6
$\begingroup$

You learn in algebra that $$\sqrt{ab}=\sqrt{a} \sqrt{b}$$ and that $$\sqrt{\frac ab}=\frac {\sqrt a}{\sqrt b}$$ You also learn to never make the fatal mistake of thinking $$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$

However, I am wondering if there is a rule for $\sqrt{a+b}$. I would think it would be pretty complex, if it exists at all.

$$\sqrt{a+b}=\text{?}$$

$\endgroup$
  • $\begingroup$ Note that the first two identities only hold when $a, b \geq 0$. As for your question, it depends on what one means by "rule", but the short answer is that there is no comparably simple expression. Since $\sqrt{\cdot}$ is concave, one can write down inequalities satisfies by $\sqrt{a + b}$. $\endgroup$ – Travis Jun 19 '15 at 2:00
  • 1
    $\begingroup$ I suppose it depends on what you mean by a "rule". If you mean "some formula that lets you express $\sqrt{a+b}$ as some sort of combination involving $\sqrt a$ and $\sqrt b$", then no, there is no rule. $\endgroup$ – mweiss Jun 19 '15 at 2:01
  • 1
    $\begingroup$ @mweiss Of course, one can write $\sqrt{a + b} = \sqrt{(\sqrt{a})^2 + (\sqrt{b})^2}$, but this is hardly illuminating and really just emphasizes your point that the answer depends on what 'rule' means. $\endgroup$ – Travis Jun 19 '15 at 2:04
  • $\begingroup$ @Travis : $\;\;\;$ More generally, they hold when $\: \operatorname{Im}\hspace{.02 in}(a) = 0 \leq \operatorname{Re}(a) \:$ or $\: \operatorname{Im}\hspace{.02 in}(b) = 0 \leq \operatorname{Re}(b) \;$. $\hspace{.88 in}$ $\endgroup$ – user57159 Jun 19 '15 at 2:19
  • $\begingroup$ I suppose you could use imaginary numbers if you wanted... $\endgroup$ – Kbot Jun 19 '15 at 2:44
12
$\begingroup$

There is no finite rule but there is the binomial series: $$ (1 + x)^\alpha = \sum_{k=0}^{\infty} \; {\alpha \choose k} \; x^k $$ where $$ {\alpha \choose k} = \frac{\alpha (\alpha-1) (\alpha-2) \cdots (\alpha-k+1)}{k!} $$ This series converges for $|x|<1$. Taking $\alpha=1/2$, we get $$ (1+x)^{1/2} = 1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5 x^4}{128}+\frac{7 x^5}{256}+\cdots $$

We can apply this to $\sqrt{a+b}$ as follows. Assume wlog that $a>b$. Then $$ \sqrt{a+b}=(a+b)^{1/2} = \sqrt{a}(1+x)^{1/2} $$ with $x=b/a<1$. We then have $$ \sqrt{a+b} = \sqrt{a}\left(1+\frac{b}{2a}-\frac{b^2}{8a^2}+\frac{b^3}{16a^3}-\frac{5 b^4}{128a^4}+\frac{7 b^5}{256a^5}+\cdots\right) $$

$\endgroup$
  • 3
    $\begingroup$ And if $b/a<1$, you can do it for with $x=a/b$. $\endgroup$ – fonini Jun 19 '15 at 2:32
2
$\begingroup$

Provided that $a>0$ and $a+b>0$, we have $$\sqrt{a+b}=\sqrt{a}\cdot\sqrt{1+\frac{b}{a}}$$

$\endgroup$
  • $\begingroup$ It's just like rewriting $a$ as $1\cdot a$ $\endgroup$ – Vim Jun 19 '15 at 10:07
  • $\begingroup$ This answer kind of reminds me of your answer here math.stackexchange.com/questions/454333/… I am a fan of your work! $\endgroup$ – graydad Jun 24 '15 at 1:35
0
$\begingroup$

The general answer is no. However, you may encounter some special occasions like: $$\sqrt{3+2\sqrt 2}=1+\sqrt 2$$ in which $\sqrt{3+2\sqrt 2}$ happens to be $(1+\sqrt 2)^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.