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Difference between revisions of "1970 IMO Problems/Problem 3"

(I found a contradiction, please check.)
 
(One intermediate revision by one other user not shown)
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<center>
 
<center>
−
<math>b_n = \sum_{k=1}^{n} \left( 1 - \frac{a_{k-1}}{a_{k}} \right)</math>
+
<math>b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}}</math>
 
</center>
 
</center>
  
Line 17: Line 17:
  
 
==Solution==
 
==Solution==
−
Contradiction to (a): Let <math>a_{k}=k</math>. Thus <math>b_n = \sum_{k=1}^{n} \left( 1 - \frac{k-1}{k} \right)=\sum_{k=1}^{n} \frac{1}{k}</math> and that sum tends to infinity as <math>k</math> tends to infinity.
+
<math>b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}} = \sum_{k=1}^{n} \frac{a_k - a_{k-1}}{a_k\sqrt{a_k}} = \sum_{k=1}^{n} (a_k - a_{k-1})\left(a_k^{-\dfrac{3}{2}}\right)</math>
  
−
{{solution}}
+
Let <math>X_k</math> be the rectangle with the verticies: <math>(a_{k-1},0)</math>; <math>(a_{k},0)</math>; <math>(a_{k},a_k^{-\dfrac{3}{2}})</math>; <math>(a_{k-1},a_k^{-\dfrac{3}{2}})</math>.
 +
 
 +
<asy>
 +
import graph;
 +
size(10cm,10cm,IgnoreAspect);
 +
Label f;
 +
f.p=fontsize(6);
 +
xaxis(0,10);
 +
yaxis(0,1);
 +
real f(real x)
 +
{
 +
return x^(-3/2);
 +
}
 +
draw(graph(f,1,10));
 +
draw((1,0)--(1,1));
 +
draw((1,0)--(2,0)--(2,2^(-3/2))--(1,2^(-3/2))--cycle);
 +
draw((2,0)--(3,0)--(3,3^(-3/2))--(2,3^(-3/2))--cycle);
 +
draw((3,0)--(4,0)--(4,4^(-3/2))--(3,4^(-3/2))--cycle);
 +
draw((6,0)--(7,0)--(7,7^(-3/2))--(6,7^(-3/2))--cycle);
 +
draw((7,0)--(8,0)--(8,8^(-3/2))--(7,8^(-3/2))--cycle);
 +
label("$X_1$",(1.5,0),0.5*N);
 +
label("$X_2$",(2.5,0),0.5*N);
 +
label("$X_3$",(3.5,0),0.5*N);
 +
label("$\cdots$",(5,0),N);
 +
label("$X_{n-1}$",(6.5,0),0.5*N);
 +
label("$X_n$",(7.5,0),0.5*N);
 +
label("$a_0$",(1,0),0.5*S);
 +
label("$a_1$",(2,0),0.5*S);
 +
label("$a_2$",(3,0),0.5*S);
 +
label("$a_3$",(4,0),0.5*S);
 +
label("$a_{n-2}$",(6,0),0.5*S);
 +
label("$a_{n-1}$",(7,0),0.5*S);
 +
label("$a_n$",(8,0),0.5*S);
 +
</asy>
 +
 
 +
For all <math>k \in \mathbb{N}</math>, the area of <math>X_k</math> is <math>(a_k - a_{k-1})\left(a_k^{-\dfrac{3}{2}}\right)</math>. Therefore, <math>b_n = \sum_{k=1}^{n} [X_k]</math>
 +
 
 +
For all sequences <math>\{ a_k \}</math> and all <math>k \in \mathbb{N}</math>, <math>X_k</math> lies above the <math>x</math>-axis, below the curve <math>f(x) = x^{-\dfrac{3}{2}}</math>, and in between the lines <math>x = 1</math> and <math>x = a_n</math>, Also, all such rectangles are disjoint.
 +
 
 +
Thus, <math>b_n = \sum_{k=1}^{n} [X_k] < \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx = \left[-\dfrac{2}{\sqrt{x}}\right]_{1}^{a_n} = 2 - \dfrac{2}{\sqrt{a_n}} < 2</math> as desired.
 +
 
 +
By choosing <math>a_k = 1 + k (\Delta x)</math>, where <math>\Delta x > 0</math>, <math>b_n</math> is a [[Riemann sum]] for <math>\int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx</math>. Thus, <math>\lim_{\Delta x \to 0^+} b_n = \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx</math>.
 +
 
 +
Therefore, <math>\lim_{n \to \infty} \left[ \lim_{\Delta x \to 0^+} b_n \right] = \lim_{n \to \infty} \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx</math> <math>= \lim_{n \to \infty} 2 - \dfrac{2}{\sqrt{a_n}} = \lim_{n \to \infty} 2 - \dfrac{2}{\sqrt{1+n(\Delta x)}} = 2</math>.
 +
 
 +
So for any <math>c \in [0,2)</math>, we can always select a small enough <math>\Delta x > 0</math> to form a sequence <math>\{ a_n \}</math>satisfying the above properties such that <math>b_n > c</math> for large enough <math>n</math> as desired.
  
 
==See also==
 
==See also==

Latest revision as of 15:10, 26 July 2009

Problem

The real numbers $a_0, a_1, \ldots, a_n, \ldots$ satisfy the condition:

$1 = a_{0} \leq a_{1} \leq \cdots \leq a_{n} \leq \cdots$.

The numbers $b_{1}, b_{2}, \ldots, b_n, \ldots$ are defined by

$b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}}$

(a) Prove that $0 \leq b_n < 2$ for all $n$.

(b) given $c$ with $0 \leq c < 2$, prove that there exist numbers $a_0, a_1, \ldots$ with the above properties such that $b_n > c$ for large enough $n$.

Solution

$b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}} = \sum_{k=1}^{n} \frac{a_k - a_{k-1}}{a_k\sqrt{a_k}} = \sum_{k=1}^{n} (a_k - a_{k-1})\left(a_k^{-\dfrac{3}{2}}\right)$

Let $X_k$ be the rectangle with the verticies: $(a_{k-1},0)$; $(a_{k},0)$; $(a_{k},a_k^{-\dfrac{3}{2}})$; $(a_{k-1},a_k^{-\dfrac{3}{2}})$.

[asy] import graph; size(10cm,10cm,IgnoreAspect); Label f; f.p=fontsize(6); xaxis(0,10); yaxis(0,1); real f(real x) { return x^(-3/2); } draw(graph(f,1,10)); draw((1,0)--(1,1)); draw((1,0)--(2,0)--(2,2^(-3/2))--(1,2^(-3/2))--cycle); draw((2,0)--(3,0)--(3,3^(-3/2))--(2,3^(-3/2))--cycle); draw((3,0)--(4,0)--(4,4^(-3/2))--(3,4^(-3/2))--cycle); draw((6,0)--(7,0)--(7,7^(-3/2))--(6,7^(-3/2))--cycle); draw((7,0)--(8,0)--(8,8^(-3/2))--(7,8^(-3/2))--cycle); label("$X_1$",(1.5,0),0.5*N); label("$X_2$",(2.5,0),0.5*N); label("$X_3$",(3.5,0),0.5*N); label("$\cdots$",(5,0),N); label("$X_{n-1}$",(6.5,0),0.5*N); label("$X_n$",(7.5,0),0.5*N); label("$a_0$",(1,0),0.5*S); label("$a_1$",(2,0),0.5*S); label("$a_2$",(3,0),0.5*S); label("$a_3$",(4,0),0.5*S); label("$a_{n-2}$",(6,0),0.5*S); label("$a_{n-1}$",(7,0),0.5*S); label("$a_n$",(8,0),0.5*S); [/asy]

For all $k \in \mathbb{N}$, the area of $X_k$ is $(a_k - a_{k-1})\left(a_k^{-\dfrac{3}{2}}\right)$. Therefore, $b_n = \sum_{k=1}^{n} [X_k]$

For all sequences $\{ a_k \}$ and all $k \in \mathbb{N}$, $X_k$ lies above the $x$-axis, below the curve $f(x) = x^{-\dfrac{3}{2}}$, and in between the lines $x = 1$ and $x = a_n$, Also, all such rectangles are disjoint.

Thus, $b_n = \sum_{k=1}^{n} [X_k] < \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx = \left[-\dfrac{2}{\sqrt{x}}\right]_{1}^{a_n} = 2 - \dfrac{2}{\sqrt{a_n}} < 2$ as desired.

By choosing $a_k = 1 + k (\Delta x)$, where $\Delta x > 0$, $b_n$ is a Riemann sum for $\int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$. Thus, $\lim_{\Delta x \to 0^+} b_n = \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$.

Therefore, $\lim_{n \to \infty} \left[ \lim_{\Delta x \to 0^+} b_n \right] = \lim_{n \to \infty} \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$ $= \lim_{n \to \infty} 2 - \dfrac{2}{\sqrt{a_n}} = \lim_{n \to \infty} 2 - \dfrac{2}{\sqrt{1+n(\Delta x)}} = 2$.

So for any $c \in [0,2)$, we can always select a small enough $\Delta x > 0$ to form a sequence $\{ a_n \}$satisfying the above properties such that $b_n > c$ for large enough $n$ as desired.

See also

1970 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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