- #1
chisigma
Gold Member
MHB
- 1,627
- 0
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
$\chi$ $\sigma$
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
$\chi$ $\sigma$