# estimate value of $\sqrt[30]{0.05}$

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?

• I don't think that there's a nice way to evaluate $$\sqrt[30]{0.05}$$ Have a look here Apr 9, 2019 at 9:31
• Are you allowed log tables or slide rules in the exam? Apr 9, 2019 at 9:32
• @Henry we are never allowed to make use of any additional sources sir. Apr 9, 2019 at 9:33
• @Dr.Mathva I tried it to see maybe wolfram gives hint to solve that, but don't. Apr 9, 2019 at 9:34
• 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 Apr 9, 2019 at 11:15

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.

• 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. Apr 9, 2019 at 9:58
• I get $a \approx 0.03$ and so $\sqrt[30]{0.05} \approx 0.97$.
– lhf
Apr 9, 2019 at 10:01
• @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. Apr 9, 2019 at 10:17

You can take it step-by-step: $$\sqrt[30]{0.05}=\sqrt[5]{\sqrt[3]{\sqrt{\frac5{100}}}}\approx\sqrt[5]{\sqrt[3]{\frac{2.2}{10}}}=\sqrt[5]{\sqrt[3]{\frac{220}{1000}}}\approx \sqrt[5]{\frac{6}{10}}=\sqrt[5]{\frac{60000}{100000}}\approx\frac9{10}.$$

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!