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This might be sort of an embarrassing question but I was reviewing real analysis and its quite common to have to use inequalities and upper/lower bounds there. So I was wondering if there was a clear set of rules of how to upper bound things?

For example, I was in particular considering:

$$ b^n - a^n = (b-a)(a^{n-1} + a^{n-2} b + \cdots + ab^{n-2} + b^{n-1})$$

and wanted to place inequalities bounds on it. When $b>a \iff b-a > 0$ then it seems that we can simply place the inequality:

$$ b^n - a^n = (b-a)(a^{n-1} + a^{n-2} b + \cdots + ab^{n-2} + b^{n-1}) < (b-a)nb^{n-1}$$

regardless of the sign of $b$ and $a$ (which seems to matter in my next example/case).

however, when $b < a \iff b-a < 0$ it seems the situation is different and I was wondering if it was something in particular to just this example of a general rule for placing upper and lower bounds that I am unaware of. In this case $b < a \iff b-a < 0 \implies b^n < a^n \iff b^n - a^n < 0$. Thus we get that on the LHS we have something negative and on the RHS we one negative term ($b-a$) multiplied by $(a^{n-1} + a^{n-2} b + \cdots + ab^{n-2} + b^{n-1})$ with an unknown sign as of yet. But since $a>b$ we do know one thing that $(a^{n-1} + a^{n-2} b + \cdots + ab^{n-2} + b^{n-1}) < n a^{n-1}$. Intuitively it means the second term increases. Thus, at least intuitively it seems to me that if I try to substitute the above inequality if $n a^{n-1} > 0$ is positive, then we actually get a lower bound (since we have something negative with something that increased being multiplied by it) i.e.

$$ b^n - a^n = (b-a)(a^{n-1} + a^{n-2} b + \cdots + ab^{n-2} + b^{n-1}) > (b-a)n a^{n-1} $$

however, this seems rather subtle and it seems this case does not happen for some reason during the first inequality I showed. What is the reason? In particular, I am interested in understanding if it just means in general I always have to be really careful when I try to bound things (since it seems inequalities my change rather subtly if one is not careful) or is there a general principle that I am missing? Or is it just case by case? Or is there something funky in this particular example due to multiplication?

In this special case it seems the inequality switches sign whenever $n$ is odd or even if $a < 0$, since for even n $ a^n >0 $ and for odd $a^n < 0$ so depending which one we have we have a different inequality. Is this true?

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The basic principle is: if $A \le B$ and $c \ge 0$, then $Ac \le Bc$. If $A \le B$ and $c \le 0$, then $Ac \ge Bc$.

In the case at hand, consider $$a^n - b^n = (a-b) (a^{n-1} + a^{n-2} b + \ldots + b^{n-1})$$

I think you want to assume $a > 0$ and $b > 0$. Then if $a > b$ you have $a^{n-1} \ge a^{n-j} b^j \ge b^{n-1}$ for all $j$ from $0$ to $n-1$, so

$$ n a^{n-1} \ge a^{n-1} + a^{n-2} b + \ldots + b^{n-1} \ge n b^{n-1}$$

and multiplying by $a - b > 0$ you get

$$ n a^{n-1} (a-b) \ge a^n - b^n \ge n b^{n-1} (a-b) $$

If $a < b$ you get the same thing with $a$ and $b$ interchanged, so in this case

$$ n b^{n-1} (b-a) \ge b^n - a^n \ge n a^{n-1} (b-a) $$

or, multiplying by $-1$, to get:

$$ n b^{n-1} (a-b) \le a^n - b^n \le n a^{n-1} (a-b) $$

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  • $\begingroup$ weird, do you have a typo on your 3rd line? $\endgroup$ – Pinocchio Jun 9 '17 at 3:24

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