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I came up with the following approximation

$$\sqrt[4]{\pi}+\frac{2}{1000}\gtrsim\frac{4}{3}$$

I don't know too much about proving an inequality like this algebraically. I was hoping for an extremely rigorous proof of this (I would definitely appreciate names of theorems). I am just starting to self study computational number theory.

I didn't know how to prove this whatsoever. I would think of using a large finite number of iterations on a Taylor series, but I really had no clue how to use that. Thanks for any help.

An similar question type to this is the following: Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$. My wording is a bit odd in this question, so please note that both questions are very similar. (Solving mine algebraically is really the basis, though)

Community
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2 Answers2

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This isn't really by hand, but here is an explanation of what OP found. We can calculate the continued fraction for $\sqrt[4]{\pi}$ as $$[1;3,55,3,1,1,2,3,37,\ldots]$$ We get a good approximation in a continued fraction by stopping just before a "big" number. Hence we get a quite good estimate as $$\sqrt[4]{\pi}\approx [1;3]=\frac{4}{3}$$ We can refine this by taking more terms: $$\sqrt[4]{\pi}\approx [1;3,55]=\frac{221}{166}\approx 1.3313253\ldots$$ This explains why $\sqrt[4]{\pi}+0.002$ is close to $\frac{4}{3}$. To prove the requested bound, we need to go a bit further. The continued fractions alternate as an overestimate, followed by an underestimate. $[1;3]$ is an overestimate (for $\sqrt[4]{\pi}$), $[1;3,55]$ is an underestimate, and so on. Just a few more terms gives us: $$\sqrt[4]{\pi}>[1;3,55,3,1,1]=\frac{1555}{1168}\approx 1.3313356>\frac{4}{3}-0.002$$ Hence $\sqrt[4]{\pi}+0.002>\frac{4}{3}$.

vadim123
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    It looks as if you didn't read the comment below the OPs question (-1) –  Nov 24 '15 at 02:13
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    Everyone's right. See, I designed this question with the wanting to solve this algebraically. As I didn't specify this very well, I will give a +1 –  Nov 24 '15 at 02:21
  • I have edited the question earlier –  Nov 24 '15 at 02:24
  • (a) The OP did not include the restriction to hand-calculation in the question itself. (I don't fault the OP, you understand, it's just an observation.) (b) This answer states upfront that it is not by hand; it is not misleading. (c) It is posted in advance of any answer that is by hand. And (d) it contributes to an understanding of the approximation. I think it is therefore worthy of a compensatory upvote. +1 – Brian Tung Nov 24 '15 at 02:42
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Maybe it is worth mentioning that this follows from the well-known approximation/inequality $\frac{22}{7} > \pi$ because $(4/3)^4 > \frac{22}{7}$.

user292958
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