Forgot password? New user? Sign up
Existing user? Log in
If x,yx,yx,y are real numbers such that x2014+y2014=1x^{2014}+y^{2014}=1x2014+y2014=1 then prove that:
(∑n=11007x2n+1x4n+1)(∑n=11007y2n+1y4n+1)<1(1−x)(1−y)\left(\sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} \right) \left(\sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} \right)< \dfrac{1}{(1-x)(1-y)}(n=1∑1007x4n+1x2n+1)(n=1∑1007y4n+1y2n+1)<(1−x)(1−y)1
Note by Shivam Jadhav 4 years, 1 month ago
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
_italics_
**bold**
__bold__
- bulleted- list
1. numbered2. list
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"
\(
\)
\[
\]
2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Sort by:
There is a typo in the question. I will explain you what the typo is.
For suppose, take x=−1x=-1x=−1 and y=0y=0y=0. Then we get
∑n=11007x2n+1x4n+1=1007∑n=11007y2n+1y4n+1=10071(1−x)(1−y)=12 \sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} = 1007 \\ \sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} = 1007 \\ \dfrac{1}{\left(1-x \right) \left(1-y \right)} = \dfrac{1}{2} n=1∑1007x4n+1x2n+1=1007n=1∑1007y4n+1y2n+1=1007(1−x)(1−y)1=21
So, according to our question it forces that 1007×1007<1/21007 \times 1007 < 1/21007×1007<1/2, which is impossible.
The question should be actually like this:
If xxx, yyyare real numbers such that x2014+y2014=1x^{2014}+y^{2014}=1x2014+y2014=1 then prove that:
(∑n=11007x2n+1x4n+1)(∑n=11007y2n+1y4n+1)<1(1−∣x∣)(1−∣y∣)\left( \sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} \right) \left( \sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} \right) < \dfrac{1}{\left(1-|x| \right) \left( 1-|y| \right)}(n=1∑1007x4n+1x2n+1)(n=1∑1007y4n+1y2n+1)<(1−∣x∣)(1−∣y∣)1
PROOF:
Let ∣x∣=a|x| = a∣x∣=a and ∣y∣=b|y| = b∣y∣=b. So, the given constraint changes into a2014+b2014=1a^{2014} + b^{2014} = 1a2014+b2014=1.
Since, a4n+1>(a2n+1)22=12×(a2n+1)×(a2n+1)>an(a2n+1) a^{4n} + 1 > \dfrac{ \left( a^{2n} + 1 \right)^2}{2} = \dfrac{1}{2} \times \left( a^{2n} + 1 \right) \times \left( a^{2n} + 1 \right) > a^{n} \left( a^{2n} + 1 \right)a4n+1>2(a2n+1)2=21×(a2n+1)×(a2n+1)>an(a2n+1). Which implies that
1+a2n1+a4n<1an∑n=110071+a2n1+a4n<∑n=110071an=(1a1007−1)×(11−a) \dfrac{1+a^{2n}}{1+a^{4n}} < \dfrac{1}{a^n} \\ \sum_{n=1}^{1007} \dfrac{1+a^{2n}}{1+a^{4n}} <\sum_{n=1}^{1007} \dfrac{1}{a^n} = \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{1-a} \right)1+a4n1+a2n<an1n=1∑10071+a4n1+a2n<n=1∑1007an1=(a10071−1)×(1−a1)
Similary ∑n=110071+b2n1+b4n<(1b1007−1)×(11−b)\sum_{n=1}^{1007} \dfrac{1+b^{2n}}{1+b^{4n}} < \left( \dfrac{1}{b^{1007}} - 1 \right) \times \left( \dfrac{1}{1-b} \right)∑n=110071+b4n1+b2n<(b10071−1)×(1−b1)
Multiplying the two, we get
∑n=110071+a2n1+a4n×∑n=110071+b2n1+b4n<(1a1007−1)×(1b1007−1)×1(1−a)(1−b) \sum_{n=1}^{1007} \dfrac{1+a^{2n}}{1+a^{4n}} \times \sum_{n=1}^{1007} \dfrac{1+b^{2n}}{1+b^{4n}} < \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{b^{1007}} - 1 \right) \times \dfrac{1}{\left(1-a \right) \left( 1-b \right)}n=1∑10071+a4n1+a2n×n=1∑10071+b4n1+b2n<(a10071−1)×(b10071−1)×(1−a)(1−b)1
So, If we prove that (1a1007−1)×(1b1007−1)<1 \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{b^{1007}} - 1 \right) < 1(a10071−1)×(b10071−1)<1, then we are done.
Since a2014+b2014=1a^{2014} + b^{2014} = 1 a2014+b2014=1 and 0≤a,b0 \leq a, b 0≤a,b. It implies a,b<1a,b < 1a,b<1. So, a1007>a2014a^{1007} > a^{2014}a1007>a2014 and b1007>b2014b^{1007} > b^{2014}b1007>b2014. Which implies that a1007+b1007>a2014+b2014=1a^{1007} + b^{1007} > a^{2014} + b^{2014} = 1a1007+b1007>a2014+b2014=1.
So,
1−a1007−b1007<0(1−a1007)×(1−b1007)<a1007×b1007(1a1007−1)×(1b1007−1)<11-a^{1007} - b^{1007} <0 \\ \left( 1- a^{1007} \right) \times \left(1- b^{1007} \right) < a^{1007} \times b^{1007} \\ \left( \dfrac{1}{a^{1007}} - 1 \right) \times \left( \dfrac{1}{b^{1007}} - 1 \right) < 1 1−a1007−b1007<0(1−a1007)×(1−b1007)<a1007×b1007(a10071−1)×(b10071−1)<1
Therefore,
(∑n=11007x2n+1x4n+1)(∑n=11007y2n+1y4n+1)=(∑n=11007a2n+1a4n+1)(∑n=11007b2n+1b4n+1)<1(1−a)(1−b)=1(1−∣x∣)(1−∣y∣)\left( \sum_{n=1}^{1007} \dfrac{x^{2n}+1}{x^{4n}+1} \right) \left( \sum_{n=1}^{1007} \dfrac{y^{2n}+1}{y^{4n}+1} \right) = \left( \sum_{n=1}^{1007} \dfrac{a^{2n}+1}{a^{4n}+1} \right) \left( \sum_{n=1}^{1007} \dfrac{b^{2n}+1}{b^{4n}+1} \right) < \dfrac{1}{\left(1-a \right) \left( 1-b \right)} = \dfrac{1}{\left(1-|x| \right) \left( 1-|y| \right)}(n=1∑1007x4n+1x2n+1)(n=1∑1007y4n+1y2n+1)=(n=1∑1007a4n+1a2n+1)(n=1∑1007b4n+1b2n+1)<(1−a)(1−b)1=(1−∣x∣)(1−∣y∣)1
Pheww!! Completed. ■\blacksquare■.
Log in to reply
Thank you very much.
Surya, explain the ∑n=110071an \sum_{n=1}^{1007} \dfrac{1}{a^n} ∑n=11007an1 part.
any hint
hint is provided
Shivam prove the below statement. If you do so, the problem is done.
i want to look its solution..
@Shivam Jadhav @Saarthak Marathe Please post a solution. My curiosity is running out and so is of some people.Thanks.
Agree....
I found another solution to this problem.This is my own solution. I did not know how to add pdf's to brilliant so I uploaded it on my google drive account. Please See the link below.
https://drive.google.com/file/d/0ByQHFlC74eP_dGxzOTBVaFZKTVE/view?usp=sharing
@Shivam Jadhav @Svatejas Shivakumar @Dev Sharma @Surya Prakash @Nihar Mahajan
@Pi Han Goh
If the following is proved then the problem is done, For x≤1x\le1x≤1,
Prove that 1−x2n1+x2n≤x\frac{1-{x}^{2n}}{1+{x}^{2n}} \le x 1+x2n1−x2n≤x where n∈Nn \in N n∈N
@Shivam Jadhav Please post a solution.
No not yet. Please wait for some more time. I think I am getting it.@Shivam Jadhav
I think we have got too much time for this. Remember that we have only half an hour in RMO for a question.
@Nihar Mahajan – But once you get such a question by yourself then you will remember the method for your lifetime.
@Saarthak Marathe – Then , please post your solution. Too curious :P
Sarthak wants to post the solution.
@Shivam Jadhav @Saarthak Marathe Please post a solution. Almost three weeks are over since this problem was posted and no one has got a solution so far.
And BTW, in the problem it should be mentioned that x and y ≠\neq= 1.
It is understood.
@Shivam Jadhav Post the solution.
@Shivam Jadhav Please post the solution!!! I am losing my curiosity!!!!
I think we are unnecessarily wasting our time asking him because he probably doesn't have any solution.
@Nihar Mahajan Yes even I think so. Can we ask Calvin sir to help us in this problem (provided this problem actually exists :D)?
@Nihar Mahajan @Calvin Lin @Daniel Liu @Joel Tan @Harsh Shrivastava @Xuming Liang @Ammar Fathin Sabili @Satyajit Mohanty
@Shivam Jadhav Please refrain from mass tagging. Please try and limit it to five people. Thanks.
Problem Loading...
Note Loading...
Set Loading...
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
Top NewestThere is a typo in the question. I will explain you what the typo is.
For suppose, take x=−1 and y=0. Then we get
n=1∑1007x4n+1x2n+1=1007n=1∑1007y4n+1y2n+1=1007(1−x)(1−y)1=21
So, according to our question it forces that 1007×1007<1/2, which is impossible.
The question should be actually like this:
If x, yare real numbers such that x2014+y2014=1 then prove that:
(n=1∑1007x4n+1x2n+1)(n=1∑1007y4n+1y2n+1)<(1−∣x∣)(1−∣y∣)1
PROOF:
Let ∣x∣=a and ∣y∣=b. So, the given constraint changes into a2014+b2014=1.
Since, a4n+1>2(a2n+1)2=21×(a2n+1)×(a2n+1)>an(a2n+1). Which implies that
1+a4n1+a2n<an1n=1∑10071+a4n1+a2n<n=1∑1007an1=(a10071−1)×(1−a1)
Similary ∑n=110071+b4n1+b2n<(b10071−1)×(1−b1)
Multiplying the two, we get
n=1∑10071+a4n1+a2n×n=1∑10071+b4n1+b2n<(a10071−1)×(b10071−1)×(1−a)(1−b)1
So, If we prove that (a10071−1)×(b10071−1)<1, then we are done.
Since a2014+b2014=1 and 0≤a,b. It implies a,b<1. So, a1007>a2014 and b1007>b2014. Which implies that a1007+b1007>a2014+b2014=1.
So,
1−a1007−b1007<0(1−a1007)×(1−b1007)<a1007×b1007(a10071−1)×(b10071−1)<1
Therefore,
(n=1∑1007x4n+1x2n+1)(n=1∑1007y4n+1y2n+1)=(n=1∑1007a4n+1a2n+1)(n=1∑1007b4n+1b2n+1)<(1−a)(1−b)1=(1−∣x∣)(1−∣y∣)1
Pheww!! Completed. ■.
Log in to reply
Thank you very much.
Log in to reply
Surya, explain the ∑n=11007an1 part.
Log in to reply
any hint
Log in to reply
hint is provided
Log in to reply
Shivam prove the below statement. If you do so, the problem is done.
Log in to reply
i want to look its solution..
Log in to reply
@Shivam Jadhav @Saarthak Marathe Please post a solution. My curiosity is running out and so is of some people.Thanks.
Log in to reply
Agree....
Log in to reply
I found another solution to this problem.This is my own solution. I did not know how to add pdf's to brilliant so I uploaded it on my google drive account. Please See the link below.
https://drive.google.com/file/d/0ByQHFlC74eP_dGxzOTBVaFZKTVE/view?usp=sharing
@Shivam Jadhav @Svatejas Shivakumar @Dev Sharma @Surya Prakash @Nihar Mahajan
Log in to reply
@Pi Han Goh
Log in to reply
If the following is proved then the problem is done, For x≤1,
Prove that 1+x2n1−x2n≤x where n∈N
Log in to reply
@Shivam Jadhav Please post a solution.
Log in to reply
No not yet. Please wait for some more time. I think I am getting it.@Shivam Jadhav
Log in to reply
I think we have got too much time for this. Remember that we have only half an hour in RMO for a question.
Log in to reply
But once you get such a question by yourself then you will remember the method for your lifetime.
Log in to reply
Then , please post your solution. Too curious :P
Log in to reply
Sarthak wants to post the solution.
Log in to reply
@Shivam Jadhav Please post a solution.
Log in to reply
@Shivam Jadhav @Saarthak Marathe Please post a solution. Almost three weeks are over since this problem was posted and no one has got a solution so far.
And BTW, in the problem it should be mentioned that x and y = 1.
Log in to reply
It is understood.
Log in to reply
@Shivam Jadhav Post the solution.
Log in to reply
@Shivam Jadhav Please post the solution!!! I am losing my curiosity!!!!
Log in to reply
I think we are unnecessarily wasting our time asking him because he probably doesn't have any solution.
Log in to reply
@Nihar Mahajan Yes even I think so. Can we ask Calvin sir to help us in this problem (provided this problem actually exists :D)?
Log in to reply
@Nihar Mahajan @Calvin Lin @Daniel Liu @Joel Tan @Harsh Shrivastava @Xuming Liang @Ammar Fathin Sabili @Satyajit Mohanty
Log in to reply
@Shivam Jadhav Please refrain from mass tagging. Please try and limit it to five people. Thanks.
Log in to reply