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I have a square root in a problem which needs to be approximated. I'm not entirely sure how to do this algebraically.

$$ \sqrt{10^2-(6.9\times 10^{-2})^2}$$ The answer the problem is proposing as the approximation is $$ 10[1-\frac 1 2(6.9)^2\times10^{-6}]$$ This hasn't exactly been the most reputable textbook, however, so it could be wrong.


My attempt:

It should be able to approximated by the square-root of 100, because the other number is so small.

$$ 10 -(6.9)^2\times 10^{-4} $$

I can see they factored a 10 out, so I go ahead and do that.

$$ 10[1- \frac 1 {10}(6.9)^2\times 10^{-4}] $$ $$ 10[1-(6.9)^2\times 10^{-5}] $$

I'm not sure where exactly I'm slipping up, or what I'm comprehending incorrectly, any help would be greatly appreciated.

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    $\begingroup$ Difference of squares... $\endgroup$
    – Frank
    Commented Sep 16, 2016 at 15:11
  • $\begingroup$ Wow, I feel pretty stupid. Sorry about that, thank you though. $\endgroup$ Commented Sep 16, 2016 at 15:12
  • $\begingroup$ I just took a glance at it and saw a difference of squares. Maybe they won't be helpful, I don't know. I haven't looked further into the problem yet. $\endgroup$
    – Frank
    Commented Sep 16, 2016 at 15:13
  • $\begingroup$ I don't think difference of squares worked it just took me back to where I was initially. $\endgroup$ Commented Sep 16, 2016 at 15:20
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    $\begingroup$ The binomial approximation . . . (1+x)^n~(1+nx) $\endgroup$
    – Kartik
    Commented Sep 16, 2016 at 16:02

3 Answers 3

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$$\begin{align} \\ & \sqrt{10^2-(6.9\times 10^{-2})^2} \\ & = \sqrt{10^2\left[1 -\frac{1}{10^{2}}\left\{(6.9)^2\times 10^{-4}\right\}\right]} \\ & = \left[10^2\{1 -(6.9)^2\times 10^{-6}\}\right]^{\frac{1}{2}} \\ & \approx 10\left[1-\frac 1 2(6.9)^2\times10^{-6}\right] \,\,\,\,\,\,\,\,\,\,\,\,\, \text{using binomial approximation} \end{align}$$ The binomial approximation is stated in this link.

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  • $\begingroup$ This is exactly what I was looking for. Thank you. $\endgroup$ Commented Sep 16, 2016 at 15:36
  • $\begingroup$ You're welcome!! $\endgroup$ Commented Sep 16, 2016 at 15:40
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\begin{align*} \sqrt{a^{2}+b} &= a\sqrt{1+\frac{b}{a^{2}}} \\ & \approx a\left( 1+\frac{b}{2a^2} \right) \\ &= a+\frac{b}{2a} \end{align*}

Now \begin{align*} \sqrt{10^{2}-(6.9)^{2} \times 10^{-4}} & \approx 10-\frac{(6.9)^{2} \times 10^{-4}}{20} \\ &= 10-\frac{(6.9)^{2} \times 10^{-5}}{2} \\ & \approx 10-0.000238 \\ & = 9.999762 \end{align*}

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Use the Taylor series for $(1+x)^{1/2}$.

First we rewrite $(A+B)^{\alpha}$ to be in a form similar to this :

$(A+B)^{\alpha} = A^{\alpha}(1 + \frac{B}{A})^{\alpha}$.

With $A \gg B$ the ratio $\frac{B}{A}$ is "small" and we do a Taylor series for $(1+x)^{\alpha}$

$$(1+x)^{\alpha} = 1 + \frac{\alpha}{1!}x + \frac{\alpha(\alpha-1)}{2!}x^{2} + \ldots$$

and now replace each $x$ with $\frac{B}{A}$ to get

$$(A+B)^{\alpha} = A^{\alpha}\left( 1 + \frac{\alpha}{1!} \frac{B}{A} + \frac{\alpha(\alpha-1)}{2!}\left( \frac{B}{A} \right)^{2} + \ldots \right)$$

Putting in $\alpha = \frac{1}{2}$ gives

$$(A+B)^{\frac{1}{2}} = A^{\frac{1}{2}}\left( 1 + \frac{1}{2} \frac{B}{A} - \frac{1}{8}\left( \frac{B}{A} \right)^{2} + \ldots \right)$$

In your case $A = 10^{2}$ and $B = -(0.69)^{2}$ so $A^{\frac{1}{2}} = 10$ and $\frac{B}{A} = -(0.069)^{2} = -(6.9 \times 10^{-3})^{2} = -0.00004761...$

Therefore

$$\begin{align} (A+B)^{\frac{1}{2}} &= \sqrt{10^{2}}\Big( 1 + \frac{1}{2}\frac{-(0.69)^{2}}{10^{2}} + \ldots \Big) = 10\left( 1 - \frac{0.00004761}{2} + \ldots \right) \\&\approx 10*(1-0.000023805) = 9.99976195 \end{align}$$

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    $\begingroup$ nice solution +1, small nitpick, you say $A/B$ when you mean $B/A$ in the 4th line. $\endgroup$ Commented Sep 16, 2016 at 15:45
  • $\begingroup$ Also, in your line with the asterisk, you mean $0.99997$ not $9.9997$. $\endgroup$
    – Jam
    Commented Sep 16, 2016 at 16:01
  • $\begingroup$ @Jam: That's a single line wrapped in an inopportune place $\endgroup$
    – user14972
    Commented Sep 17, 2016 at 0:55

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