A joke in approximating numbers raised to irrational powers
How it started
The story behind this post surprisingly starts on Facebook, on a group dedicated to mathematics. A math teacher asked audience to find the \(\lfloor 3^{\sqrt{3}}\rfloor\) without using a calculator. Of course, it was a matter of proving \(\lfloor 3^{\sqrt{3}}\rfloor=6\), or that \(\lfloor 3^{\sqrt{3}}\rfloor=7\) (my intuition said it’s 7).
After a few back and forth discussions, everyone agreed the following solution (by Mihai Cris) is the most acceptable:
\[3^{7}>2^{10} \Rightarrow 3^{7}*3^{10}>2^{10}*3^{10} \Rightarrow 3^{17} > 6^{10} \Rightarrow 3^{\frac{17}{10}}>6 \Rightarrow 3^{1.7} > 6\]In the same time:
\[3^7=2187<2401=7^4 \Rightarrow 3^{7/4} < 7 \Rightarrow 3^{1.75}<7\]Because \(\sqrt{3}\approx 1.73 \Rightarrow \lfloor 3^{\sqrt{3}}\rfloor=6\).
Later edit:
After posting on reddit, /u/JiminP came with a different solution:
In the same time:
\(48<49 \Rightarrow 3*2^4 < 7^2\) and \(3^7<256*3^2=2^8*3^2<7^4 \Rightarrow 7^4 > 3^{\frac{7}{4}}=3^{\sqrt{\frac{49}{16}}} > 3^{\sqrt{\frac{48}{16}}} > 3^{\sqrt{3}}\)
The decision on which solution is the best is left to the reader.
The new problem
The previous problem was cute, but it made me wonder if I can find a way to approximate (small) numbers raised to (small) irrational powers without using a calculator (no logarithms and radicals), by relying solely on addition and multiplication. Basically I was looking for a solution where pen, paper and patience are enough.
This is primarily a joke, so if you don’t want to continue reading, here is the answer:
\[a^{\sqrt{c}} \approx \frac{-120-60[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]-12[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{2}-[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{3}}{-120+60[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]-12[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{2}+[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{3}}\](… as long as \(\frac{1}{2} < \ln a * \sqrt{c} < 3\))
The first attempt
The Logarithmic and the Exponential functions came to the rescue.
- We start by writing \(3^{\sqrt{3}}=(e^{\ln 3})^{\sqrt{3}}=e^{\sqrt{3}\ln 3}\). We know this is true because \(3=e^{\ln 3}=3^{\ln e}=3\).
- We use the Taylor Series expansion of \(e^x=1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots\) .
- So approximating \(3^{\sqrt{3}}\) is just a matter of computing \(1+\frac{(\sqrt{3}\ln 3)^1}{1!}+\frac{(\sqrt{3}\ln 3)^2}{2!}+\frac{(\sqrt{3}\ln 3)^3}{3!}+\frac{(\sqrt{3}\ln 3)^4}{4!}+\dots\).
To get even closer to the actual result, I computed the first 8 terms of the series expansion (and no, it wasn’t by hand, as I’ve used WolframAlpha do it for me):
1+Sqrt[3]*Ln[3]+(Sqrt[3]*Ln[3])^2/2!+(Sqrt[3]*Ln[3])^3/3!+(Sqrt[3]*Ln[3])^4/4!+(Sqrt[3]*Ln[3])^5/5!+(Sqrt[3]*Ln[3])^6/6!+(Sqrt[3]*Ln[3])^7/7!+(Sqrt[3]*Ln[3])^8/8!
The eureka!
At this point I had a few flashbacks from my school days.
Firstly, there was something called Small Angle Approximations, that happen when the angle is, as you would expect, (very) very small:
\[\sin(x) \approx x \\ \cos(x) \approx 1-\frac{x^2}{2} \\ \tan(x) \approx x\]So I’ve wondered if there isn’t something similar for \(e^x, \ln x\) and \(\sqrt{x}\), anything that works for small numbers.
Secondly, I remember watching a few months ago a video from Michael Penn, about something called Padé Approximations: Pade Approximation – unfortunately missed in most Caclulus courses. It was a subject worth exploring.
Pade approximations are “smooth”
So, without going into too much details, a Padé approximation is a rational function (the ratio of two polynomials) used to represent a given function. The smart idea behind this technique is to distribute the control points of a polynomial between the denominator and the numerator of the rational function.
A Padé approximation of order [m/n] for a function \(f(x)\) is expressed as:
Compared to Taylor Series, Padé seem to handle exponential behaviors and discontinuities better, plus in a lot of cases it converges faster.
Needless to say, I was curious how well those approximations actually work, so I’ve computed the Padé aproximation of order [1/1], \(P_{[1/1]}(x)\) for \(e^x\). (For in depth tutorial follow the Professor’s Penn tutorial, linked above).
To find \(a_0, a_1, b_1\) we need to solve the following system of equations:
\[\begin{cases} e^0=P(0)=a_0=1 \\ e^0=P'(0)=a_1-b_1=1 \\ e^0=P''(0)=2b_1(b_1-a_1)=1 \\ \end{cases}\]After solving this, we will get \(P_{[1/1]}=\frac{2+x}{2-x}\).
We can already see that the \(P_{[1/1]}\) follows nicely \(e^x\) for the numbers around \(0\).
\(P_{[2/2]}=\frac{x^2+6x+12}{x^2-6x+12}\) is even better at approximating \(e^x\), at least up to a certain point.
So, in a way, \(e^x \approx \frac{x^2+6x+12}{x^2-6x+12}\) is true for small numbers, just like \(\sin(x) \approx x\) is true for small angles.
Or, even better, \(e^x \approx \frac{-120-60x-12x^{2}-x^{3}}{-120+60x-12x^{2}+x^{3}}\) is true for small small numbers, just like \(\sin(x) \approx x\) is true for small angles.
And it’s get better, for the \(\ln x\) function the equivalent [2/2] Padé approximation is: \(\frac{3(x-1)(x+1)}{x^{2}+4x+1}\). Notice how the approximation is bad if \(x < 0.5\). For this we should find a better function.
And for \(\sqrt{x}\) the [2/2] approximation will be \(\frac{5x^{2}+10x+1}{x^{2}+10x+5}\).
The “monster” function
Putting all together.
- To compute \(e^x\) for \(x \in (0,4)\), we use: \(\frac{-120-60x-12x^{2}-x^{3}}{-120+60x-12x^{2}+x^{3}}\).
- To compute \(\ln x\) for \(x \in (0.5, 4)\) we use: \(\frac{3(x-1)(x+1)}{x^{2}+4x+1}\).
- To compute \(\sqrt{x}\) for \(x \in (0, 4)\) we use: \(\frac{5x^{2}+10x+1}{x^{2}+10x+5}\).
So for “small numbers” we can approximate \(a^b\) using the following expression:
\[a^b \approx \frac{-120-60[b*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]-12[b*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{2}-[b*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{3}}{-120+60[b*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]-12[b*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{2}+[b*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{3}}\]Or we can compute:
\[a^{\sqrt{c}} \approx \frac{-120-60[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]-12[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{2}-[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{3}}{-120+60[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]-12[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{2}+[\frac{5c^{2}+10c+1}{c^{2}+10c+5}*\frac{3(a-1)(a+1)}{a^{2}+4a+1}]^{3}}\]Testing the “monster” function
Now, to test our marvelous approximation, let’s compute \(2^{\sqrt{2}}\):
- We start approximating the logarithm: \(\ln 2=\frac{3(2-1)(2+1)}{2^2+8+1}=\frac{3*1*3}{4+8+1}=\frac{9}{13} \approx 0.69\).
- We continue approximating the radical: \(\sqrt{2}=\frac{5*2^2+20+1}{4+20+5}=\frac{20+20+1}{29}=\frac{41}{29} \approx 1.41\)
- We compute the product: \(0.69*1.41 \approx 0.97\).
- We actually compute the power: \(2^{\sqrt{2}}=e^{\sqrt{2}\ln 2} \approx e^{0.97}=\frac{-120-60*(0.97)-12*(0.97)^{2}-(0.97)^{3}}{-120+60*(0.97)-12*(0.97)^{2}+(0.97)^{3}} \approx 2.6612\)
The actual result should’ve been: \(2.6651\). Not bad.
Computing \(3^{\sqrt{3}}\) with our function yields: \(6.58\) instead of \(6.70\). Not terrible.
The deviation for \(0.1^{\sqrt{x}}\), \(2^{\sqrt{x}}\) and \(3^{\sqrt{x}}\)
Final thoughts
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The monster function is impractical, especially after the discovery of the transistor and the logic gates. Even so, the article’s true purpose was to find a goofy pretext to try Padé Approximations by myself.
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I was considering applying the same concept that Padé used with polynomials to Fourier series. While others have attempted this, the results aren’t as remarkable. For instance, an expression like \(S_N(x)=a_{0}+\sum_{n=1}{N}(a_n\cos(nx)+b_n\sin(nx))\) is certainly polynomial-like enough to be represented as a rational function.Doing this seems to reduce the Gibbs phenomenon near discontinuities. However, the main drawback is that Padé approximations are not periodic in nature.
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