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Is it possible to calculate and find the solution of $ \; \large{105^{1/5}} \; $ without using a calculator? Could someone show me how to do that, please?

Well, when I use a Casio scientific calculator, I get this answer: $105^{1/5}\approx " 2.536517482 "$. With WolframAlpha, I can an even more accurate result.

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    $\begingroup$ $(105)^{1/5}$ is not square root of $105$. $\endgroup$ Apr 2, 2012 at 17:37
  • $\begingroup$ Yes, sorry about that, you're right. I'm editing it right now... $\endgroup$ Apr 2, 2012 at 17:39
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    $\begingroup$ I think it's better now, right...? :) $\endgroup$ Apr 2, 2012 at 17:41
  • $\begingroup$ Perfect! +1 to the question! $\endgroup$ Apr 2, 2012 at 17:47
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    $\begingroup$ @KerimAtasoy Good for you then! $\endgroup$
    – Pedro
    Apr 30, 2012 at 16:43

5 Answers 5

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I go back to the days BC (before calculators). We did have electricity, but you had to rub a cat's fur to get it.

We also had slide rules, from which a $2$ to $3$ place answer could be found quickly, with no battery to go dead in the middle of an exam. Engineering students wore theirs in a belt holster. Unfortunately, slide rules were expensive, roughly the equivalent of two meals at a very good restaurant. For higher precision work, everyone had a book of tables.

My largish book of tables has the entry $021189$ beside $105$. This means that $\log(105)=2.021189$ (these are logarithms to the base $10$, and of course the user supplies the $2$). Divide by $5$, which is trivial to do in one's head (multiply by $2$, shift the decimal point). We get $0.4042378$.

Now use the tables backwards. The log entry for $2536$ is $404149$, and the entry for $2537$ is $414320$. Note that our target $0.4042378$ is about halfway between these. We conclude that $(105)^{1/5}$ is about $2.5365$.

The table also has entries for "proportional parts," to make interpolation faster. As for using the table backwards, that is not hard. Each page of the $27$ page logarithms section has in a header the range of numbers, and the range of logarithms. The page I used for reverse lookup is headed "Logs $.398\dots$ to $.409\dots$."

There are other parts of the book of tables that deal with logarithms, $81$ pages of logs of trigonometric functions (necessary for navigation, also for astronomy, where one really wants good accuracy). And of course there are natural logarithms, only $17$ pages of these. And exponential and hyperbolic functions, plus a few odds and ends.

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    $\begingroup$ Very interesting... Glad that you made this far and sharing your experiences world wide... :) $\endgroup$ Apr 2, 2012 at 20:37
  • $\begingroup$ ...but are not these numbers coming from answers like the others?! $\endgroup$ Apr 21, 2015 at 15:25
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    $\begingroup$ The tables, sure, they depend on the patient work of generations past. I should take this opportunity to mention that good "ballpark" mental estimates were part of the culture in all scientific disciplines. They still are, though I think much less than they were, particularly at the student level. $\endgroup$ Apr 21, 2015 at 16:18
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You can try using binomial theorem for real exponents.

You can write this as

$$ ((2.5)^5 + (105 - (2.5)^5))^{1/5} = 2.5 \left(1 + \frac{105 - (2.5)^5}{2.5^5}\right)^{1/5} = \frac{5}{2} \left(1 + \frac{47}{625}\right)^{1/5}$$

Taking first three terms of the binomial series

$$(1+x)^r = 1 + rx + \frac{r(r-1)x^2}{2!} + \frac{r(r-1)(r-2)x^3}{3!} + \dots$$

using $r = \frac{1}{5}$ and $x = \frac{47}{625}$ gives us

$$ \frac{5}{2} \left(1 + \frac{47}{5*625} - \frac{4 * 47^2}{2*5^2*625^2}\right) = \frac{4954041}{1953125} \approx 2.5365$$

If you need a better approximation, you can include more terms.

All this can be done by hand using integer arithmetic, but is tedious.

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    $\begingroup$ VERY GOOD!... :) Thank you very much!... I should better study about these terms and subjects also... :) Thank you again... $\endgroup$ Apr 2, 2012 at 18:55
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    $\begingroup$ Notice that using the 1st term of the binomial expansion is similar to a Newton update. To wit, suppose $x$ is an approximation to $105^{\frac{1}{5}}$, then we can write $105^{\frac{1}{5}} = (x^5+(105-x^5))^{\frac{1}{5}} = x(1+\frac{105-x^5}{x^5})^{\frac{1}{5}}$. Expanding the term in parentheses to the first term gives $x(1+\frac{1}{5}\frac{105-x^5}{x^5})$ which simplifies to the Newton update. $\endgroup$
    – copper.hat
    Apr 2, 2012 at 19:27
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    $\begingroup$ @KerimAtasoy: Yes, there is a lot of theory involved which can be quite interesting. Good luck with your studies! $\endgroup$
    – Aryabhata
    Apr 2, 2012 at 19:28
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    $\begingroup$ @MonK: You can compute it by hand. For instance: en.wikipedia.org/wiki/Long_division. $\endgroup$
    – Aryabhata
    Aug 26, 2014 at 18:01
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    $\begingroup$ Where did you get the $2.5$ number? Is this just a "starting guess" for the root, or is there some other reason for this choice? $\endgroup$
    – apnorton
    Oct 19, 2014 at 12:25
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I'm not exactly sure what you mean by 'without a calculator'.

You could try Newton's method to solve $f(x) = 0$, where $f(x) = x^5-105$. The Newton update is then $x_{n+1} = \frac{4}{5}x_n + \frac{1}{5} \frac{105}{{x_n}^4}$. This converges very quickly.

Of course, this involves computing the 4th power, and dividing...

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  • $\begingroup$ :) Well, How about "... without using a calculator" ..? OK, I'm editing it again... :) $\endgroup$ Apr 2, 2012 at 17:53
  • $\begingroup$ I didn't get this method... Could you explain and show it more, please? For example, where that $ \; \large{\frac{4}{5}} \; $ comes from...? $\endgroup$ Apr 2, 2012 at 18:20
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    $\begingroup$ Newton's method is based on a linear approximation of a function using the Maclaurin series: $$ f(x)=f(x_0)+f'(x_0)(x-x_0)+O(x-x_0)^2\tag{1} $$ where $O(x-x_0)^2$ is considered insignificant. Trying to find the $x$ so that $f(x)=0$, $(1)$ says $$ x-x_0=-\frac{f(x_0)}{f'(x_0)}+O(x-x_0)^2\tag{2} $$ which, ignoring $O(x-x_0)^2$, leads to the iteration $$ x=x_0-\frac{f(x_0)}{f'(x_0)}\tag{3} $$ Plugging $f(x)=x^5-a$ into $(3)$ yields $$ \begin{align} x &=x_0-\frac{x_0^5-a}{5x_0^4}\\ &=\frac45x_0+\frac{a}{5x_0^4}\tag{4} \end{align} $$ $\endgroup$
    – robjohn
    Apr 2, 2012 at 19:02
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    $\begingroup$ My advice if using this and other formulae like it would be to rewrite $x_n$ as $p_n/q_n$, then plug these into the formula to get an expression for $x_{n+1}=p_{n+1}/q_{n+1}$ - you can then divide this into two series for the $p$s and $q$s. $\endgroup$ Apr 3, 2012 at 19:11
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    $\begingroup$ In our case we would have $$ \frac{p_{n+1}}{q_{n+1}}=\frac{4}{5}\frac{p_n}{q_n} + \frac{a}{5} \frac{q_n^4}{p_n^4} = \frac{4 p_n^5 + aq_n^5}{5p_n q_n^4} $$ Giving the two series $$ p_{n+1} = 4 p_n^5 + aq_n^5 ,\;\; q_{n+1} = 5p_n q_n^4 $$ Doing this has the advantage that all the subsequent calculations will involve integers (as the $p$s and $q$s are integers) and not messy fractions. You can then long divide at any time to yield a decimal answer $\endgroup$ Apr 3, 2012 at 19:11
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Another way of doing this would be to use logarithm, just like Euler did: $$ 105^{1/5} = \mathrm{e}^{\tfrac{1}{5} \log (105)} = \mathrm{e}^{\tfrac{1}{5} \log (3)} \cdot \mathrm{e}^{\tfrac{1}{5} \log (5)} \cdot \mathrm{e}^{\tfrac{1}{5} \log (7)} $$ Use $$\log(3) = \log\left(\frac{2+1}{2-1}\right) = \log\left(1+\frac{1}{2}\right)-\log\left(1-\frac{1}{2}\right) = \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{2^{2k+1}} = 1 + \frac{1}{12} + \frac{1}{80} + \frac{1}{448} = 1.0.83333+0.0125 + 0.0022 = 1.09803$$ $$ \log(5) = \log\frac{4+1}{4-1} + \log(3) = \log(3) + \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{4^{2k+1}} = \log(3) + \frac{1}{2} + \frac{1}{96} +\frac{1}{2560} $$ $$ \log(7) = \log\frac{8-1}{8+1} + 2 \log(3) = 2 \log(3) - \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{8^{2k+1}} = 2 \cdot \log(3) - \frac{1}{4} - \frac{1}{768} $$ Thus $$ \frac{1}{5} \left( \log(3) + \log(5) + \log(7)\right) = \frac{4}{5} \log(3) + \frac{1}{5} \left( \frac{1}{2} - \frac{1}{4} + \frac{1}{96} - \frac{1}{768} + \frac{1}{2560} \right) = \frac{4}{5} \log(3) + \frac{1993}{38400}= 0.9303 = 1-0.0697 $$ Now $$ \exp(0.9303) = \mathrm{e} \cdot \left( 1 - 0.0697 \right) = 2.71828 \cdot 0.9303 = 2.5288 $$

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  • $\begingroup$ :) AMAZING!... I hadn't even noticed about these approachings untill now, sir. Thank you very much!... :) $\endgroup$ Apr 2, 2012 at 18:59
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You can just do it by trial, but it gets tiring: $2^5\lt 105 \lt 3^5$ so it is between $2$ and $3$. You might then try $2.5^5 \approx 98$ so the true value is a bit higher and so on.

An alternate is to use the secant method. If you start with $2^5=32, 3^5=243$, your next guess is $2+\frac {243-105}{243-32}=2.654$ Then $2.654^5=131.68$ and your next guess is $2.654-\frac {131.68-105}{131.68-32}=2.386$ and so on. Also a lot of work.

Added: if you work with RF engineers who are prone to use decibels, you can do this example easily. $105^{0.2}=100^{0.2}\cdot 1.05^{0.2}=10^{0.4}\cdot 1.01=4 dB \cdot 1.01= (3 dB + 1 dB)1.01=2 \cdot 1.25 \cdot 1.01=2.525$, good to $\frac 12$%, where $1.05^{0.2}\approx 1.01$ comes from the binomial $(1+x)^n\approx 1+nx$ for $x \ll 1$

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  • $\begingroup$ Interesting... I'm wondering how do you get these results... :) You guys have very good skills actually... :) $\endgroup$ Apr 2, 2012 at 18:13
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    $\begingroup$ @KerimAtasoy: The strategies are well known in numerical analysis. In a sense, I cheated, as I used Wolfram Alpha as a calculator to get the numerics. But I believe they are doable by hand if you are determined enough. $\endgroup$ Apr 2, 2012 at 19:17
  • $\begingroup$ @RossMillikan please see my recent question too on mertens theorem ! $\endgroup$
    – Shivanshu
    Jan 2, 2014 at 0:24

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