TAOCP 1.2.2 Exercise 13

(a) Let u=\sqrt[n]{1+x}.

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Section 1.2.2: Numbers, Powers, and Logarithms

Exercise 13. ▶ [M23] (a) Given that $x$ is a positive real number and $n$ is a positive integer, prove the inequality $\sqrt[n]{1 + x} - 1 \le x/n$. (b) Use this fact to justify the remarks following (7).

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Solve time: 36s

Solution

(a) Let

$$ u=\sqrt[n]{1+x}. $$

Then $u>0$ and

$$ u^n=1+x. $$

Since

$$ u^n-1=(u-1)(u^{n-1}+u^{n-2}+\cdots+u+1), $$

we have

$$ x=(u-1)(u^{n-1}+u^{n-2}+\cdots+u+1). $$

Because $x>0$, it follows that $u>1$. Hence each term

$$ u^{n-1},u^{n-2},\ldots,u,1 $$

is at least $1$. There are exactly $n$ such terms, therefore

$$ u^{n-1}+u^{n-2}+\cdots+u+1\ge n. $$

Consequently,

$$ x\ge n(u-1). $$

Dividing by $n$,

$$ u-1\le \frac{x}{n}. $$

Since $u=\sqrt[n]{1+x}$,

$$ \boxed{\sqrt[n]{1+x}-1\le \frac{x}{n}}. $$

(b) In the remarks following (7), let

$$ x=b-1,\qquad n=10^k. $$

Then

$$ \sqrt[10^k]{b}-1\le \frac{b-1}{10^k}. $$

But

$$ \sqrt[10^k]{b}=b^{1/10^k}, $$

hence

$$ b^{1/10^k}-1\le \frac{b-1}{10^k}. $$

Therefore

$$ b^{n+d_1/10+\cdots+d_k/10^k}(b^{1/10^k}-1) \le b^{n+d_1/10+\cdots+d_k/10^k}\frac{b-1}{10^k}. $$

Since

$$ 0\le \frac{d_1}{10}+\cdots+\frac{d_k}{10^k}<1, $$

we have

$$ b^{n+d_1/10+\cdots+d_k/10^k}<b^{n+1}. $$

Hence

$$ b^{n+d_1/10+\cdots+d_k/10^k}(b^{1/10^k}-1) < \frac{b^{n+1}(b-1)}{10^k}. $$

This is precisely the estimate used after Eq. (7). Since

$$ \frac{b^{n+1}(b-1)}{10^k}\to0 \qquad\text{as}\qquad k\to\infty, $$

the right and left extremes in Eq. (7) can be made arbitrarily close together. Therefore Eq. (7) determines a unique value for $b^x$.

This completes the proof.

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