SMO 2004 Question 21
Let $frac{1}{a_1}$, $frac{1}{a_2}$, $frac{1}{a_3}$.... be a sequence of positive numbers defined by: $$a_1=1, a_{n+1}=a_n+frac{1}{a_n}$$ Find the integer part of $a_{100}$.
This question was given in the Singapore Mathematics Olympiad in 2004 and it doesn't follow any of the typical recursion functions. Does anyone know how to even start approaching this question and what kind of motivation would make you use such an approach?
recurrence-relations
asked Dec 21 at 19:06
Matthew Tan
395
add a comment |
Let $frac{1}{a_1}$, $frac{1}{a_2}$, $frac{1}{a_3}$.... be a sequence of positive numbers defined by: $$a_1=1, a_{n+1}=a_n+frac{1}{a_n}$$ Find the integer part of $a_{100}$.
This question was given in the Singapore Mathematics Olympiad in 2004 and it doesn't follow any of the typical recursion functions. Does anyone know how to even start approaching this question and what kind of motivation would make you use such an approach?
recurrence-relations
asked Dec 21 at 19:06
Matthew Tan
395
See also this question.
– Dietrich Burde
Dec 21 at 19:17
BTW, research brings out this tighter inequality which can potentially give the integral part of $a_n$ for a much bigger $n$. artofproblemsolving.com/community/c6h1167601p5587849
– Lance
Dec 23 at 0:46
add a comment |
Let $frac{1}{a_1}$, $frac{1}{a_2}$, $frac{1}{a_3}$.... be a sequence of positive numbers defined by: $$a_1=1, a_{n+1}=a_n+frac{1}{a_n}$$ Find the integer part of $a_{100}$.
This question was given in the Singapore Mathematics Olympiad in 2004 and it doesn't follow any of the typical recursion functions. Does anyone know how to even start approaching this question and what kind of motivation would make you use such an approach?
recurrence-relations
asked Dec 21 at 19:06
Matthew Tan
395
Let $frac{1}{a_1}$, $frac{1}{a_2}$, $frac{1}{a_3}$.... be a sequence of positive numbers defined by: $$a_1=1, a_{n+1}=a_n+frac{1}{a_n}$$ Find the integer part of $a_{100}$.
This question was given in the Singapore Mathematics Olympiad in 2004 and it doesn't follow any of the typical recursion functions. Does anyone know how to even start approaching this question and what kind of motivation would make you use such an approach?
recurrence-relations
recurrence-relations
asked Dec 21 at 19:06
Matthew Tan
395
asked Dec 21 at 19:06
Matthew Tan
395
asked Dec 21 at 19:06
Matthew Tan
395
asked Dec 21 at 19:06
Matthew Tan
395
asked Dec 21 at 19:06
Matthew Tan
395
395
See also this question.
– Dietrich Burde
Dec 21 at 19:17
BTW, research brings out this tighter inequality which can potentially give the integral part of $a_n$ for a much bigger $n$. artofproblemsolving.com/community/c6h1167601p5587849
– Lance
Dec 23 at 0:46
add a comment |
See also this question.
– Dietrich Burde
Dec 21 at 19:17
BTW, research brings out this tighter inequality which can potentially give the integral part of $a_n$ for a much bigger $n$. artofproblemsolving.com/community/c6h1167601p5587849
– Lance
Dec 23 at 0:46
See also this question.
– Dietrich Burde
Dec 21 at 19:17
See also this question.
– Dietrich Burde
Dec 21 at 19:17
BTW, research brings out this tighter inequality which can potentially give the integral part of $a_n$ for a much bigger $n$. artofproblemsolving.com/community/c6h1167601p5587849
– Lance
Dec 23 at 0:46
BTW, research brings out this tighter inequality which can potentially give the integral part of $a_n$ for a much bigger $n$. artofproblemsolving.com/community/c6h1167601p5587849
– Lance
Dec 23 at 0:46
add a comment |
2 Answers
2
active
oldest
votes
This was given in a competition? I am quite surprised, since it is a very classical problem.
Let us define $b_n$ as $a_n^2$. Then
$$ b_{n+1} = b_n + 2 + frac{1}{b_n} $$
easily leads (by induction) to $b_ngeq 2n-1$. By plugging back this approximation in the above recursion, we get $b_n leq 2n-1+left(frac{1}{1}+frac{1}{3}+ldots+frac{1}{2n-3}right)$. In particular $a_{100}$ is bounded between $sqrt{199}$ and
$$ sqrt{199+H_{198}-frac{H_{99}}{2}}leq sqrt{205}, $$
so $lfloor a_{100}rfloor = color{red}{14}.$
answered Dec 21 at 19:17
Jack D'Aurizio
286k33279655
How did you get b_n is less than or equal to 2n-1?
– Matthew Tan
Dec 22 at 10:02
@Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
– Lance
Dec 22 at 22:12
add a comment |
The growth rate of this sequence can be approximately modeled by the differential equation $y' = frac {1}{y}$
$a_napprox sqrt{2n}\
a_{100}approx 14.14$
answered Dec 21 at 19:18
Doug M
43.9k31854
In numerical algorithms for (P)DE we usually replace $frac{d}{dx}$ with the forward difference operator. Here we are (reasonably) doing the opposite (since the associated DE is very simple to solve), but how to be sure such approximation does not change the leading term of the asymptotic behaviour of $a_n$ (and ultimately leads to the same value for $lfloor a_nrfloor$)?
– Jack D'Aurizio
Dec 21 at 20:38
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This was given in a competition? I am quite surprised, since it is a very classical problem.
Let us define $b_n$ as $a_n^2$. Then
$$ b_{n+1} = b_n + 2 + frac{1}{b_n} $$
easily leads (by induction) to $b_ngeq 2n-1$. By plugging back this approximation in the above recursion, we get $b_n leq 2n-1+left(frac{1}{1}+frac{1}{3}+ldots+frac{1}{2n-3}right)$. In particular $a_{100}$ is bounded between $sqrt{199}$ and
$$ sqrt{199+H_{198}-frac{H_{99}}{2}}leq sqrt{205}, $$
so $lfloor a_{100}rfloor = color{red}{14}.$
answered Dec 21 at 19:17
Jack D'Aurizio
286k33279655
How did you get b_n is less than or equal to 2n-1?
– Matthew Tan
Dec 22 at 10:02
@Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
– Lance
Dec 22 at 22:12
add a comment |
This was given in a competition? I am quite surprised, since it is a very classical problem.
Let us define $b_n$ as $a_n^2$. Then
$$ b_{n+1} = b_n + 2 + frac{1}{b_n} $$
easily leads (by induction) to $b_ngeq 2n-1$. By plugging back this approximation in the above recursion, we get $b_n leq 2n-1+left(frac{1}{1}+frac{1}{3}+ldots+frac{1}{2n-3}right)$. In particular $a_{100}$ is bounded between $sqrt{199}$ and
$$ sqrt{199+H_{198}-frac{H_{99}}{2}}leq sqrt{205}, $$
so $lfloor a_{100}rfloor = color{red}{14}.$
answered Dec 21 at 19:17
Jack D'Aurizio
286k33279655
How did you get b_n is less than or equal to 2n-1?
– Matthew Tan
Dec 22 at 10:02
@Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
– Lance
Dec 22 at 22:12
add a comment |
This was given in a competition? I am quite surprised, since it is a very classical problem.
Let us define $b_n$ as $a_n^2$. Then
$$ b_{n+1} = b_n + 2 + frac{1}{b_n} $$
easily leads (by induction) to $b_ngeq 2n-1$. By plugging back this approximation in the above recursion, we get $b_n leq 2n-1+left(frac{1}{1}+frac{1}{3}+ldots+frac{1}{2n-3}right)$. In particular $a_{100}$ is bounded between $sqrt{199}$ and
$$ sqrt{199+H_{198}-frac{H_{99}}{2}}leq sqrt{205}, $$
so $lfloor a_{100}rfloor = color{red}{14}.$
answered Dec 21 at 19:17
Jack D'Aurizio
286k33279655
This was given in a competition? I am quite surprised, since it is a very classical problem.
Let us define $b_n$ as $a_n^2$. Then
$$ b_{n+1} = b_n + 2 + frac{1}{b_n} $$
easily leads (by induction) to $b_ngeq 2n-1$. By plugging back this approximation in the above recursion, we get $b_n leq 2n-1+left(frac{1}{1}+frac{1}{3}+ldots+frac{1}{2n-3}right)$. In particular $a_{100}$ is bounded between $sqrt{199}$ and
$$ sqrt{199+H_{198}-frac{H_{99}}{2}}leq sqrt{205}, $$
so $lfloor a_{100}rfloor = color{red}{14}.$
answered Dec 21 at 19:17
Jack D'Aurizio
286k33279655
answered Dec 21 at 19:17
Jack D'Aurizio
286k33279655
answered Dec 21 at 19:17
Jack D'Aurizio
286k33279655
answered Dec 21 at 19:17
Jack D'Aurizio
286k33279655
286k33279655
How did you get b_n is less than or equal to 2n-1?
– Matthew Tan
Dec 22 at 10:02
@Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
– Lance
Dec 22 at 22:12
add a comment |
How did you get b_n is less than or equal to 2n-1?
– Matthew Tan
Dec 22 at 10:02
@Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
– Lance
Dec 22 at 22:12
How did you get b_n is less than or equal to 2n-1?
– Matthew Tan
Dec 22 at 10:02
How did you get b_n is less than or equal to 2n-1?
– Matthew Tan
Dec 22 at 10:02
@Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
– Lance
Dec 22 at 22:12
@Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
– Lance
Dec 22 at 22:12
add a comment |
The growth rate of this sequence can be approximately modeled by the differential equation $y' = frac {1}{y}$
$a_napprox sqrt{2n}\
a_{100}approx 14.14$
answered Dec 21 at 19:18
Doug M
43.9k31854
In numerical algorithms for (P)DE we usually replace $frac{d}{dx}$ with the forward difference operator. Here we are (reasonably) doing the opposite (since the associated DE is very simple to solve), but how to be sure such approximation does not change the leading term of the asymptotic behaviour of $a_n$ (and ultimately leads to the same value for $lfloor a_nrfloor$)?
– Jack D'Aurizio
Dec 21 at 20:38
add a comment |
The growth rate of this sequence can be approximately modeled by the differential equation $y' = frac {1}{y}$
$a_napprox sqrt{2n}\
a_{100}approx 14.14$
answered Dec 21 at 19:18
Doug M
43.9k31854
In numerical algorithms for (P)DE we usually replace $frac{d}{dx}$ with the forward difference operator. Here we are (reasonably) doing the opposite (since the associated DE is very simple to solve), but how to be sure such approximation does not change the leading term of the asymptotic behaviour of $a_n$ (and ultimately leads to the same value for $lfloor a_nrfloor$)?
– Jack D'Aurizio
Dec 21 at 20:38
add a comment |
The growth rate of this sequence can be approximately modeled by the differential equation $y' = frac {1}{y}$
$a_napprox sqrt{2n}\
a_{100}approx 14.14$
answered Dec 21 at 19:18
Doug M
43.9k31854
The growth rate of this sequence can be approximately modeled by the differential equation $y' = frac {1}{y}$
$a_napprox sqrt{2n}\
a_{100}approx 14.14$
answered Dec 21 at 19:18
Doug M
43.9k31854
answered Dec 21 at 19:18
Doug M
43.9k31854
answered Dec 21 at 19:18
Doug M
43.9k31854
answered Dec 21 at 19:18
Doug M
43.9k31854
43.9k31854
In numerical algorithms for (P)DE we usually replace $frac{d}{dx}$ with the forward difference operator. Here we are (reasonably) doing the opposite (since the associated DE is very simple to solve), but how to be sure such approximation does not change the leading term of the asymptotic behaviour of $a_n$ (and ultimately leads to the same value for $lfloor a_nrfloor$)?
– Jack D'Aurizio
Dec 21 at 20:38
add a comment |
In numerical algorithms for (P)DE we usually replace $frac{d}{dx}$ with the forward difference operator. Here we are (reasonably) doing the opposite (since the associated DE is very simple to solve), but how to be sure such approximation does not change the leading term of the asymptotic behaviour of $a_n$ (and ultimately leads to the same value for $lfloor a_nrfloor$)?
– Jack D'Aurizio
Dec 21 at 20:38
In numerical algorithms for (P)DE we usually replace $frac{d}{dx}$ with the forward difference operator. Here we are (reasonably) doing the opposite (since the associated DE is very simple to solve), but how to be sure such approximation does not change the leading term of the asymptotic behaviour of $a_n$ (and ultimately leads to the same value for $lfloor a_nrfloor$)?
– Jack D'Aurizio
Dec 21 at 20:38
In numerical algorithms for (P)DE we usually replace $frac{d}{dx}$ with the forward difference operator. Here we are (reasonably) doing the opposite (since the associated DE is very simple to solve), but how to be sure such approximation does not change the leading term of the asymptotic behaviour of $a_n$ (and ultimately leads to the same value for $lfloor a_nrfloor$)?
– Jack D'Aurizio
Dec 21 at 20:38
add a comment |
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See also this question.
– Dietrich Burde
Dec 21 at 19:17
BTW, research brings out this tighter inequality which can potentially give the integral part of $a_n$ for a much bigger $n$. artofproblemsolving.com/community/c6h1167601p5587849
– Lance
Dec 23 at 0:46