AuEN Ratio

Algebra Level 4

Let $x \in R^{+}$ such that $\{x\} , [x] , x$ are in a geometric progression. What is the common ratio of the G.P. ?

Notations: $[x]$ denotes the greatest integer less than $x$ while $\{x\}$ is defined to be $x-[x]$ . You may refer to the Wikipedia link for more information about these notations.

Try this sweet problem - Simple yet enjoyable

The answer is 1.618.

3 solutions

Krishna Sharma
Dec 3, 2014

As terms are in G.P. we can write

$[x]^{2}$ = $x\{x\}$

$[x]^{2} = x^{2} - x[x]$

$\displaystyle x^{2} - x[x] - [x]^{2} = 0$

Divide the equation by $[x]^{2}$

Then put $\frac{x}{[x]} = t$

$t^{2} - t - 1 = 0$

From here we get

$t = \frac{ 1 + \sqrt{5}}{2}$

Negative root rejected as $x \in \mathbb R^{+}$

$\displaystyle r = \frac{x}{[x]} = \boxed{\frac{1 + \sqrt{5}}{2}}$

Note that only solution to this equation is

$x = \frac{1 + \sqrt{5}}{2}$

Because as you put $x = n\phi$ , n > 1 , the floor function would yield a value greater than 'n' hence we can say there is only 1 non - zero solution to the equation.

Note: We do not have a true "quadratic" equation, because $[x]$ is not a constant. As such, we cannot blindly apply the quadratic formula. Can you explain how to solve that equation properly?

Also, if you are applying the quadratic formula, you should explain why the negative root is ignored.

Calvin Lin Staff - 6 years, 6 months ago

$x \in R^{+}$ so negative root rejected.

Krishna Sharma - 6 years, 6 months ago

The question is, how do you know that there are no further solutions? Even though it "looks like" a quadratic equation, that does not mean that it only has 2 solutions. For example, the equation $x^2 - \lfloor x \rfloor x = 1$ has infinitely many solutions.

Calvin Lin Staff - 6 years, 6 months ago

@Calvin Lin – Note that ' $x$ ' may vary but $\frac{x}{[x]}$ can be constant. In your quadratic we cannot apply quadratic formula because it would become more complicated. (no common factor comes out). Quadratic formula helps sometimes(like in original problem) if not then go with I-F method.

Krishna Sharma - 6 years, 6 months ago

@Krishna Sharma – Right. So the point is "how do you know that there are no further solutions, other than the class that you listed?"

Furthermore, Is every value of $\frac{ n + n \sqrt{5} } { 2}$ a solution? E.g. with $n = 2$ , we get $x = 3 . 236 \ldots$ , and $0.236, 3, 3.236$ is not a geometric progression, so it doesn't satisfy the original condition in the question. How was this extraneous solution introduced?

Calvin Lin Staff - 6 years, 6 months ago

@Calvin Lin – Clearing everything

Q1:- How can you apply quadratic formula when [x] is variable?

A:- yes, quadratic formula can be applied if you are not satisfied divide the quadratic by $[x]^{2}$ then put $\frac{x}{[x]} = t$ .

Q2 :- How can you say there is only one solution of the problem? (x= 0 not included)

A:- let $\displaystyle x = \frac{k(1 + \sqrt{5})}{2}$ where $k \in Z$

We will solve $\displaystyle \frac{x}{[x]} = \phi$

Put k= 1 we will get $x = \phi$ (only solution)

Put k= 2 we will get $\frac{x}{[x]} = \frac{2\phi}{3}$ which would not satisfy,

As you go on substituting 'k' the difference of numerator and denominator of coefficient of $\phi$ will go on increasing hence you can say there is only 1 solution.

I hope this clarification is enough.

Krishna Sharma - 6 years, 6 months ago

@Krishna Sharma – Great! Using the approach of $t = \frac{ x} { \lfloor x \rfloor }$ was what I was thinking of. We can then solve it as a quadratic equation, and are only interested in $t > 0$ . This give us $t = \phi$ as the solution.

We then have to verify that there is an $x$ value which yields this corresponding $t$ value. In this case, setting $x = \phi$ , we get $\lfloor x \rfloor = 1$ and thus $t = \phi$ .

Note that interestingly, there are no other solutions. Because, if $\lfloor x \rfloor x = n \geq 2$ , then we must have $x = n \phi > n + 1$ . But this would contradict $\lfloor x \rfloor = n$ .

Can you add these to your solution for completeness?

Calvin Lin Staff - 6 years, 6 months ago

Sir if u check [x] is. Natural no.. Hence its logically correct we know some part of its v gud sold..or u can check mine

Rohit Singh - 6 years, 6 months ago

The fact that $\lfloor x \rfloor$ is a natural number doesn't matter.

We are not guaranteed that there are only solutions which are "derived" from the quadratic equation. Generally, such problems must be solved in a slightly different way, to ensure that we have found all solutions. How do you know that no other solutions exist?

For example, he should also have found the solution $x = 0$ , and then rejected it because it is not out of range.

Calvin Lin Staff - 6 years, 6 months ago

@Calvin Lin – Sir, then please explain the actual way of solving these type of pseudo quadratic..

Akhil Bansal - 5 years, 8 months ago

Can you explain me the very first step?

Snehal Shekatkar - 6 years, 6 months ago

Well they are in G.P so the product of the first and the third is equal to the square of the second. In the second step we write fractionalpart(x)=x-floor(x)

Keshav Tiwari - 6 years, 6 months ago

Soutrik Bandyopadhyay
Dec 14, 2014

Golden Ratio Rocks!!!

Rohit Singh
Dec 3, 2014

Let {x} = a & [x]=z ==>x = z+a From gp we can say z/a =z+a/z=r= common ratio

So( 1/r= a/z)

Let's take r = z+a/z. ( squaring both side ) r^2=(z+a)^2/r^2 and solve and incest r every where inplace of z and a then u will get a equation of degree four r^4 - r^2 - 2r. - 1=0 solve equation by farari method or interpretation of newtonraphanson method you will get only one R+ no that is 1.61803 like golden ratio if u notice like follow me if u like

Note that the quartic could be factorized as $(r^2 - r - 1 ) ( r^2 + r + 1) = 0$ , which is why it gives us 1 positive root, 1 negative root and 2 complex roots.

Calvin Lin Staff - 6 years, 6 months ago

You are correct but can we .can't take those value which r out of our interval and complex is out of question initial values restrictions

Rohit Singh - 6 years, 6 months ago

Can u tell me m I correct while solving it in this way

Rohit Singh - 6 years, 6 months ago

AuEN Ratio

The answer is 1.618.

3 solutions

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