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Recursion: computing $f(3^100)$
Recursion function question againcomputer theory recursionMaths/Programming recursion questionA Problem about RecursionHow to solve recursion?Lin Alg 100-Level Recursion ProblemSolve this recursionA positive integer is equal to the sum of digits of a multiple of itself.Recursion RelationsKilling Flies with a Checkerboard Flyswatter
$begingroup$
Given the function $f:N^+ rightarrow N$ such that: $f(2^n)=n^2$, $f(3n)=f(2n)+5$, Find the value of $f(3^100)$:
I tried calculating some smaller $f(3^k)$ for integer $k$, to look for a pattern, If I haven't done silly mistakes I got:$$f(3)=6,f(3^2)=9,f(3^3)=24,f(3^4)=36$$
I noticed that $f(3)cdot4=f(3^3)$ and $f(3^2)cdot4=f(3^4)$, nonetheless, I don't think this kind of pattern holds, I wrote in general that: $$f(3cdot2^n)=n^2+2n+6$$
Hints or partial help is much appreciated!
contest-math recursion
asked Mar 20 at 19:54
Spasoje DurovicSpasoje Durovic
40811
$endgroup$
add a comment |
$begingroup$
Given the function $f:N^+ rightarrow N$ such that: $f(2^n)=n^2$, $f(3n)=f(2n)+5$, Find the value of $f(3^100)$:
I tried calculating some smaller $f(3^k)$ for integer $k$, to look for a pattern, If I haven't done silly mistakes I got:$$f(3)=6,f(3^2)=9,f(3^3)=24,f(3^4)=36$$
I noticed that $f(3)cdot4=f(3^3)$ and $f(3^2)cdot4=f(3^4)$, nonetheless, I don't think this kind of pattern holds, I wrote in general that: $$f(3cdot2^n)=n^2+2n+6$$
Hints or partial help is much appreciated!
contest-math recursion
asked Mar 20 at 19:54
Spasoje DurovicSpasoje Durovic
40811
$endgroup$
$begingroup$
Let me know if I made some silly mistake!
$endgroup$
– Spasoje Durovic
Mar 20 at 19:55
add a comment |
$begingroup$
Given the function $f:N^+ rightarrow N$ such that: $f(2^n)=n^2$, $f(3n)=f(2n)+5$, Find the value of $f(3^100)$:
I tried calculating some smaller $f(3^k)$ for integer $k$, to look for a pattern, If I haven't done silly mistakes I got:$$f(3)=6,f(3^2)=9,f(3^3)=24,f(3^4)=36$$
I noticed that $f(3)cdot4=f(3^3)$ and $f(3^2)cdot4=f(3^4)$, nonetheless, I don't think this kind of pattern holds, I wrote in general that: $$f(3cdot2^n)=n^2+2n+6$$
Hints or partial help is much appreciated!
contest-math recursion
asked Mar 20 at 19:54
Spasoje DurovicSpasoje Durovic
40811
$endgroup$
Given the function $f:N^+ rightarrow N$ such that: $f(2^n)=n^2$, $f(3n)=f(2n)+5$, Find the value of $f(3^100)$:
I tried calculating some smaller $f(3^k)$ for integer $k$, to look for a pattern, If I haven't done silly mistakes I got:$$f(3)=6,f(3^2)=9,f(3^3)=24,f(3^4)=36$$
I noticed that $f(3)cdot4=f(3^3)$ and $f(3^2)cdot4=f(3^4)$, nonetheless, I don't think this kind of pattern holds, I wrote in general that: $$f(3cdot2^n)=n^2+2n+6$$
Hints or partial help is much appreciated!
contest-math recursion
contest-math recursion
asked Mar 20 at 19:54
Spasoje DurovicSpasoje Durovic
40811
asked Mar 20 at 19:54
Spasoje DurovicSpasoje Durovic
40811
asked Mar 20 at 19:54
Spasoje DurovicSpasoje Durovic
40811
asked Mar 20 at 19:54
Spasoje DurovicSpasoje Durovic
40811
asked Mar 20 at 19:54
Spasoje DurovicSpasoje Durovic
40811
40811
$begingroup$
Let me know if I made some silly mistake!
$endgroup$
– Spasoje Durovic
Mar 20 at 19:55
add a comment |
$begingroup$
Let me know if I made some silly mistake!
$endgroup$
– Spasoje Durovic
Mar 20 at 19:55
$begingroup$
Let me know if I made some silly mistake!
$endgroup$
– Spasoje Durovic
Mar 20 at 19:55
$begingroup$
Let me know if I made some silly mistake!
$endgroup$
– Spasoje Durovic
Mar 20 at 19:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can easily show by induction that for all $n in lbrace 0, ..., 100 rbrace$,
$$f(3^100)=f(2^n times 3^100-n) + 5n$$
Using that with $n=100$, you get
$$f(3^100) = f(2^100) + 500 = 100^2+ 500 = 10500$$
answered Mar 20 at 20:02
TheSilverDoeTheSilverDoe
5,066215
$endgroup$
$begingroup$
$f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
$endgroup$
– TheSilverDoe
Mar 20 at 20:10
$begingroup$
Right! I had deleted my comment because I was trying exactly that! great, thanks
$endgroup$
– Spasoje Durovic
Mar 20 at 20:11
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can easily show by induction that for all $n in lbrace 0, ..., 100 rbrace$,
$$f(3^100)=f(2^n times 3^100-n) + 5n$$
Using that with $n=100$, you get
$$f(3^100) = f(2^100) + 500 = 100^2+ 500 = 10500$$
answered Mar 20 at 20:02
TheSilverDoeTheSilverDoe
5,066215
$endgroup$
$begingroup$
$f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
$endgroup$
– TheSilverDoe
Mar 20 at 20:10
$begingroup$
Right! I had deleted my comment because I was trying exactly that! great, thanks
$endgroup$
– Spasoje Durovic
Mar 20 at 20:11
add a comment |
$begingroup$
You can easily show by induction that for all $n in lbrace 0, ..., 100 rbrace$,
$$f(3^100)=f(2^n times 3^100-n) + 5n$$
Using that with $n=100$, you get
$$f(3^100) = f(2^100) + 500 = 100^2+ 500 = 10500$$
answered Mar 20 at 20:02
TheSilverDoeTheSilverDoe
5,066215
$endgroup$
$begingroup$
$f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
$endgroup$
– TheSilverDoe
Mar 20 at 20:10
$begingroup$
Right! I had deleted my comment because I was trying exactly that! great, thanks
$endgroup$
– Spasoje Durovic
Mar 20 at 20:11
add a comment |
$begingroup$
You can easily show by induction that for all $n in lbrace 0, ..., 100 rbrace$,
$$f(3^100)=f(2^n times 3^100-n) + 5n$$
Using that with $n=100$, you get
$$f(3^100) = f(2^100) + 500 = 100^2+ 500 = 10500$$
answered Mar 20 at 20:02
TheSilverDoeTheSilverDoe
5,066215
$endgroup$
You can easily show by induction that for all $n in lbrace 0, ..., 100 rbrace$,
$$f(3^100)=f(2^n times 3^100-n) + 5n$$
Using that with $n=100$, you get
$$f(3^100) = f(2^100) + 500 = 100^2+ 500 = 10500$$
answered Mar 20 at 20:02
TheSilverDoeTheSilverDoe
5,066215
answered Mar 20 at 20:02
TheSilverDoeTheSilverDoe
5,066215
answered Mar 20 at 20:02
TheSilverDoeTheSilverDoe
5,066215
answered Mar 20 at 20:02
TheSilverDoeTheSilverDoe
5,066215
5,066215
$begingroup$
$f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
$endgroup$
– TheSilverDoe
Mar 20 at 20:10
$begingroup$
Right! I had deleted my comment because I was trying exactly that! great, thanks
$endgroup$
– Spasoje Durovic
Mar 20 at 20:11
add a comment |
$begingroup$
$f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
$endgroup$
– TheSilverDoe
Mar 20 at 20:10
$begingroup$
Right! I had deleted my comment because I was trying exactly that! great, thanks
$endgroup$
– Spasoje Durovic
Mar 20 at 20:11
$begingroup$
$f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
$endgroup$
– TheSilverDoe
Mar 20 at 20:10
$begingroup$
$f(3^100)=f(3.3^99)=f(2.3^99)+5 = f(3.(2.3^98))+5 = f(2.2.3^98) + 5+5 = f(2^2.3^98)+2times5 = ...$ and so on.
$endgroup$
– TheSilverDoe
Mar 20 at 20:10
$begingroup$
Right! I had deleted my comment because I was trying exactly that! great, thanks
$endgroup$
– Spasoje Durovic
Mar 20 at 20:11
$begingroup$
Right! I had deleted my comment because I was trying exactly that! great, thanks
$endgroup$
– Spasoje Durovic
Mar 20 at 20:11
add a comment |
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Let me know if I made some silly mistake!
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– Spasoje Durovic
Mar 20 at 19:55