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Two interesting problems...

chisigma

Well-known member
Feb 13, 2012
1,704
An interesting pair of problems has been proposed in…

http://mathhelpforum.com/pre-calculus/203756-geometric-sequences.html

Now we try to solve #74 that looks like a ‘yellow roman’…


Consider the sequence…

4 , 20.4 , 104.04 , 531.6444 , …

What is the error in using $a_{277}= 4 \cdot 5.1^{276}$ to findthe 277 th term?...


The hypothesis that the sequence is the solutionof the difference equation...


$\displaystyle a_{n+1}= 5.1\ a_{n}\ ,\ a_{0}=4$(1)

... is true for n=1 and n=2 but not for n=3because is $4 \cdot 5,1^{3} = 530.604 \ne 531.6444$. That suggests that thesequence is solution of the difference equation...

$\displaystyle a_{n=1}= 5.1\ a_{n}\ + .2601\ a_{n-3}\ , a_{0}=4\ ,\ a_{1}=20.4\ ,\ a_{2}=104.04$ (2)


The characteristic equation corresponding to(2) is…


$\displaystyle x^{3} – 5.1\ x^{2} - .2601=0$(3)


… and its solution are $x \sim 5.10996$ and $x\sim -.00498053 \pm i\ .225557$, so that the solution of (2) is…


$\displaystyle a_{n}= c_{0}\ 5.10996^{n} +(-.00498053)^{n}\ (c_{1}\ \cos .225557\ n + c{2}\ \sin .225557\ n)$ (4)


The constants $c_{0}$ , $c_{1}$ and $c_{2}$ canbe found from the initial conditions and that is left to the reader. Apreliminary valuation of the error however can be made observing the ratio…


$\displaystyle r =(\frac{5.10996}{5.1})^{277}\sim 1.71$

Kind regards

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
Very interesting is the second problem too...

Consider the sequence...

$x+3\ ,\ x+7\ ,\ 4x-2\ ,...$

a) if the sequence is arithmetic find x and then determine each of the terms and the 4-th term...

b) if the sequence is geometric find x and then determine each of the terms and the 4-th term...


a) if the sequence is arithmetic is $\displaystyle a_{n+1}=a_{n}+y$ and we can find x and y solving the pair of equations...

$\displaystyle x+7-x-3=y \\ 4x-2-x-7=y$

... that means $x=\frac{13}{3}\ ,\ y=4$...

b) if the sequence is geometric is $\displaystyle a_{n+1}=a_{n}\ y$ and we can find x and y solving the pair of equations...

$\displaystyle \frac{x+7}{x+3}=y \\ \displaystyle \frac{4x-2}{x+7}=y$

In this case we have two solutions and one of them is $x=5\ ,\ y= \frac{3}{2}$...

Kind regards

$\chi$ $\sigma$