B Determine a fractional square root without calculator

[vcsharp2003]

TL;DR Summary

Solving a numerical problem without a calculator

I have to solve a certain numerical problem without using calculator and furthermore, there is a time limit for solving this problem.

The answer I have got so far is $\sqrt{\frac{100}{99}}$

I know I can reduce the numerator to 10 but then I am stuck with square root of denominator which is not a perfect square.

Question: How would I determine in shortest possible time the above square root without any calculator or tables?

 

[jedishrfu]

There are several choices outlined here:

https://en.wikipedia.org/wiki/Methods_of_computing_square_roots

 

[vcsharp2003]

There are several choices outlined here:

https://en.wikipedia.org/wiki/Methods_of_computing_square_roots

Is it possible to use Binomial theorem to do a quick approximation of the fractional square root? This might be better than finding square root of denominator since then I will need to do two approximations, first for denominator square root and then find decimal equivalent for fraction. Binomial theorem might involve a single operation.

 

[PeroK]

Is it possible to use Binomial theorem to do a quick approximation of the fractional square root?

How could that not be possible?

PS note that $(\frac{100}{99})^{1/2} = (\frac{99}{100})^{-1/2}$

 

[vcsharp2003]

How could that not be possible?

That might be easier. I will try expanding the binomial expression on RHS in equation below and maybe take only the first two terms.

$$\sqrt{\frac{100}{99}} = {(1 +\frac{1}{99})}^{\frac{1}{2}}$$

 

[PeroK]

That might be easier. I will try expanding the binomial expression below and maybe take only the first two terms.

$${(1 +\frac{1}{99})}^{\frac{1}{2}}$$

Okay, but see above for a useful idea.

 

[vcsharp2003]

Okay, but see above for a useful idea.

I saw your last post. Once I get $(\frac{99}{100})^{-1/2}$, then how would I proceed?

 

[PeroK]

I saw your last post. Once I get $(\frac{99}{100})^{-1/2}$, then how would I proceed?

Use the binomial theorem. This time you have $(1 - 0.01)^{-1/2}$

 

[PeroK]

Of course, you could always approximate $\frac 1 {99} \approx 0.01$, but inverting the fraction feels neater to me.

 

[vcsharp2003]

Of course, you could always approximate $\frac 1 {99} \approx 0.01$, but inverting the fraction feels neater to me.

Yes, that's more easier than what I suggested since getting integer powers of .01 is very easy as opposed to getting integer powers of $\frac{1}{99}$. After taking only first two terms of binomial expansion I get an approximation of 1.005.

$$\sqrt{\frac{100}{99}} = (1 - 0.01)^{-1/2} \approx 1.005$$

First two terms of binomial expression are obtained as below.
$$ (1 - 0.01)^{-1/2} = 1^{-\frac{1}{2}} + (-\frac{1}{2}) 1^ {(-\frac{1}{2} - 1)} (-.01)^1 + \frac{(-\frac{1}{2}) (-\frac{1}{2} -1)}{(1)(2)} 1^ {(-\frac{1}{2} - 2)} (-.01)^2 + ...$$

Taking the first two terms in binomial expression since the terms starting from third term are going to very small ( even smaller than .0001) we get $1+.005 = 1.005$ which seems correct. My initial approximation of .995 was not correct.

 

[PeroK]

Yes, that's more easier than what I suggested since getting integer powers of .01 is very easy as opposed to getting integer powers of 199. After taking only first two terms of binomial expansion I get an approximation of 0.995.

10099=(1−0.01)−1/2≈.995

That can't be right. The answer must be greater than $1$.

In general $$(1 \pm \epsilon)^{1/2} \approx 1 \pm \frac 1 2 \epsilon$$ and $$(1 \pm \epsilon)^{-1/2} \approx 1 \mp \frac 1 2 \epsilon$$

 

[vcsharp2003]

That can't be right. The answer must be greater than $1$.

In general $$(1 \pm \epsilon)^{1/2} \approx 1 \pm \frac 1 2 \epsilon$$ and $$(1 \pm \epsilon)^{-1/2} \approx 1 \mp \frac 1 2 \epsilon$$

You are correct. I have edited my last post to reflect this. Thanks.

 

[vcsharp2003]

That can't be right. The answer must be greater than $1$.

In general $$(1 \pm \epsilon)^{1/2} \approx 1 \pm \frac 1 2 \epsilon$$ and $$(1 \pm \epsilon)^{-1/2} \approx 1 \mp \frac 1 2 \epsilon$$

Your formulas are awesome. I can memorize these formulas for quicker calculations and not even bother to use Binomial expansion. Thanks.

 

[PeroK]

Yes, that's more easier than what I suggested since getting integer powers of .01 is very easy as opposed to getting integer powers of $\frac{1}{99}$. After taking only first two terms of binomial expansion I get an approximation of 0.995.

$$\sqrt{\frac{100}{99}} = (1 - 0.01)^{-1/2} \approx .995$$

First two terms of binomial expression are obtained as below.
$$ (1 - 0.01)^{-1/2} = 1^{-\frac{1}{2}} + (-\frac{1}{2}) 1^ {(-\frac{1}{2} - 1)} (-.01)^1 + \frac{(-\frac{1}{2}) (-\frac{1}{2} -1)}{(1)(2)} 1^ {(-\frac{1}{2} - 2)} (-.01)^2 + ...$$

Taking the first two terms in binomial expression since the terms starting from third term are going to very small ( even smaller than .0001) we get $1+.005 = 1.005$ which seems correct.

The usual binomial expansion has a $1$ in the first position. I would always take the factor out first. E.g.
$$(a + \epsilon)^{1/2} = a^{1/2}(1 + \frac{\epsilon}{a})^{1/2}$$ In any case, if you already have $1$ here then: $$(1 + \epsilon)^{1/2} = 1 + \frac 1 2 \epsilon + \frac{1}{2!}(\frac 1 2)(-\frac 1 2)\epsilon^2 + \frac{1}{3!}(\frac 1 2)(-\frac 1 2)(-\frac 3 2)\epsilon^3 \dots$$ Also, to avoid missing minus signs, I would tend to do: $$(1 - \epsilon)^{1/2} = (1 + (-\epsilon))^{1/2} = 1 + \frac 1 2 (-\epsilon) + \frac{1}{2!}(\frac 1 2)(-\frac 1 2)(-\epsilon)^2 + \frac{1}{3!}(\frac 1 2)(-\frac 1 2)(-\frac 3 2)(-\epsilon)^3 \dots$$

 

[vcsharp2003]

That can't be right. The answer must be greater than $1$.

In general $$(1 \pm \epsilon)^{1/2} \approx 1 \pm \frac 1 2 \epsilon$$ and $$(1 \pm \epsilon)^{-1/2} \approx 1 \mp \frac 1 2 \epsilon$$

I think these approximation formulas are only valid if $\epsilon$ is very small i.e. as close to 0 as possible on the real number line. Is that correct?

 

[PeroK]

I think these approximation formulas are only valid if $\epsilon$ is very small i.e. as close to 0 as possible on the real number line. Is that correct?

Yes, that's why I used $\epsilon$. The series converges for any $-1 < \epsilon < 1$, but it's only useful as an approximation for small $\epsilon$.