"Find f(n) when n=10k, where ff satisfies the..." was answered by students like you

Wribreeminsl

Answered question

2021-01-08

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

Answer & Explanation

delilnaT

Skilled2021-01-09Added 94 answers

Since we need to find $f (n)$ when $n = 10 k$ , we need to determine a pattern when nn is a power of 10.
Substituting $n = 10$ into $f (n) = f (\frac{n}{10}) g i v e s f (10) = f (\frac{10}{10} = f (1))$ .

Since $f (1) = 10$ , then $f (10) = 10$ .
Substituting $n = 100$ into $f (n) = f (\frac{n}{10})$ gives $f (100) = f (\frac{100}{10} = f (10))$ .

Since $f (10) = 10$ , then $f (100) = 10$ .
Substituting $n = 1000$ into $f (n) = f (\frac{n}{10} g i v e s f (1000) = f (\frac{1000}{10} = f (100)))$ .

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 $n = 10 k$ we have $f (n) = 10.$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Algebra I

Didn't find what you were looking for?