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Rosie · « **on:** May 10, 2008, 08:56:53 am »

1. The following are general first-order difference equations. Solve:
t_n+1 = 1.5t_n-8, t₁=32

I'm a little confused about the meaning of a first-order difference equation. Are they just equations that have a t_n and t_n+1, that is, they differ by one step.

Collin Li · « **Reply #1 on:** May 11, 2008, 01:19:35 am »

A first-order difference equation is a difference equation that has the smallest step differing with the highest step by 1.

In this case, the highest step term is $n+1$ , while the smallest step term is $n$ , hence the difference equation is first order.

So yes, your definition is correct. I know nothing of the Further Mathematics course, so I'm not sure whether you're required to know what a "difference equation" is in general. If you don't know, ask!

Rosie · « **Reply #2 on:** May 11, 2008, 08:32:36 am »

Can anyone solve the question I have above, this is a number pattern question

Mao · « **Reply #3 on:** May 11, 2008, 08:54:14 am »

what am I solving for....?

am i writing the first 10 numbers of the sequence down, or what... =\

(its not even an equation!)

bigtick · « **Reply #4 on:** May 11, 2008, 09:23:21 am »

first order means t(n) to the power of 1.
solve means find t(n) given the difference equation.
solution is t(n)=32x1.5^(n-1)+(-8)(1.5^(n-1) -1)/(1.5 -1).

Rosie · « **Reply #5 on:** May 11, 2008, 10:45:08 am »

Yes, you have that question on the top and you have to use the rule/equation for solving an equation that is both an arithmetic and geometric sequence.
It is t_n=ar^n-1+d $\frac {(r^{n-1}-1)} {r-1}$

You have to solve using this equation by putting in all teh required values that are given in the equation but I can't seem to get the correct answer.

Mao · « **Reply #6 on:** May 11, 2008, 10:59:06 am »

a=32, r=1.5, d=-8, n=variable

Rosie · « **Reply #7 on:** May 11, 2008, 11:18:40 am »

Ok, I'll show you the steps I did to solve this

t_n=32*1.5^n-1-8 $\frac{1.5^{n-1}-1}{1.5-1}$
t_n=32*1.5^n-1-8 $\frac{1.5^{n-1}-1}{0.5}$
I'm not sure what to do next. When simplifying teh fraction, should I divide -1 by 0.5 or do I bring 0.5 to the -8 to simplify the equation.

Collin Li · « **Reply #8 on:** May 11, 2008, 11:51:15 am »

Quote from: Rosie on May 11, 2008, 11:18:40 am

I'm not sure what to do next. When simplifying teh fraction, should I divide -1 by 0.5 or do I bring 0.5 to the -8 to simplify the equation.

$8\times\frac{1.5^{n-1}-1}{0.5} = \frac{8}{0.5}\left(1.5^{n-1}-1\right) = 16\left(1.5^{n-1}-1\right)$

Now you can just expand it and collect like terms.

Rosie · « **Reply #9 on:** May 11, 2008, 12:37:24 pm »

Hey, I've got it now. What I was doing wrong, was multiplying those numbers together instead of dividing.
However, I'm not sure why you divide it.

Mao · « **Reply #10 on:** May 11, 2008, 01:44:15 pm »

because:

$a \cdot \frac{b}{c} = \frac{a}{1} \cdot \frac{b}{c} = \frac{a\cdot b}{c} = \frac{a}{c} \cdot \frac{b}{1} = (a \div c) \times b$

dcc · « **Reply #11 on:** May 11, 2008, 01:53:51 pm »

"Do not use a cannon to kill a mosquito." - Confucius

$a_{n + 1} = 1.5a_{n} - 8, a_{1} = 32$

Substituting $n \mapsto n + 1$ , we arrive at another form for this equation:

$a_{n + 2} = 2.5a_{n + 1} - 1.5a_{n}$

$\mbox{Let }A(x) = \sum\limits_{n=0}^{\infty} a_{n} x^n$

Now multiply our new equation by $x^n$ :

$a_{n + 2} \cdot x^n = 2.5a_{n +1} \cdot x^n - 1.5a_{n} \cdot x^n$

Now, this really defines an infinite set of equations, one for each possible value of n, so we can write this with a summation of all the different variations of n:

$\implies \sum\limits_{n=1}^{\infty} a_{n + 2} \cdot x^n = 2.5 \sum\limits_{n=1}^{\infty} a_{n + 1} \cdot x^n - 1.5 \sum\limits_{n=1}^{\infty} a_{n} \cdot x^n$

$\implies \sum\limits_{n=3}^{\infty} a_{n} \cdot x^{n - 2} = 2.5 \sum\limits_{n=2}^{\infty} a_{n} \cdot x^{n-1} - 1.5 \sum\limits_{n=1}^{\infty} a_{n} \cdot x^n$

First Consider:
$1.5 \sum\limits_{n=1}^{\infty} a_{n} \cdot x^n = 1.5 \sum\limits_{n=0}^{\infty} a_{n} \cdot x^n - 1.5 a_{0} \cdot x^{0} = 1.5 \sum\limits_{n=0}^{\infty} a_{n} \cdot x^n - 40 = 1.5A(x) - 40$

Next:
$2.5 \sum\limits_{n=2}^{\infty} a_{n} \cdot x^{n-1} = 2.5x^{-1} \sum\limits_{n=2}^{\infty} a_{n} \cdot x^{n} = 2.5x^{-1} (\sum\limits_{n=0}^{\infty} a_{n} \cdot x^{n} - a_{0}x^{0} - a_{1}x) = 2.5x^{-1} (A(x) - \dfrac{40}{1.5} - 32x)$

Then:

$\sum\limits_{n=3}^{\infty} a_{n} \cdot x^{n - 2} = x^{-2} \sum\limits_{n=3}^{\infty} a_{n} \cdot x^{n} = x^{-2} (\sum\limits_{n=0}^{\infty} a_{n} \cdot x^{n} - a_{0}x^0 - a_{1}x - a_{2}x^2) = x^{-2} (A(x) - \dfrac{40}{1.5} - 32x - 40x^2)$

Combining these:

$\implies x^{-2} (A(x) - \dfrac{40}{1.5} - 32x - 40x^2) = 2.5x^{-1} (A(x) - \dfrac{40}{1.5} - 32x) - (1.5A(x) - 40)$

$\implies A(x) = \dfrac{\dfrac{40}{1.5} - \dfrac{104x}{3}}{1.5x^2 + 1 - 2.5x}$

Now expanding using the method of partial fractions:

$\implies A(x) = \dfrac{80}{1 - 1.5x} - \dfrac{\frac{160}{3}}{1-x} + \dfrac{\frac{208}{3}}{1 - x} - \dfrac{\frac{208}{3}}{1 - 1.5x} = \dfrac{\frac{32}{3}}{1 - 1.5x} + \dfrac{16}{1 - x}$

Now these are variations on the inifinite geometric sequence (which have a nice little series we can work with):

$\implies A(x) = \dfrac{32}{3} \cdot \dfrac{1}{1 - 1.5x} + 16 \cdot \dfrac{1}{1 - x} = \frac{32}{3} \sum\limit_{n=0}^{\infty} (1.5x)^n + 16 \sum\limit_{n=0}^{\infty} (x)^n = \dfrac{32}{3}(\sum\limit_{n=0}^{\infty} (1.5^n + 1.5)x^n) = \dfrac{\sum\limit_{n=0}^{\infty} (1.5^n + 1.5)x^n)}{\frac{3}{32}}$

BUT REMEMBER:

$A(x) = \sum\limits_{n=0}^{\infty} a_{n} x^n = \dfrac{\sum\limit_{n=0}^{\infty} (1.5^n + 1.5)x^n)}{\frac{3}{32}}$

Comparing Coefficients, it becomes apparent:

$a_{n} = \dfrac{32}{3} (1.5^n + 1.5)$

Mao · « **Reply #12 on:** May 11, 2008, 01:58:38 pm »

lol, rob you have WAY~ too much time on your hands

(and btw, further people, ignore this post. it does not affect you AT ALL)

clinton_09 · « **Reply #13 on:** May 11, 2008, 01:59:26 pm »

wow who would have thought this is further maths

Mao · « **Reply #14 on:** May 11, 2008, 02:01:46 pm »

Quote from: clinton_09 on May 11, 2008, 01:59:26 pm

wow who would have thought this is further maths

dcc, Ahmad II - self_control.

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