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

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

Question
Functions
asked 2021-01-08
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.

Answers (1)

2021-01-09
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(n/10) 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 \(\displaystyle{f{{\left({n}\right)}}}={f{{\left({\frac{{{n}}}{{{10}}}}\ {g}{i}{v}{e}{s}\ {P}{S}{K}{f{{\left({1000}\right)}}}={f{{\left({\frac{{{1000}}}{{{10}}}}={f{{\left({100}\right)}}}\right.}}}\right.}}}\). 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.
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