Is it possible to reuse computation while dividing?

Question

Suppose I wanted to divide $1$ by $x$, and I get the answer as $\frac{1}{x}$. Is there any way to go from this to the answer to $\frac{1}{x+1}$ without having to restart the division? I know the answer's going to be pretty close, but is it possible to quantify this without a harder division? Another example of this could be going from $\frac{1}{x}$ to $\frac{10}{10x+1}$

I know that $\frac{1}{73} \approx 0.0136986$, could I use that to find $\frac{1}{74}$? Or, I know that $\frac{1}{73} \approx 0.0136986$, can I find $\frac{1}{73.1}$ easily?

If multiplication is easier than division and $x \gg 1$ then you can say $\frac{1}{x+1}=\frac1x-\left(\frac1x\right)^2+\left(\frac1x\right)^3-\left(\frac1x\right)^4+\cdots$ — Henry, Nov 08 '22 at 08:37

score 5 · Accepted Answer · answered Nov 08 '22 at 08:36

5

If $x$ is large, then

$$\frac1x - \frac1{x+1} = \frac1{x(x+1)} \approx \frac1{x^2}$$

so that

$$\frac1{x+1} \approx \frac1x - \frac1{x^2}$$

answered Nov 08 '22 at 08:36

JMP

21,771

and $\frac1{x+1} \approx \frac1x - \frac1{x^2}+\frac1{x^3}$ would be even closer – Henry Nov 08 '22 at 08:38
Yes, I recommend you add that to the answer! – avighnac Nov 08 '22 at 08:39
Sometimes using this approximation will not be worth. Take for instance this trivial example: $\frac{1}{999+1}$. It is better to just calculate $\frac{1}{1000}$ instead of $\frac{1}{999}+\frac{1}{999^2}$. But it is true that in many cases it would be really useful. – ordptt Nov 19 '22 at 01:03

Anne Bauval · Answer 2 · 2022-11-08T10:00:23.100

5

Since $\frac1{1-z}=\sum_{n=0}^\infty z^n$ for every number $z$ such that $|z|<1,$ $$\begin{align}\frac1{x+\epsilon}&=\frac1x\frac1{1+\frac\epsilon x}\\&=\frac1x\sum_{n=0}^\infty\left(\frac{-\epsilon}x\right)^n\\&\approx\frac1x\left(1-\frac\epsilon x\right), \end{align}$$ the conditions being $|\epsilon|<|x|$ for the convergence of the geometric power series, and $|\epsilon|\ll|x|$ for its approximation.

edited Nov 08 '22 at 10:00

answered Nov 08 '22 at 08:39

Anne Bauval

34,650

1

I feel like some text/explanation would be good. (even if this is quite simple.) – Naitik Mundra Nov 08 '22 at 09:08
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Done, hopefully sufficiently. (I initially hoped the OP wouldn't need more or would ask if necessary. It is not easy to guess what people need.) – Anne Bauval Nov 08 '22 at 09:19
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@AnneBauval although additional explanation would be nicer, I perfectly understood why this works, which is why I didn't ask – avighnac Nov 08 '22 at 09:47
I already added some text. What kind additional explanation do you think would be nicer? – Anne Bauval Nov 08 '22 at 09:51
1

@AnneBauval maybe you could add "since $\frac{1}{1-x} = \displaystyle\sum_{n=0}^{\infty} x^n$" – avighnac Nov 08 '22 at 09:56
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This is plenty of explanation for a math site, thanks. – Fattie Nov 08 '22 at 16:49

score 1 · Answer 3 · answered Nov 19 '22 at 02:30

Generalizing the problem, here's an expression to calculate $\frac{1}{x+c}$: $$\frac{1}{x+c}=\frac{1}{x}-4\sum_{k=0}^{\infty}\frac{c^{2k+1}}{(2x+c)^{2k+2}}$$ Using the first term gives $$\frac{1}{x+c}\approx\frac{1}{x}-\frac{4c}{(2x+c)^2}$$ in which, for large $x$, the $+c$ term in the denominator will be negligible $$\frac{1}{x+c}\approx\frac{1}{x}-\frac{4c}{(2x)^2}$$ $$\frac{1}{x+c}\approx\frac{1}{x}-\frac{c}{x^2}$$

Is it possible to reuse computation while dividing?

3 Answers3