I was reading page 25 of this document. They present a strange form of the series expansion of the Modified Bessel Function of the second kind. But I do not know how can the author get this form ?
${K_1}(x) = {\left( {\frac{\pi }{{2x}}} \right)^{1/2}}{e^{ – x}}\left[ {1 + \frac{3}{{8x}}\left( {1 – \frac{5}{{2(8x)}}\left( {1 – \frac{{21}}{{3(8x)}}} \right)} \right)} \right]$.
I myself observed that (subjectively) using only the first term provide a pretty good approximation.
My problem is to try to use this form to approximate ${K_1}(2\sqrt x )$.
Please help me with this.
Thank you !
Best Answer
The expression$${K_1}(x) = {\left( {\frac{\pi }{{2x}}} \right)^{1/2}}{e^{ - x}}\left[ {1 + \frac{3}{{8x}}\left( {1 - \frac{5}{{2(8x)}}\left( {1 - \frac{{21}}{{3(8x)}}} \right)} \right)} \right]$$ is nothing else than the Horner method.
SInce you use Mathematica, ask for the series expansion of $K_1(x)$ for infinitely large values of $x$. After factoring and simplifying, you have $${K_1}(x) = {\left( {\frac{\pi }{{2x}}} \right)^{1/2}}{e^{ - x}}\left[1+\frac{3}{8 x}-\frac{15}{128 x^2}+\frac{105}{1024 x^3}-\frac{4725}{32768 x^4}+O\left(\frac{1}{x^5}\right)\right]$$ For the expression inside backets, let $t=\frac 1{8x}$ to make $$1+3 t-\frac{15 t^2}{2}+\frac{105 t^3}{2}-\frac{4725 t^4}{8}+O\left(t^5\right)$$ which is $$1+3t\left(1-\frac 5 2t(1-7t(1+ \cdots\right)$$
S0 for your case $${K_1}(2\sqrt x )=\frac 12 {\left( {\frac{\pi }{{\sqrt x}}} \right)^{1/2}}e^{-2 \sqrt{x}}\left[1+3 t-\frac{15 t^2}{2}+\frac{105 t^3}{2}-\frac{4725 t^4}{8}+O(t^5)\right]$$ with $t=\frac 1{16\sqrt x}$.
Now use Horner's method to write the polynomial it $t$. $$1+t \left(t \left(\left(\frac{105}{2}-\frac{4725 t}{8}\right) t-\frac{15}{2}\right)+3\right)$$