MHB (2+sqrt(3))^50 close to being an integer

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The discussion centers on proving that (2 + √3)^50 is close to an integer by constructing a sequence of integers a_n defined by a recurrence relation similar to the Fibonacci sequence. The characteristic equation derived from the roots 2 ± √3 leads to the recurrence relation a_{n+1} = 4a_n - a_{n-1}, with initial values a_0 = 2 and a_1 = 4. This sequence shows that a_n is always an integer, and the difference between (2 + √3)^n and an integer is given by (2 - √3)^n. As n increases, (2 - √3)^n approaches zero, confirming that (2 + √3)^50 is indeed very close to an integer. The clarity in forming the recurrence relation was highlighted as a key aspect of the proof.
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Q:Explain this phenomenon by finding a sequence of integers a_i, defined by a recursion relation similar to the Fibonacci sequence, such that a_n = (2 + √
3)^n + (2 −√3)^n.
Make sure to explain why constructing such a sequence proves (2 + √3)^50 is close to an integer, though you needn’t analyze precisely how close it is.

I managed to come up with a proof of why the number is so close to an integer but I have not been able to explain it in the way the question is asking me. Can somebody please help?
 
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To find the recursion having the given closed form, we should look for the minimal polynomial having the roots:

$$2\pm\sqrt{3}$$

So, we could write:

$$x=2\pm\sqrt{3}$$

$$x-2=\pm\sqrt{3}$$

$$(x-2)^2=3$$

$$x^2-4x+4=3$$

$$x^2-4x+1=0$$

So, using this as your characteristic equation, and the initial values of the given closed form, can you construct the recursion?
 
Thanks for your reply! This was actually as far as I got, because I was told to look at the Golden Ratios and how they are the solutions to x^2= x+1, and apply this to 2 +- sqrt3
 
Okay, using the characteristic equation we found, along with the fact that:

$$a_0=2,\,a_1=4$$

we may state:

$$a_{n+1}=4a_{n}-a_{n-1}$$

Does this make sense?
 
Yes, this makes sense! It's analogous to the one for the Fibonacci sequence
 
Yes, the method for finding the closed form of a linear recurrence is to raise each of the roots of the corresponding characteristic equation by a power of $n$ and multiply each by a parameter which you can determine from the given initial conditions.

So, knowing that $a_n$ is an integer for all $n$ (which we can see from the recurrence), then we know that $$\left(2+\sqrt{3}\right)^n$$ will differ from an integer by the value of $$\left(2-\sqrt{3}\right)^n$$. Since $2-\sqrt{3}<1$, we know that as $n$ increases, this value will approach zero.
 
Thank you very much indeed! It was the forming of the recurrence relation that got me; you made it very clear indeed.
 
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