Question

# Find f(n) when n=10k, where ff satisfies the recurrence relation f(n)=ffrac{n}{10}) with f(1)=10.

Functions

Find $$f(n)$$ when $$n=10k$$, where ff satisfies the recurrence relation $$\displaystyle{f{{\left({n}\right)}}}={f}{\frac{{{n}}}{{{10}}}}$$ with $$f(1)=10$$.

Since we need to find $$f(n)$$ when $$n=10k$$, we need to determine a pattern when nn is a power of 10.
Substituting $$n=10$$ into $$\displaystyle{f{{\left({n}\right)}}}={f{{\left({\frac{{{n}}}{{{10}}}}\right)}}}\ {g}{i}{v}{e}{s}{f{{\left({10}\right)}}}={f{{\left({\frac{{{10}}}{{{10}}}}={f{{\left({1}\right)}}}\right.}}}$$. Since $$f(1)=10$$, then $$f(10)=10$$.
Substituting $$n=100$$ into $$f(n)=f\left(\frac{n}{10}\right)$$ gives $$\displaystyle{f{{\left({100}\right)}}}={f{{\left({\frac{{{100}}}{{{10}}}}={f{{\left({10}\right)}}}\right.}}}$$. Since $$f(10)=10$$, then $$f(100)=10$$.
Substituting $$n=1000$$ into $$\displaystyle{f{{\left({n}\right)}}}={f{{\left({\frac{{{n}}}{{{10}}}}\ {g}{i}{v}{e}{s}\ {f{{\left({1000}\right)}}}={f{{\left({\frac{{{1000}}}{{{10}}}}={f{{\left({100}\right)}}}\right.}}}\right.}}}$$. Since $$f(100)=10$$, then $$f(1000)=10$$.
Based on these results, then nn is a power of 10, then the value of f(n) is 10. Therefore, when $$n=10k$$ we have $$f(n)=10.$$