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Define the Fibonacci sequence via $f_1 = f_2 = 1$, and if $n > 2$, then $f_n = f_{n−1} + f_{n−2}$ and show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$

I have been solving this problem using strong induction. This is my attemp.

For $n=1$.

$f_1=1$ and,

$\displaystyle \frac{\left ( \frac{1+\sqrt5}{2} \right )-\left ( \frac{1-\sqrt5}{2} \right )}{\sqrt5}=1$

For $n=2$

$f_1=2$ and,

$\displaystyle \frac{\left ( \frac{1+\sqrt5}{2} \right )^2-\left ( \frac{1-\sqrt5}{2} \right )^2}{\sqrt5}=1$ as well.

the statement is also true for $n = 3$

Now, assume that the statement is true for some $n-1$ and $n-2$. We get

$\displaystyle f_n =f_{n-1}+f_{n-2}=\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-1}-\left ( \frac{1-\sqrt5}{2} \right )^{n-1}}{\sqrt5}+\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-2}-\left ( \frac{1-\sqrt5}{2} \right )^{n-2}}{\sqrt5}$

From here, my intention is to show that the statement for $f_n$ is true. But I couldn't move on.

User8976
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Hopmaths
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4 Answers4

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Don't work so hard. Define some notation to simplify the algebra.

Let $$\varphi = \frac{1 + \sqrt{5}}{2}, \quad \bar \varphi = \frac{1 - \sqrt{5}}{2}.$$ Then we can easily see that $$\varphi + \bar \varphi = 1, \quad \varphi \bar \varphi = \frac{1^2 - (\sqrt{5})^2}{2^2} = -1, \quad \varphi - \bar \varphi = \sqrt{5}.$$ Consequently, $$\varphi = \varphi(\varphi + \bar \varphi) = \varphi^2 - 1, \quad \bar \varphi = \bar \varphi (\varphi + \bar \varphi) = -1 + \bar \varphi^2,$$ or $$\varphi^2 = 1 + \varphi, \quad \bar \varphi^2 = 1 + \bar \varphi.$$ Now we have $$f_n = \frac{\varphi^n - \bar \varphi^n}{\varphi - \bar \varphi},$$ and $$\begin{align} f_{n-1} + f_{n-2} &= \frac{\varphi^{n-1} - \bar \varphi^{n-1} + \varphi^{n-2} - \bar \varphi^{n-2}}{\varphi - \bar \varphi} \\ &= \frac{\varphi^{n-2}(1 + \varphi) - \bar \varphi^{n-2}(1 + \bar \varphi)}{\varphi - \bar \varphi} \\ &= \frac{\varphi^{n-2} \varphi^2 - \bar \varphi^{n-2} \bar \varphi^2}{\varphi - \bar \varphi} \\ &= \frac{\varphi^n - \bar \varphi^n}{\varphi - \bar \varphi} \\ &= f_n. \end{align}$$

This concludes the proof.

heropup
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2

For homogeneous second order linear difference equations there is a shortcut. I wonder why it is not well known and almost everybody is using the "standard way" of solving a linear equation system. BTW it just so happened somebody else applied it earlier today in this post: https://math.stackexchange.com/a/3898819/843178

This technique can be used in higher order equations (along with a treatment of duplicate roots) but it gets messy. So for higher order LDEs it's best to use the "standard solution".

The characteristic equation $^2−−1=0$ has two roots $\varphi=\frac{1+\sqrt 5}{2}, \psi=\frac{1-\sqrt 5}{2}$. Vieta’s formulas give $\varphi+\psi=1, \varphi \psi = -1$.

Therefore $$f_{+2}−(\varphi+\psi)f_{n+1}+\varphi \psi f_=0.$$

Rearranging the terms, we get $$f_{n+2}−\psi f_{+1}=\varphi(f_{+1}−\psi f_) $$

$$f_{n+2}−\varphi f_{+1}=\psi (f_{+1}−\varphi f_) $$

Both are geometric sequences, so $$f_{+1}−\psi f_=\varphi^ (f_1−\psi f_0 )=\varphi^ \tag 1$$ $$f_{+1}−\varphi f_=\psi^ (f_1− \varphi f_0 )=\psi^ \tag 2$$

(1)-(2), $$(−) f_=\varphi^−\psi^⇒ f_=(\varphi^−\psi^)/(\varphi−\psi).\blacksquare$$

Neat Math
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$\displaystyle f_n =f_{n-1}+f_{n-2}=\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-1}-\left ( \frac{1-\sqrt5}{2} \right )^{n-1}}{\sqrt5}+\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-2}-\left ( \frac{1-\sqrt5}{2} \right )^{n-2}}{\sqrt5}$

Hint: Consider only the numerator part:

$$\left ( \frac{1+\sqrt5}{2} \right )^{n-1}-\left ( \frac{1-\sqrt5}{2} \right )^{n-1} + \left ( \frac{1+\sqrt5}{2} \right )^{n-2}-\left ( \frac{1-\sqrt5}{2} \right )^{n-2}$$

$$= \left ( \frac{1+\sqrt5}{2} \right )^{n-2} \left( \frac{1+\sqrt5}{2} +1\right) - \left ( \frac{1-\sqrt5}{2} \right )^{n-2} \left( \frac{1-\sqrt5}{2} +1\right)$$

$$= \left ( \frac{1+\sqrt5}{2} \right )^{n-2} \left( \frac{3+\sqrt5}{2} \right) - \left ( \frac{1-\sqrt5}{2} \right )^{n-2} \left( \frac{3-\sqrt5}{2} \right)$$

User8976
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$f_n=f_{n-1}+f_{n-2}$ is a difference equation whiose solution is found by putting $f_n=t^n$, then $t_{1,2}=\frac{1\pm \sqrt{5}}{2}$. The solution is $$f_n=A t_1^n+ B t_2^n$$. By putting $f_1=f_2=1$, we get $$A=-B=\frac{1}{\sqrt{5}}.$$

Z Ahmed
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