Methodology for Solving Recursive Functions Problems :
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0
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Given that $f(x) = f(x+3)+ x^2 +x -3$ for all real numbers ,
and $f(1)=2$. Find $f(400)$ .
What would be the general approach for these sorts of problems ?
sequences-and-series algebra-precalculus functions arithmetic recursion
asked Nov 22 at 10:21
A.S.O
32
add a comment |
up vote
0
down vote
favorite
Given that $f(x) = f(x+3)+ x^2 +x -3$ for all real numbers ,
and $f(1)=2$. Find $f(400)$ .
What would be the general approach for these sorts of problems ?
sequences-and-series algebra-precalculus functions arithmetic recursion
asked Nov 22 at 10:21
A.S.O
32
Did you invent this problem ? I doubt there is a closed-form solution...
– Yves Daoust
Nov 23 at 9:47
@YvesDaoust No, It was in a Russian Math. Olympiad .
– A.S.O
Nov 23 at 9:48
@YvesDaoust I Checked it again. and corrected $+x^2$ instead of $-x^2$ .
– A.S.O
Nov 23 at 9:53
My bad, it is an easy recurrence.
– Yves Daoust
Nov 23 at 9:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given that $f(x) = f(x+3)+ x^2 +x -3$ for all real numbers ,
and $f(1)=2$. Find $f(400)$ .
What would be the general approach for these sorts of problems ?
sequences-and-series algebra-precalculus functions arithmetic recursion
asked Nov 22 at 10:21
A.S.O
32
Given that $f(x) = f(x+3)+ x^2 +x -3$ for all real numbers ,
and $f(1)=2$. Find $f(400)$ .
What would be the general approach for these sorts of problems ?
sequences-and-series algebra-precalculus functions arithmetic recursion
sequences-and-series algebra-precalculus functions arithmetic recursion
asked Nov 22 at 10:21
A.S.O
32
asked Nov 22 at 10:21
A.S.O
32
edited Nov 23 at 9:51
asked Nov 22 at 10:21
A.S.O
32
asked Nov 22 at 10:21
A.S.O
32
asked Nov 22 at 10:21
A.S.O
32
32
Did you invent this problem ? I doubt there is a closed-form solution...
– Yves Daoust
Nov 23 at 9:47
@YvesDaoust No, It was in a Russian Math. Olympiad .
– A.S.O
Nov 23 at 9:48
@YvesDaoust I Checked it again. and corrected $+x^2$ instead of $-x^2$ .
– A.S.O
Nov 23 at 9:53
My bad, it is an easy recurrence.
– Yves Daoust
Nov 23 at 9:55
add a comment |
Did you invent this problem ? I doubt there is a closed-form solution...
– Yves Daoust
Nov 23 at 9:47
@YvesDaoust No, It was in a Russian Math. Olympiad .
– A.S.O
Nov 23 at 9:48
@YvesDaoust I Checked it again. and corrected $+x^2$ instead of $-x^2$ .
– A.S.O
Nov 23 at 9:53
My bad, it is an easy recurrence.
– Yves Daoust
Nov 23 at 9:55
Did you invent this problem ? I doubt there is a closed-form solution...
– Yves Daoust
Nov 23 at 9:47
Did you invent this problem ? I doubt there is a closed-form solution...
– Yves Daoust
Nov 23 at 9:47
@YvesDaoust No, It was in a Russian Math. Olympiad .
– A.S.O
Nov 23 at 9:48
@YvesDaoust No, It was in a Russian Math. Olympiad .
– A.S.O
Nov 23 at 9:48
@YvesDaoust I Checked it again. and corrected $+x^2$ instead of $-x^2$ .
– A.S.O
Nov 23 at 9:53
@YvesDaoust I Checked it again. and corrected $+x^2$ instead of $-x^2$ .
– A.S.O
Nov 23 at 9:53
My bad, it is an easy recurrence.
– Yves Daoust
Nov 23 at 9:55
My bad, it is an easy recurrence.
– Yves Daoust
Nov 23 at 9:55
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Hint:
From the given equation,
$$f(x+3)=f(x)-(x^2+x-3)$$
and by induction,
$$f(400)=f(1)-sum_{n=0}^{132}((3n+1)^2+(3n+1)-3)=f(1)-sum_{n=0}^{132}(9n^2+9n-1).$$
The terms of the summation are easily found using the Faulhaber formulas.
edited Nov 23 at 10:24
Yannick Neyt
33
answered Nov 23 at 10:01
Yves Daoust
123k668219
@yannickneyt: right, thanks for the fix.
– Yves Daoust
Nov 23 at 10:25
One needs to know $f(x)$ for all $xin[0,3)$ (or some equivalent set mod $3$) to know the values for all $mathbb{R}$. Since the question only asked for $f(400)$ given $f(1)$ and $400equiv1pmod3$, we're okay.
– robjohn♦
Nov 23 at 14:26
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Hint:
From the given equation,
$$f(x+3)=f(x)-(x^2+x-3)$$
and by induction,
$$f(400)=f(1)-sum_{n=0}^{132}((3n+1)^2+(3n+1)-3)=f(1)-sum_{n=0}^{132}(9n^2+9n-1).$$
The terms of the summation are easily found using the Faulhaber formulas.
edited Nov 23 at 10:24
Yannick Neyt
33
answered Nov 23 at 10:01
Yves Daoust
123k668219
@yannickneyt: right, thanks for the fix.
– Yves Daoust
Nov 23 at 10:25
One needs to know $f(x)$ for all $xin[0,3)$ (or some equivalent set mod $3$) to know the values for all $mathbb{R}$. Since the question only asked for $f(400)$ given $f(1)$ and $400equiv1pmod3$, we're okay.
– robjohn♦
Nov 23 at 14:26
add a comment |
up vote
0
down vote
accepted
Hint:
From the given equation,
$$f(x+3)=f(x)-(x^2+x-3)$$
and by induction,
$$f(400)=f(1)-sum_{n=0}^{132}((3n+1)^2+(3n+1)-3)=f(1)-sum_{n=0}^{132}(9n^2+9n-1).$$
The terms of the summation are easily found using the Faulhaber formulas.
edited Nov 23 at 10:24
Yannick Neyt
33
answered Nov 23 at 10:01
Yves Daoust
123k668219
@yannickneyt: right, thanks for the fix.
– Yves Daoust
Nov 23 at 10:25
One needs to know $f(x)$ for all $xin[0,3)$ (or some equivalent set mod $3$) to know the values for all $mathbb{R}$. Since the question only asked for $f(400)$ given $f(1)$ and $400equiv1pmod3$, we're okay.
– robjohn♦
Nov 23 at 14:26
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint:
From the given equation,
$$f(x+3)=f(x)-(x^2+x-3)$$
and by induction,
$$f(400)=f(1)-sum_{n=0}^{132}((3n+1)^2+(3n+1)-3)=f(1)-sum_{n=0}^{132}(9n^2+9n-1).$$
The terms of the summation are easily found using the Faulhaber formulas.
edited Nov 23 at 10:24
Yannick Neyt
33
answered Nov 23 at 10:01
Yves Daoust
123k668219
Hint:
From the given equation,
$$f(x+3)=f(x)-(x^2+x-3)$$
and by induction,
$$f(400)=f(1)-sum_{n=0}^{132}((3n+1)^2+(3n+1)-3)=f(1)-sum_{n=0}^{132}(9n^2+9n-1).$$
The terms of the summation are easily found using the Faulhaber formulas.
edited Nov 23 at 10:24
Yannick Neyt
33
answered Nov 23 at 10:01
Yves Daoust
123k668219
edited Nov 23 at 10:24
Yannick Neyt
33
edited Nov 23 at 10:24
Yannick Neyt
33
edited Nov 23 at 10:24
Yannick Neyt
33
33
answered Nov 23 at 10:01
Yves Daoust
123k668219
answered Nov 23 at 10:01
Yves Daoust
123k668219
answered Nov 23 at 10:01
Yves Daoust
123k668219
123k668219
@yannickneyt: right, thanks for the fix.
– Yves Daoust
Nov 23 at 10:25
One needs to know $f(x)$ for all $xin[0,3)$ (or some equivalent set mod $3$) to know the values for all $mathbb{R}$. Since the question only asked for $f(400)$ given $f(1)$ and $400equiv1pmod3$, we're okay.
– robjohn♦
Nov 23 at 14:26
add a comment |
@yannickneyt: right, thanks for the fix.
– Yves Daoust
Nov 23 at 10:25
One needs to know $f(x)$ for all $xin[0,3)$ (or some equivalent set mod $3$) to know the values for all $mathbb{R}$. Since the question only asked for $f(400)$ given $f(1)$ and $400equiv1pmod3$, we're okay.
– robjohn♦
Nov 23 at 14:26
@yannickneyt: right, thanks for the fix.
– Yves Daoust
Nov 23 at 10:25
@yannickneyt: right, thanks for the fix.
– Yves Daoust
Nov 23 at 10:25
One needs to know $f(x)$ for all $xin[0,3)$ (or some equivalent set mod $3$) to know the values for all $mathbb{R}$. Since the question only asked for $f(400)$ given $f(1)$ and $400equiv1pmod3$, we're okay.
– robjohn♦
Nov 23 at 14:26
One needs to know $f(x)$ for all $xin[0,3)$ (or some equivalent set mod $3$) to know the values for all $mathbb{R}$. Since the question only asked for $f(400)$ given $f(1)$ and $400equiv1pmod3$, we're okay.
– robjohn♦
Nov 23 at 14:26
add a comment |
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Did you invent this problem ? I doubt there is a closed-form solution...
– Yves Daoust
Nov 23 at 9:47
@YvesDaoust No, It was in a Russian Math. Olympiad .
– A.S.O
Nov 23 at 9:48
@YvesDaoust I Checked it again. and corrected $+x^2$ instead of $-x^2$ .
– A.S.O
Nov 23 at 9:53
My bad, it is an easy recurrence.
– Yves Daoust
Nov 23 at 9:55