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Yesterday I got an exam in which there was a problem and its solution results in $$\sqrt[30]{0.05}$$

I didn't go further calculation. Still I can't.

My lecturer said, even I'm still not sure if he made ironic humor, that if there is a math operation requiring higher than mid-school level math you to do, not do that, let it leave as it is. As a note, my department is not math.

Did he really make a humor?

Is there way/methods to figure out/estimate its result without calculator w.r.t thinking in exam(time limit) and not in exam?

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    $\begingroup$ I don't think that there's a nice way to evaluate $$\sqrt[30]{0.05}$$ Have a look here $\endgroup$
    – Dr. Mathva
    Apr 9, 2019 at 9:31
  • $\begingroup$ Are you allowed log tables or slide rules in the exam? $\endgroup$
    – Henry
    Apr 9, 2019 at 9:32
  • $\begingroup$ @Henry we are never allowed to make use of any additional sources sir. $\endgroup$ Apr 9, 2019 at 9:33
  • $\begingroup$ @Dr.Mathva I tried it to see maybe wolfram gives hint to solve that, but don't. $\endgroup$ Apr 9, 2019 at 9:34
  • $\begingroup$ There are methods to calculate this such as Taylor Series expansion, or using the n-th root formula or using Newton's approximation or using Log function (or its expansion), but non results-in a simple calculation. I don't think in high school you know about any of this with the exception of Log function which may be what you should use. Some reference is here: en.wikipedia.org/wiki/Nth_root $\endgroup$
    – NoChance
    Apr 9, 2019 at 11:15

3 Answers 3

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We know that $\sqrt[30]{0.05}$ is a number a little smaller than $1$ because $\sqrt[n]{0.05}$ converges to $1$ for $n$ to infinity. So set $\sqrt[30]{0.05}=1-a$ and then try to estimate $a$. $a$ satisfies the equation $$(1-a)^{30}=\frac{1}{20}$$ Writing out the first few terms gives $$1-30a+\frac{30\cdot 29}{2}a^2 + ...= \frac{1}{20}$$ Note that because $a$ is small, the further coefficients are decreasing quickly.

Using just $1-30a\sim \frac{1}{20}$ yields $\sqrt[30]{0.05} \sim 0.93$ without calculator. If you use more terms you should get better approximations.

Edit: It turns out this doesn't work quite as well as I thought. While it is true that the later coefficients with higher powers of $a$ are decreasing quickly, the biggest coefficient in the series is at $a^3$. So in order to get something that is actually an approximation of $a$ one would have to compute at least until $a^4$ or $a^5$. This leads to a polynomial which is not really easy to solve by hand. Computing further terms would increase accuracy but I'm not sure whether this is helpful in a no calculator scenario.

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    $\begingroup$ However, if you try to use the quadratic term, we do not get a solution because the discriminant $\Delta < 0$. So, you should use more terms which entails very complicated computations compared to the original one. Conclusion: your method "works" only for a linear approximation. $\endgroup$
    – Alex Silva
    Apr 9, 2019 at 9:58
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    $\begingroup$ I get $a \approx 0.03$ and so $\sqrt[30]{0.05} \approx 0.97$. $\endgroup$
    – lhf
    Apr 9, 2019 at 10:01
  • $\begingroup$ @lhf, since the the real value is around $0.9049661$, even the linear approximation does not work very well . So, I do not understand the massive upvotes. $\endgroup$
    – Alex Silva
    Apr 9, 2019 at 10:17
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You can take it step-by-step: $$\sqrt[30]{0.05}=\sqrt[5]{\sqrt[3]{\sqrt{\frac5{100}}}}\approx\sqrt[5]{\sqrt[3]{\frac{2.2}{10}}}=\sqrt[5]{\sqrt[3]{\frac{220}{1000}}}\approx \sqrt[5]{\frac{6}{10}}=\sqrt[5]{\frac{60000}{100000}}\approx\frac9{10}.$$

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This is how the ancients might have attempted this, using the method of false position.

Solve $x^2=0.05$, starting with an initial guess of $x=0.2$.

$$0.05=(0.2+h)^2\approx0.04+0.4h\implies h=0.02$$ $$0.05=(0.22+h)^2\approx0.0484+0.44h\implies h=0.0036$$ So $x=0.2236$.

Now solve $y^3=0.2236$, starting with $0.6$.

$$0.2236=(0.6+h)^3\approx0.216+1.08h\implies h=0.007$$ $$0.2236=(0.607+h)^3\approx0.2236+1.1h\implies h=0$$

Finally solve $z^5=0.607$, starting with $0.9$.

$$0.607=(0.9+h)^5\approx0.5905+3.28h\implies h=0.005$$ so $z\approx0.905$.

Disclaimer: I used a calculator!

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