7

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.

Asaf Karagila
  • 393,674
Mr Prof
  • 393
  • 1
    How can it be wrong? When $n=1$, the inequality is trivially true since $b-a=b-a$. – Clayton Apr 05 '18 at 00:15
  • 4
    @Clayton : Since it was stated as a strict inequality, it is trivially false when $n=1. \qquad$ – Michael Hardy Apr 05 '18 at 00:19
  • From what I can see, when n=1, b - a < b - a . – Mr Prof Apr 05 '18 at 00:21
  • Sorry @MichaelHardy: the text would have been easier to read if the OP had used $\LaTeX$. I mistook the written $<$ for $\leq$. – Clayton Apr 05 '18 at 00:24
  • @Clayton : True. But you shouldn't call it LaTeX. It's MathJax. LaTeX is immensely more elaborate than MathJax. – Michael Hardy Apr 05 '18 at 00:33
  • The strict inequality does hold for $n>1$; it is interesting to trace where the case $n=1$ breaks down. The assumption made is that there is at least one term strictly less than $b^{n-1}$, and that assumption is not valid when $b^{n-1}=b^0$ is the only term. – Maxim Apr 05 '18 at 00:44
  • @Michael: I’m not well-enough educated on the differences to distinguish one from the other. I know MathJax has some quirks compared to the TeX that I write, but I always just assumed it was the underlying compiler (or something like that). – Clayton Apr 05 '18 at 00:54
  • @MichaelHardy It's a strict subset of LaTeX, isn't it? So one can certainly say it's valid LaTeX syntax. And it's important to realize that it is, since that means any existing knowledge about LaTeX you have is going to be useful. – jpmc26 Apr 05 '18 at 07:08
  • @jpmc26 : It's not a strict subset of LaTeX. For one thing, you can freely use many non-ASCII characters in MathJax. – Michael Hardy Apr 05 '18 at 10:52
  • @Clayton : LaTeX does all sorts of things that don't involve mathematical notation. Making tables of contents, formatting chapters and sections, including footnotes, etc. etc. And in LaTeX you can use style files and \input and many other things. – Michael Hardy Apr 05 '18 at 10:54
  • When stating things like "It does not hold for other values of $n$ too." in the question, it would help if you gave an example. – JiK Apr 05 '18 at 11:03

4 Answers4

15

\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}

Michael Hardy
  • 1
5

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.

DeepSea
  • 77,651
  • 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. – Mr Prof Apr 05 '18 at 00:32
  • 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. – Btzzzz Apr 05 '18 at 00:33
  • 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 – Mr Prof Apr 05 '18 at 00:44
  • @ Btzzzz. Ok I see – Mr Prof Apr 05 '18 at 00:48
  • Perhaps something in the context of that section of the textbook implies $n>1$? – Michael Hardy Apr 05 '18 at 10:56
3

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.

Btzzzz
  • 1,113
1

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}

Michael Hardy
  • 1
fleablood
  • 124,253