Polynmials

meli3000 · #1 $Old$ April 11th, 2008, 06:18 PM

I have a grade 12 Advanced functions questions.

If f(x+1)=(3f(x)-1)/2 and f(0)=5 find the value of f(n) where n is a positive

integer.

TheEmptySet · #2 $Old$ April 11th, 2008, 06:24 PM

Quote:

Originally Posted by meli3000 $View Post$

I have a grade 12 Advanced functions questions.

If f(x+1)=(3f(x)-1)/2 and f(0)=5 find the value of f(n) where n is a positive

integer.

$f(x+1)=\frac{3f(x)-1}{2}$ eval at x=0 we get

$f(0+1)=\frac{3f(0)-1}{2} \iff f(1)=\frac{14}{2}=7$

now again with x=1

$f(1+1)=\frac{3f(1)-1}{2} \iff f(2)=\frac{21-1}{2}=10$

sorry I misread the problem I thought is wanted f(2)

I will look for a pattern

Soroban · #3 $Old$ April 11th, 2008, 09:10 PM

Hello, meli3000!

Grade 12? . . . Just how advanced are you?

Quote:

If $f(x+1)\:=\:\frac{3f(x)-1}{2}\:\text{ and }\:f(0)\,=\,5$

find the value of $f(n)$ where $n$ is a positive integer.

I'll skip the tedious derivation . . .

. . $f(n) \;=\;4\left(\frac{3}{2}\right)^n + 1$

meli3000 · #4 $Old$ April 12th, 2008, 09:55 AM

Quote:

Originally Posted by Soroban $View Post$

Hello, meli3000!

Grade 12? . . . Just how advanced are you?

I'll skip the tedious derivation . . .

. . $f(n) \;=\;4\left(\frac{3}{2}\right)^n + 1$

Thanks for the help!
How did you derive that equation?

Soroban · #5 $Old$ April 20th, 2008, 10:17 AM

Hello, meli3000!

We have the recurrence relation: . $f(n+1) \:=\:\frac{3}{2}\!\cdot\!f(n) - \frac{1}{2}$

Assume that $f(n)$ is of the form: $X^n$

. . . . Then we have: . $X^{n+1} \:=\:\frac{3}{2}X^n - \frac{1}{2}\quad\;\;{\color{blue}[1]}$

The "next" equation: . $X^{n+2} \:=\:\frac{3}{2}X^{n+1} - \frac{1}{2}\;\;{\color{blue}[2]}$

Subtract [1] from [2]: . $X^{n+2} - X^{n+1} \:=\:\frac{3}{2}X^{n+1} - \frac{3}{2}X^n \quad\Rightarrow\quad X^{n+2} - \frac{5}{2}X^{n+1} + \frac{3}{2}X^n \:=\:0$

Multiply by $\frac{2}{X^n}\!:\;\;2x^2 - 5X + 3 \:=\:0\quad\Rightarrow\quad (2X - 3)(X - 1) \:=\:0$

. . And we have two roots: . $X \;=\;\frac{3}{2},\;1$

Form a linear combination with the two roots: . $f(n) \;=\;A\left(\frac{3}{2}\right)^n + B\!\cdot\!1^n\;\;{\color{red}(Q)}$

We know the first two terms: . $f(0) = 5,\;f(1) = 7$

. . $\begin{array}{ccccccccc}f(0) = 5: & A\left(\frac{3}{2}\right)^0 + B & = & 5 & \Rightarrow & A + B &=& 5 & {\color{blue}[3]}\\ f(1) = 7: & A\left(\frac{3}{2}\right)^1 + B &=& 7 & \Rightarrow & \frac{3}{2}A + B &=& 7 & {\color{blue}[4]}\end{array}$

Subtract [3] from [4]: . $\frac{1}{2}A \:=\:2\quad\Rightarrow\quad \boxed{A \,=\,4}$

Substitute into [3]: . $4 + B \:=\:5 \quad\Rightarrow\quad \boxed{B \,=\,1}$

Then (Q) becomes: . $\boxed{f(n) \;=\;4\left(\frac{3}{2}\right)^n + 1}\quad\hdots\quad \text{ta-}DAA!$

Mathstud28 · #6 $Old$ April 20th, 2008, 10:27 AM

Quote:

Originally Posted by Soroban $View Post$

Hello, meli3000!

Grade 12? . . . Just how advanced are you?

I'll skip the tedious derivation . . .

. . $f(n) \;=\;4\left(\frac{3}{2}\right)^n + 1$

just out of curiosity when you said "avanced" did you mean that sardonically or just saying that this is difficult?

Soroban · #7 $Old$ April 20th, 2008, 12:04 PM

Hello, Mathstud28!

Quote:

Just out of curiosity when you said "avanced",
did you mean that sardonically or just saying that this is difficult?

I mean that it is difficult.

If I were trying to be funny, it would have been obvious.
. . (Maybe not very funny, but certainly obvious.)

I didn't learn about Recurrences until years after I began teaching.
I found some publications from The Fibonacci Society which taught
me about Recurrences and Iterative Relations.