Determine $(3.12)^5$ to one decimal place.

Question

How many terms in the binomial expansion would be needed to determine $(3.12)^5$ to one decimal place?

No calculators allowed by the way. Not really sure how to go about doing it, I could do all the calculations if nessecary, but I was wondering whether there is a more general way of doing things.

Thanks in advance for any contributions.

Answer 1

We solve $243*0.\underbrace{00...00}_{\text{k zeores}}9\le0.1$ or $(2187/10^k)/10\le0.1$ so $k=4$. There's a 9 at the end for the worst case. Pay attention to the new 10 at the power of k+1 $$\newcommand{\c}[2]{{}^{#1}{\mathbb C}_{#2}} (3.12)^5=(3\times 1.04)^5=3^5(1+0.04)^5\\=243(\c500.04^0+\c510.04+\c520.04^2+\c530.04^3+\c540.04^4+\c550.04^5)\\=243(1+0.2+0.016+\underbrace{0.00064}_{\substack{\text{last term with a}\\\text{ non-zero digit on 4th decimal place}}}+...)\approx295.6$$

Answer 2

In these same spirit as in ADG's answer, suppose that you have to compute $x^n$ where $x$ write $x=a.b$, $a$ being an integer and $b$ being the decimal part. So $$x=a\times (1+\frac ba)$$ $$x^n=a^n\times (1+\frac ba)^n$$ Let us call $c=\frac ba<1$ and apply the binomial expansion $$(1+c)^n=1+n c+\frac{n(n-1) }{2} c^2 +\frac{n(n-1)(n-2)}{6} c^3 +\frac{n(n-1)(n-2)(n-3)}{24} c^4+\cdots $$ Then $x^n$.

As a short cut (since $n>n-1>n-2,\cdots$), if you want the error to be less than $\epsilon$, you can such for $k$ such that $$a^n\frac{n^k c^k}{k!} <\epsilon$$ For you case $a=3$, $c=0.04$, $n=5$ so $4$ terms will be sufficient for $\epsilon=0.1$ as ADG showed.