Simplifying logarithm of a product

Question

My textbook reads (without explanation, naturally):

$\log \prod_{y=0}^{9}{(1+\frac{1}{10x+y})} = \log{(1+\frac{1}{x})}$

Wondering how this was achieved...

Answer 1

Notice that $1+\frac{1}{10x+y}=\frac{10x+y+1}{10x+y}$, so the product is: $$\prod_{y=0}^{9}{\left(1+\frac{1}{10x+y}\right)}=\frac{10x+1}{10x}\cdot\frac{10x+2}{10x+1}\cdot\frac{10x+3}{10x+2}\dots\frac{10x+10}{10x+9}.$$ Everything cancels out except the first denominator and the last numerator, giving $$\frac{10x+10}{10x}=\frac{x+1}{x}=1+\frac{1}{x}.$$

Answer 2

HINT:

Note that

$$\prod_{y=1}^9\left(1+\frac{1}{10x+y}\right)=\left(\frac{10x+1}{10x}\right)\left(\frac{10x+2}{10x+1}\right)\cdots \left(\frac{10x+9}{10x+8}\right)\left(\frac{10x+10}{10x+9}\right)=1+\frac{1}{x}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ For an 'automatic derivation', you can use the Pochhammer Symbol $\ds{\pars{a} _{n}}$ and its relation to the Gamma Function $\ds{\Gamma}$. Namely, $\ds{\pars{a}_{n} = \Gamma\pars{a + n}/\Gamma\pars{a}}$: \begin{align} \prod_{y = 0}^{9}\pars{1 + {1 \over 10x + y}} & = {\prod_{y = 0}^{9}\pars{y + 10x + 1} \over \prod_{y = 0}^{9}\pars{y + 10x}} = {\pars{10x + 1}_{\ 10} \over \pars{10x}_{\ 10}} = {\Gamma\pars{10x + 11} \over \Gamma\pars{10x + 1}}\, {\Gamma\pars{10x} \over \Gamma\pars{10x + 10}} \\[5mm] & = {\Gamma\pars{10x + 11} \over \Gamma\pars{10x + 10}}\, {\Gamma\pars{10x} \over \Gamma\pars{10x + 1}} = \pars{10x + 10}{1 \over 10x} = \bbox[10px,#ffe,border:1px solid navy]{\ds{1 + {1 \over x}}} \end{align}