \end{align}

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  • $\begingroup$ $(b-a)\sum_{k=1}^{n-1} b^{n-k}a^k$ should be $=(b-a)\sum_{k=0}^{n-1} b^{n-1-k}a^k$. Then $n-1$ becomes $n$. $\endgroup$ – Jens Schwaiger Apr 5 '18 at 3:22

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