Introducing the Chikoritaylor Approximation

Tired of the Taylor/Maclaurin approximation? Want something more rational? Well, if you are here in this blog post, welcome to my recreational mathematics blog post. In my sub-blog, we talk about recreational mathematical topics that might appeal to Robloxiapedia readers. Other sub-blogs I do are physics in Roblox games. Now let us get started:

We are now going to define a Chikoritaylor Expansion. But first, let us sum a very special geometric series; it will come back later (chikorita promises).

We are going to sum a geometric series which for ${\displaystyle n=0}$, the term is ${\displaystyle a}$ and for ${\displaystyle n=1}$, the term is ${\displaystyle b}$ .

The geometric progression is then: ${\displaystyle c_{n}=a\left({\frac {b}{a}}\right)^{n}}$.

The finite sum is then: ${\displaystyle \sum _{n=0}^{k}a\left({\frac {b}{a}}\right)^{n}={\frac {a\left(a-b\left({\frac {b}{a}}\right)^{k}\right)}{a-b}}}$

The infinite sum is then: ${\displaystyle \lim _{k\to \infty }{\frac {a\left(a-b\left({\frac {b}{a}}\right)^{k}\right)}{a-b}}={\frac {a^{2}}{a-b}}}$

However, there are caveats: ${\displaystyle a\ne0}$ and ${\displaystyle \left|{\frac {b}{a}}\right|<1}$.

Now we are ready to create a general form:

We define ${\displaystyle a={\frac {1}{n!}}f^{n}\left(0\right)x^{n}}$ and ${\displaystyle b={\frac {1}{\left(n+1\right)!}}f^{\left(n+1\right)}\left(0\right)x^{\left(n+1\right)}}$

To help us clean up the expression further, we will define ${\displaystyle c=f^{n}\left(0\right)}$ and ${\displaystyle d=f^{n+1}\left(0\right)}$.

So now we have: ${\displaystyle a={\frac {1}{n!}}cx^{n}}$ and ${\displaystyle b={\frac {1}{\left(n+1\right)!}}dx^{\left(n+1\right)}}$.

The sub-sum is then ${\displaystyle s={\frac {c^{2}\left(1+n\right)x^{n}}{\left(c\left(1+n\right)-dx\right)n!}}}$, which in raw form becomes: ${\displaystyle s={\frac {f^{n}\left(0\right)^{2}\left(1+n\right)x^{n}}{\left(f^{n}\left(0\right)\left(1+n\right)-f^{\left(n+1\right)}\left(0\right)x\right)n!}}}$.

We are now ready to define the approximation as ${\displaystyle S\left(x\right)=\sum _{n=0}^{k-1}{\frac {1}{n!}}f^{k}\left(0\right)x^{n}+{\frac {f^{k}\left(0\right)^{2}\left(1+k\right)x^{k}}{\left(f^{k}\left(0\right)\left(1+k\right)-f^{\left(k+1\right)}\left(0\right)x\right)k!}}}$.

Example: Approximating ${\displaystyle y\ =\ \sin \left(x\right)}$:

${\displaystyle y\ =\ \sum _{n=0}^{a}{\frac {\left(-1\right)^{n}x^{\left(2n+1\right)}}{\left(2n+1\right)!}}}$

Let ${\displaystyle a={\frac {\left(-1\right)^{k}x^{1+2k}}{\left(1+2k\right)!}}}$. Similarly, let ${\displaystyle b={\frac {\left(-1\right)^{1+k}x^{3+2k}}{\left(3+2k\right)!}}}$.

Then ${\displaystyle s={\frac {\left(-1\right)^{k}x^{1+2k}\left(3+2k\right)!}{\left(1+2k\right)\left(2k\right)!\left(x^{2}\left(1+2k\right)!+\left(3+2k\right)!\right)}}}$

So now the series becomes: ${\displaystyle \sin \left(x\right)\approx \sum _{n=0}^{k-1}{\frac {\left(-1\right)^{n}x^{1+2n}}{\left(1+2n\right)!}}+{\frac {\left(-1\right)^{k}x^{1+2k}\left(3+2k\right)!}{\left(1+2k\right)\left(2k\right)!\left(x^{2}\left(1+2k\right)!+\left(3+2k\right)!\right)}}}$

Convergence:

Convergence of the sequence is guaranteed if: ${\displaystyle \left|f^{n+1}\left(0\right)x\right|<\left|f^{n}\left(0\right)\left(n+1\right)\right|}$ for all integral n. As in the case of the above example, ${\displaystyle \left|{\frac {x^{2\left(k+1\right)-2k}\left(2k+1\right)!}{\left(2\left(k+1\right)+1\right)!}}\right|<1}$ for all integral k. As k approaches infinity, the LHS goes to zero, of which its absolute value is clearly less than one.

Scrutiny:

Similar to the original Taylor, the series is prone to more over/under estimation as factorials outperform exponential functions.

Reality:

Virtually everything can be turned into one.