Arbitrary decimal value of $A(n)=\left(\frac{11}{10}\right)^n$

Question

For $n\in\mathbb{Z}$ consider the number $$A(n)=\left(1+x\right)^n\bigg{|}_{x=\frac{1}{10}}=\sum_{k=0}^\infty\binom{n}{k}10^{-k}$$ which we have expanded by the Taylor series. It is found that $$a_1=\binom{n}{1},\,a_2=\binom{n}{2},\,\dots ,\,a_k=\binom{n}{k},\dots$$ which I thought gives the explicit formula for the $k$th digit below the decimal point of any number with the form $$A(n)=\left(\frac{11}{10}\right)^n$$ but take for example $A(5)=(11/10)^5=1.\color{red}{6}{1}051$ and say we want to find the $1$st digit below the decimal, highlighted red. We see $a_1={}_5C_{1}=5$ which isn't $6$ as $a_2={}_5C_2=10$ contributes $1$ to the next digit. My question is, is it possible to find the $k$th decimal value of $A(n)$ by means of coefficient extraction from some generating function?

Edit 1: To clarify, I am not looking for a generating function, but some explicit formula for the $k$th digit below the decimal point of $A(n)$. Now, using generating functions to find the formula is of course fine.

Answer 1

Well, if you just want the formula to get to the $r$th digit from starting

Let $$S=\sum_{i=0}^{n-r+1} (10)^{i} \binom{n}{i}$$

Let $P=S\mod{(10^{[n\log_{10}(11)]-r+2})}$

Let $\log_{10}P=Q$

let $\{Q\}$=R ..............$($where $\{Q\}$ denotes fractional part of $Q$$)$

Let $10^R=V$

Let $[V]=U$........(Where $[V]$ denotes greatest integer less than or equal to $V$)

then the $r$th digit from beginning of decimals of $(1.1)^n =U$

finally $$U(n,r)=\left[10^{\left \{{\log_{10}((\sum_{i=0}^{n-r+1}\binom{n}{i}10^i)\mod10^{[nlog_{10}(11)]-r+2})} \right \}}\right]$$

where

$[x]=$ Greatest integer less than or equal to $x$

$\{x\}=$ Fractional part of $x$

Answer 2

Consider the series expansion of the map $$f := z \mapsto \frac{1}{1-11z}$$ around $z=0$. You should find $$f(z) = \sum_{k=0}^\infty \left ( 11z \right) ^ k$$

Therefore, the $m$th digit of $(11/10)^\ell$ is the $(m-1)$th digit of the coefficient of $z^\ell$ in $f$. Can you figure out how to compute that? You should note that after $\ell \ge 5$, you start to have overlapping effects from the binomial sum across digits.

Answer 3

First I focus on finding the first $k$ th digit of $11^n$.

Before you start this exercise, it is helpful to note that you can evaluate the last $p$ digits of $11^n$ in a much easier way, it is simply the last $p$ digits of:

$$\sum_{i=0}^{p-1}{n \choose i} 10^i$$

(Consider the binomial expansion $(10+1)^n$).

Now, recall from number theory that the number of digits in radix $10$ in some number $a$ is given by $\lfloor \log_{10}(a)\rfloor + 1$

So the k-th digit from the start is the $\lfloor n\log_{10}(11)\rfloor + 2 - k$ th digit from the end. So in order to find the first $k$-th digit from the left of $11^n$, you need to evaluate the last digit of the integral part of:

$$ \frac{\sum_{i=0}^{\lfloor n\log_{10}(11)\rfloor + 1 - k} {n\choose i}10^i}{10^{\lfloor n\log_{10}(11)\rfloor + 1 - k }} $$

An approximation for $\log_{10}{11}$ is given by $1.0413$. So in the case given in your question, you just want to find the second digit of $11^5$, so it evaluates to finding:

$$ \sum_{i=0}^{4} {5\choose i}10^i = 61051 $$

Hence, the second digit being $6$, not $5$.

Now if we consider $(1.1)^n$, the $k$ th digit after the decimal is simply the $n - k+ 1$ the digit from the end.

So the final answer to your problem: the first $k$ th digit after the decimal of $1.1^n$ is the last digit from left of:

$$ \lfloor\frac{\sum_{i=0}^{n - k} {n\choose i}10^i}{10^{n-k}}\rfloor $$

or simply:

$$ \lfloor\frac{\sum_{i=0}^{n - k} {n\choose i}10^i}{10^{n-k}}\rfloor \bmod 10 $$

Although I do not find to be very useful, I think it answers your problem.