SMO 2004 Question 21












4












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    See also this question.
    $endgroup$
    – Dietrich Burde
    Dec 21 '18 at 19:17












  • $begingroup$
    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
    $endgroup$
    – Lance
    Dec 23 '18 at 0:46


















4












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    See also this question.
    $endgroup$
    – Dietrich Burde
    Dec 21 '18 at 19:17












  • $begingroup$
    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
    $endgroup$
    – Lance
    Dec 23 '18 at 0:46
















4












4








4





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 21 '18 at 19:06









Matthew TanMatthew Tan

555




555












  • $begingroup$
    See also this question.
    $endgroup$
    – Dietrich Burde
    Dec 21 '18 at 19:17












  • $begingroup$
    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
    $endgroup$
    – Lance
    Dec 23 '18 at 0:46




















  • $begingroup$
    See also this question.
    $endgroup$
    – Dietrich Burde
    Dec 21 '18 at 19:17












  • $begingroup$
    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
    $endgroup$
    – Lance
    Dec 23 '18 at 0:46


















$begingroup$
See also this question.
$endgroup$
– Dietrich Burde
Dec 21 '18 at 19:17






$begingroup$
See also this question.
$endgroup$
– Dietrich Burde
Dec 21 '18 at 19:17














$begingroup$
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
$endgroup$
– Lance
Dec 23 '18 at 0:46






$begingroup$
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
$endgroup$
– Lance
Dec 23 '18 at 0:46












2 Answers
2






active

oldest

votes


















6












$begingroup$

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 $lfloor a_{100}rfloor = color{red}{14}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How did you get b_n is less than or equal to 2n-1?
    $endgroup$
    – Matthew Tan
    Dec 22 '18 at 10:02










  • $begingroup$
    @Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
    $endgroup$
    – Lance
    Dec 22 '18 at 22:12





















5












$begingroup$

The growth rate of this sequence can be approximately modeled by the differential equation $y' = frac {1}{y}$



$a_napprox sqrt{2n}\
a_{100}approx 14.14$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $lfloor a_nrfloor$)?
    $endgroup$
    – Jack D'Aurizio
    Dec 21 '18 at 20:38













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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 $lfloor a_{100}rfloor = color{red}{14}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How did you get b_n is less than or equal to 2n-1?
    $endgroup$
    – Matthew Tan
    Dec 22 '18 at 10:02










  • $begingroup$
    @Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
    $endgroup$
    – Lance
    Dec 22 '18 at 22:12


















6












$begingroup$

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 $lfloor a_{100}rfloor = color{red}{14}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How did you get b_n is less than or equal to 2n-1?
    $endgroup$
    – Matthew Tan
    Dec 22 '18 at 10:02










  • $begingroup$
    @Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
    $endgroup$
    – Lance
    Dec 22 '18 at 22:12
















6












6








6





$begingroup$

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 $lfloor a_{100}rfloor = color{red}{14}.$






share|cite|improve this answer









$endgroup$



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 $lfloor a_{100}rfloor = color{red}{14}.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 21 '18 at 19:17









Jack D'AurizioJack D'Aurizio

1




1












  • $begingroup$
    How did you get b_n is less than or equal to 2n-1?
    $endgroup$
    – Matthew Tan
    Dec 22 '18 at 10:02










  • $begingroup$
    @Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
    $endgroup$
    – Lance
    Dec 22 '18 at 22:12




















  • $begingroup$
    How did you get b_n is less than or equal to 2n-1?
    $endgroup$
    – Matthew Tan
    Dec 22 '18 at 10:02










  • $begingroup$
    @Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
    $endgroup$
    – Lance
    Dec 22 '18 at 22:12


















$begingroup$
How did you get b_n is less than or equal to 2n-1?
$endgroup$
– Matthew Tan
Dec 22 '18 at 10:02




$begingroup$
How did you get b_n is less than or equal to 2n-1?
$endgroup$
– Matthew Tan
Dec 22 '18 at 10:02












$begingroup$
@Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
$endgroup$
– Lance
Dec 22 '18 at 22:12






$begingroup$
@Mattew Tan $b_{n}>2+b_{n-1}implies b_n>2(n-1)+b_1=2n-1\b_n=b_{n-1}+2+cfrac 1{b_{n-1}}<b_{n-1}+2+cfrac 1{2n-3}, \ cdots \ b_2<b_1+2+cfrac 1{2cdot 2-1},\ text{add above and cancel out }b_{n-1} text{ to } b_2, text{ to arrive at the 2nd inequality.}$
$endgroup$
– Lance
Dec 22 '18 at 22:12













5












$begingroup$

The growth rate of this sequence can be approximately modeled by the differential equation $y' = frac {1}{y}$



$a_napprox sqrt{2n}\
a_{100}approx 14.14$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $lfloor a_nrfloor$)?
    $endgroup$
    – Jack D'Aurizio
    Dec 21 '18 at 20:38


















5












$begingroup$

The growth rate of this sequence can be approximately modeled by the differential equation $y' = frac {1}{y}$



$a_napprox sqrt{2n}\
a_{100}approx 14.14$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 $lfloor a_nrfloor$)?
    $endgroup$
    – Jack D'Aurizio
    Dec 21 '18 at 20:38
















5












5








5





$begingroup$

The growth rate of this sequence can be approximately modeled by the differential equation $y' = frac {1}{y}$



$a_napprox sqrt{2n}\
a_{100}approx 14.14$






share|cite|improve this answer









$endgroup$



The growth rate of this sequence can be approximately modeled by the differential equation $y' = frac {1}{y}$



$a_napprox sqrt{2n}\
a_{100}approx 14.14$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 21 '18 at 19:18









Doug MDoug M

44.7k31854




44.7k31854












  • $begingroup$
    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 $lfloor a_nrfloor$)?
    $endgroup$
    – Jack D'Aurizio
    Dec 21 '18 at 20:38




















  • $begingroup$
    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 $lfloor a_nrfloor$)?
    $endgroup$
    – Jack D'Aurizio
    Dec 21 '18 at 20:38


















$begingroup$
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 $lfloor a_nrfloor$)?
$endgroup$
– Jack D'Aurizio
Dec 21 '18 at 20:38






$begingroup$
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 $lfloor a_nrfloor$)?
$endgroup$
– Jack D'Aurizio
Dec 21 '18 at 20:38




















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