Too complicated

Algebra Level 5

Consider the recurrence relation f n + 1 ( x ) = x + 2 f n ( x ) f_{n+1}(x) = \sqrt{x+ 2 f_{n}(x)} for n = 1 , 2 , 3 , n=1,2,3,\ldots and f 1 ( x ) = 3 x f_1 (x) = \sqrt{3x} .

Find the sum of all possible roots of x = f n ( x ) x = f_n (x) for all positive integer n n .


The answer is 3.

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1 solution

Anubhav Tyagi
Dec 26, 2016

x = f n ( x ) = x + 2 f n 1 ( x ) = x + 2 x + 2 f n 2 ( x ) Hence we write as: x = x + 2 x + 2 x + 2 + 2 3 x ( 1 ) ( n times ) x = x + 2 x + 2 x + 2 + 2 x + 2 x ( n times ) Replace underlined x from equation (1) x = x + 2 x + 2 x + 2 + 2 x + 2 x ( 2 n times ) Replace underlined x from equation (1) x = x + 2 x + 2 x + 2 + 2 x + 2 x ( 3 n times ) Replace underlined x from equation (1) x = x + 2 x + 2 x + 2 + 2 x + ( lim k k times ) x = x + 2 x x = 0 , 3 \begin{aligned} &&x= f_n (x) = \sqrt{x+ 2 f_{n-1}(x)} =\sqrt{x+ 2\sqrt{x+2 f_{n-2}(x)}} \\ &&\text{Hence we write as:} \\ &&x=\sqrt{ x + 2\sqrt{x + 2\sqrt{x + 2\sqrt{\cdots +2\sqrt{3x}}}}}&&(1) \\ &&(n \text{ times}) \\\\ &&x=\sqrt{ x + 2\sqrt{x + 2\sqrt{x + 2\sqrt{\cdots +2\sqrt{x+2\underline{x}}}}}} \\ &&(n \text{ times}) \\\\ &&\text{Replace underlined x from equation (1)}\\ &&x=\sqrt{ x + 2\sqrt{x + 2\sqrt{x + 2\sqrt{\cdots +2\sqrt{x+2\underline{x}}}}}} \\ &&(2n \text{ times}) \\\\ &&\text{Replace underlined x from equation (1)}\\ &&x=\sqrt{ x + 2\sqrt{x + 2\sqrt{x + 2\sqrt{\cdots +2\sqrt{x+2\underline{x}}}}}} \\ &&(3n \text{ times}) \\\\ &&\text{Replace underlined x from equation (1)}\\ &&x=\sqrt{ x + 2\sqrt{x + 2\sqrt{x + 2\sqrt{\cdots +2\sqrt{x+\cdots}}}}} \\ &&( \lim_{k\to\infty}k \text{ times}) \\\\ &\Rightarrow x = \sqrt{x+2x} \\ &\Rightarrow x = 0,3 \\ \end{aligned}

Write easyly 1 2 ( x 2 x ) = x \frac{1}{2}(x^{2}-x)=x following x=0 and x=3.

Andreas Wendler - 4 years, 5 months ago

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i did the same!

Prakhar Bindal - 4 years, 5 months ago

@Prakhar Bindal - Try this one out

Anubhav Tyagi - 4 years, 5 months ago

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Nice problem!!

Prakhar Bindal - 4 years, 5 months ago

@Pi Han Goh - You changed my problem language now what about my solution. It does not go well with the new language

Anubhav Tyagi - 4 years, 5 months ago

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It's wrong from the start. You have only shown that the answer is 3 when n n\to\infty .

Pi Han Goh - 4 years, 5 months ago

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It was correct according to previous language

Anubhav Tyagi - 4 years, 5 months ago

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@Anubhav Tyagi No, it's not. Tell me how you've shown (in your solution) that the answer is 3 regardless of the value of n n .

Pi Han Goh - 4 years, 5 months ago

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@Pi Han Goh You may check by putting for n= any integer

Anubhav Tyagi - 4 years, 5 months ago

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@Anubhav Tyagi Are you saying you've done proof by exhaustion for all positive integers n n ?

Pi Han Goh - 4 years, 5 months ago

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@Pi Han Goh Yes any doubts

Anubhav Tyagi - 4 years, 5 months ago

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@Anubhav Tyagi First of all, you have not shown that in your solution. And second of all, you can't perform "proof by exhaustion" for an infinitely many number of cases because it's an impossible task.

Pi Han Goh - 4 years, 5 months ago

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@Pi Han Goh It has got 3 upvotes which means that people might have understood it.

Anubhav Tyagi - 4 years, 5 months ago

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@Anubhav Tyagi I could upvote it too, does it mean that your solution is correct?

Pi Han Goh - 4 years, 5 months ago

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@Pi Han Goh You won't cause you will never learn to appreciate

Anubhav Tyagi - 4 years, 5 months ago

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@Anubhav Tyagi This is a very fruitful discussion.

Pi Han Goh - 4 years, 5 months ago

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@Pi Han Goh What do you mean?

Anubhav Tyagi - 4 years, 5 months ago

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@Anubhav Tyagi He is being sarcastic.

Please be respectful of moderators, they have to take decisions in behalf of the community, which is not an easy task.

Agnishom Chattopadhyay - 4 years, 5 months ago

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@Agnishom Chattopadhyay Are you saying that I am being sarcastic?

Anubhav Tyagi - 4 years, 5 months ago

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@Anubhav Tyagi

This is a very fruitful discussion

That is a sarcastic comment.

Agnishom Chattopadhyay - 4 years, 5 months ago

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@Agnishom Chattopadhyay Yes I do agree .

Anubhav Tyagi - 4 years, 5 months ago

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