• How it started: The story began on Facebook in a math group when a teacher asked to determine the value of $\\lfloor 3^{\\sqrt{3}}\\rfloor$ without a calculator. The challenge was to prove whether it's 6 or 7 (intuition leaned toward 7). Mihai Cris proposed the solution $3^{7}>2^{10} \Rightarrow 3^{17} > 6^{10} \Rightarrow 3^{1.7} > 6$ and $3^7<7^4 \Rightarrow 3^{1.75}<7$. Since $\\sqrt{3}\\approx 1.73$, $\\lfloor 3^{\\sqrt{3}}\\rfloor=6$. Later on Reddit, user /u/JiminP proposed an alternative method.
• The new problem: This problem led to the question of whether small numbers raised to irrational powers could be approximated using only addition and multiplication without calculators, logs, or radicals. A final formula was provided: $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: Logarithmic and exponential functions were used. $3^{\\sqrt{3}}=(e^{\\ln 3})^{\\sqrt{3}}=e^{\\sqrt{3}\\ln 3}$. The Taylor Series expansion of $e^x$ was used to approximate $3^{\\sqrt{3}}$ by computing $1+\\frac{(\\sqrt{3}\\ln 3)^1}{1!}+\\frac{(\\sqrt{3}\\ln 3)^2}{2!}+\\frac{(\\sqrt{3}\\ln 3)^3}{3!}+\\frac{(\\sqrt{3}\\ln 3)^4}{4!}+\\dots$. The first 8 terms were computed using WolframAlpha.
• The eureka!: Flashbacks from school days led to the idea of small angle approximations for $e^x$, $\\ln x$, and $\\sqrt{x}$. Also, watching a video on Padé Approximations by Michael Penn sparked interest.
• Pade approximations are “smooth”: Padé approximation is a rational function used to represent a given function. It distributes control points between the denominator and numerator. For $e^x$, the $[1/1]$ Padé approximation is $\\frac{2+x}{2-x}$, and the $[2/2]$ approximation is $\\frac{x^2+6x+12}{x^2-6x+12}$, which approximates $e^x$ better. For $\\ln x$, the $[2/2]$ Padé approximation is $\\frac{3(x-1)(x+1)}{x^{2}+4x+1}$, but it's bad for $x < 0.5$. For $\\sqrt{x}$, the $[2/2]$ approximation is $\\frac{5x^{2}+10x+1}{x^{2}+10x+5}$.
• The “monster” function: By combining the approximations for $e^x$, $\\ln x$, and $\\sqrt{x}$, an approximation for $a^b$ was derived: $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}}$.
• Testing the “monster” function: When computing $2^{\\sqrt{2}}$, the approximation was $2.6612$ (actual result $2.6651$). For $3^{\\sqrt{3}}$, the approximation was $6.58$ (actual result $6.70$).
• The deviation for \(0.1^{\sqrt{x}}\), \(2^{\sqrt{x}}\) and \(3^{\sqrt{x}}\): Graphs showed the deviation of the approximation for these values.
• Final thoughts: The "monster" function is impractical in modern times but was just a fun way to explore Padé Approximations. There was a brief consideration of extending the technique to Fourier series.