# set solutions

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• May 3rd 2011, 04:40 PM
hazeleyes
set solutions
the question is
''Find the solution to the equations.
(i) $25T_{n+2}=T_{n}\\T_{0}=125\\T_{1}=25$

(ii) $25T_{n+2}=-T_{n}\\T_{0}=125\\T_{1}=25$

for i) i have got
If $T_{n}=x^n$, so $25x^{n+2}-x^n=0\\x^n(25x^2-1)=0\\x=0.2$
general solution: $T_{n}=A(1/5)^n\\T_{0}=A=125\\A=125$
therefore actual solution is $T_{n}=125*(1/5)^n$

for ii) i have got
If
$T_{n}=x^n$,
so
$25x^{n+2}+x^n=0\\x^n(25x^2+1)=0\\x=\pm 0.2i$

general solution:
$T_{n}=A\cos(0.2n)+B\sin(0.2n)$

where A,B are constants.

after using
$T_{0}=125$,
I've got A=125

I'm wondering is my general solution incorrect and how can I find the constant B?
• May 4th 2011, 09:06 AM
FernandoRevilla
The general solutions for (i) and (ii) are

$(i)\quad T_n=A\left(\dfrac{1}{5}\right)^n+B\left(-\dfrac{1}{5}\right)^n$

$(ii)\quad T_n=\left(\dfrac{1}{5}\right)^n\left(A\cos \dfrac{n\pi}{2}+B\sin \dfrac{n\pi}{2}\right)$

In both cases substitute

$T_0=125,\;T_1=25$

and you'll find A and B by means of a 2 x 2 system.
• May 4th 2011, 01:54 PM
hazeleyes
Quote:

Originally Posted by FernandoRevilla
The general solutions for (i) and (ii) are

$(i)\quad T_n=A\left(\dfrac{1}{5}\right)^n+B\left(-\dfrac{1}{5}\right)^n$

$(ii)\quad T_n=\left(\dfrac{1}{5}\right)^n\left(A\cos \dfrac{m\pi}{2}+B\sin \dfrac{m\pi}{2}\right)$

In both cases substitute

$T_0=125,\;T_1=25$

and you'll find A and B by means of a 2 x 2 system.

Hi thanks for your replied, but what is m stand for?
also,
$(i)\quad T_n=A\left(\dfrac{1}{5}\right)^n+B\left(-\dfrac{1}{5}\right)^n$
why does the second term has a negative sign? thanks
• May 4th 2011, 04:54 PM
FernandoRevilla
Quote:

Originally Posted by hazeleyes
Hi thanks for your replied, but what is m stand for?

A typo. I've just corrected it ( n instead of m ).

Quote:

also,
$(i)\quad T_n=A\left(\dfrac{1}{5}\right)^n+B\left(-\dfrac{1}{5}\right)^n$
why does the second term has a negative sign? thanks

The roots of the characteristic equation are 1/5 and -1/5.
• May 4th 2011, 09:06 PM
hazeleyes
sorry for (i) I'm still confused about the roots, $25x^{n+2}-x^n=0\\x^n(25x^2-1)=0\\then\\x=\sqrt{1/25}\\x=\frac{1}{5}$
therefore why does second term has negative sign. thanks
• May 5th 2011, 12:05 AM
FernandoRevilla
Quote:

Originally Posted by hazeleyes
sorry for (i) I'm still confused about the roots, $25x^{n+2}-x^n=0\\x^n(25x^2-1)=0\\then\\x=\sqrt{1/25}\\x=\frac{1}{5}$
therefore why does second term has negative sign. thanks

The characteristic equation of

$T_{n+2}=T_n\quad [1]$

is

$25x^2-1=0$

whose solutions are

$x=\sqrt {1/25}=\pm 1/5$

According to a well known theorem, the general solution of [1] is

$T_n=C_1\;(1/5)^n+C_2\;(-1/5)^n$