Difference between revisions of "1970 IMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | {{ | + | <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> |
+ | |||
+ | 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 16:10, 26 July 2009
Problem
The real numbers satisfy the condition:
.
The numbers are defined by
(a) Prove that for all .
(b) given with , prove that there exist numbers with the above properties such that for large enough .
Solution
Let be the rectangle with the verticies: ; ; ; .
For all , the area of is . Therefore,
For all sequences and all , lies above the -axis, below the curve , and in between the lines and , Also, all such rectangles are disjoint.
Thus, as desired.
By choosing , where , is a Riemann sum for . Thus, .
Therefore, .
So for any , we can always select a small enough to form a sequence satisfying the above properties such that for large enough 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 |