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I have to show that $\left(\frac{1+\sqrt{5}}{2}\right)^5 > 10$.

I've already proven that if $f_n$ is the nth Fibonnaci number, then,

$$f_{n+1} > \left(\frac{1+\sqrt{5}}{2}\right)^{n-1} $$

But I really don't get how to go from this statement to what I need to prove.

Any advices on how to take from here? Any help is welcome.

Thanks.

nonuser
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zairhenrique
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  • do you know how to expand that $(\dots)^5$ expression - brute force, but it works – Nick Pavlov Oct 01 '17 at 18:03
  • You want to show that your expression is larger than $10$, which if you don't want to do it directly means that you have to find something which is smaller than your expression, but at the same time larger than $10$. A Fibonacci number which is larger than your expression doesn't help. – Arthur Oct 01 '17 at 18:03
  • @Arthur oddly, this previous statement is the previous step on the exercise. – zairhenrique Oct 01 '17 at 18:07
  • I would recommend that if you use the brute force method that you multiply both sides by a particular number to get rid of any fractions. ... Fractions are really annoying. – user357980 Oct 01 '17 at 18:07
  • $${F_n} = \frac{{{\varphi ^n} - {{( - \varphi )}^{ - n}}}}{{\sqrt 5 }} = \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^n} - {{( - \varphi )}^{ - n}}}}{{\sqrt 5 }}\\begin{array}{l} \varphi = \frac{{1 + \sqrt 5 }}{2} \approx 1.61803{\mkern 1mu} 39887 \cdots \
    • \frac{1}{\varphi } \approx - 0.61803{\mkern 1mu} 39887 \cdots

    \end{array}$$ when $n$ go larger $${F_n} = \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^n} - {{( - \varphi )}^{ - n}}}}{{\sqrt 5 }}\sim \frac{{{{(\frac{{1 + \sqrt 5 }}{2})}^n} }}{{\sqrt 5 }}$$

     – Khosrotash Oct 01 '17 at 18:07
  • $5\log_{10}\left(\dfrac{1+\sqrt{5}}{2}\right)\approx 1.044 >1$

    Thus $\left(\dfrac{1+\sqrt{5}}{2}\right)^5>10$

     – Raffaele Oct 01 '17 at 18:50

4 Answers4

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One option is to expand the power. This is made easier by the fact that $\phi^2 = \phi+1$. $$ \phi^5 = (\phi+1)(\phi+1)\phi = (3\phi + 2)\phi = 5\phi+3 $$ and since it's easily shown that $\phi > 3/2$, the inequality follows.

If you want to relate to the Fibonacci numbers directly, you can also use $\phi^n = F_n\phi + F_{n-1}$ to get $\phi^5 = 5\phi+3$.

eyeballfrog
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Say $x=\sqrt{5}$. Then we have to prove:

$$1+5x+10x^2+10x^3+5x^4+x^5>64x^2 $$

or

$$1+5x+50x+25x> 29x^2$$

or $ 80x> 144$ or $5x>9$ or if we square $125>81$ which is true.

nonuser
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Binomially expand \begin{eqnarray*} \left( \frac{1+ \sqrt{5}}{2} \right)^5 &=& \frac{1+5 \sqrt{5} +10 \times 5 +10 \times 5 \sqrt{5}+5 \times 25 +25 \sqrt{5}}{32} \\ &=&\frac{11+5 \sqrt{5}}{2}. \end{eqnarray*} Now $125>121 $ square root this and we have the stronger result that \begin{eqnarray*} \left( \frac{1+ \sqrt{5}}{2} \right)^5 > 11. \\ \end{eqnarray*}

Donald Splutterwit
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0

The light brute force way is sure fire...Just using the elemetary algebra math level: $LHS > RHS \iff \left(1+\sqrt{5}\right)^5 > 2^5\cdot 10\iff(1+\sqrt{5})\left((1+\sqrt{5})^2\right)^2> 2^5\cdot 10\iff(1+\sqrt{5})(6+2\sqrt{5})^2> 2^5\cdot 10\iff (1+\sqrt{5})(56+24\sqrt{5}) > 320\iff 176+80\sqrt{5}> 320\iff \sqrt{5} > \dfrac{144}{80} = 1.8\iff 5 > 1.8^2 = 3.24$ which is clear. If the exponent gets sufficiently big, then the other methods will be much better for sure.

DeepSea
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