Question

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

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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 \(f(n)=f\left(\frac{n}{10}\right)\) 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 \(f(n)=10.\)

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