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September 22nd, 2015, 09:44 PM   #1
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Second order difference equation

Sorry, need help to solve this urgently
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September 22nd, 2015, 11:06 PM   #2
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Without any knowledge of Mathcad, I suspect you want a loop structure to repeatedly calculate values using that formula until you hit $L_43$.

It is possible to solve the sequence exactly for all $n$ quicker than I could write something in Mathcad, but since this is presumably a problem to get you familiar with the software, I'd do it that way.
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September 23rd, 2015, 01:39 AM   #3
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Originally Posted by v8archie View Post
Without any knowledge of Mathcad, I suspect you want a loop structure to repeatedly calculate values using that formula until you hit $L_43$.

It is possible to solve the sequence exactly for all $n$ quicker than I could write something in Mathcad, but since this is presumably a problem to get you familiar with the software, I'd do it that way.
Sorry, I'm actually asking without using mathcad, how do to solve this problem step by step. Thanks
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September 23rd, 2015, 04:25 AM   #4
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For $\displaystyle n = 3:$
$\displaystyle L_3 = 2L_2 - \tfrac{3}{2}L_1 = 2 \cdot \tfrac{1}{5} - \tfrac{3}{2} \cdot \tfrac{1}{10} = -\tfrac{1}{10}$

Repeat until $\displaystyle n = 43$.
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September 23rd, 2015, 06:03 AM   #5
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How about if I solve a different but similar problem and you apply the method to this one?
Let $\displaystyle 8a_n = 2a_{n-1} + 3a_{n-2} \qquad a_0 = 0, \; a_1 = 1$

Letcus suppose (without justification) that $\displaystyle a_n = Aa^n$ for constants $\displaystyle a$ and $\displaystyle A$. We can put this formula into the given difference equation to get $\displaystyle 8Aa^n = 2Aa^{n-1} + 3Aa^{n-2}$ which can be rearranged to get $\displaystyle Aa^{n-2}\left(8a^2-2a-3\right)=0$.

This is satisfied if $\displaystyle A=0$ or $\displaystyle a=0$ but that is not interesting, because then every $\displaystyle a_n=0$. So the interesting solution comes when the quadratic factor is equal to zero. The quadratic is called the characteristic polynomial of the equation, and we can solve it for $\displaystyle a$ by factorising or by using the quadratic formula. We always find two possible values for $\displaystyle a$, in this case $\displaystyle a \in \{-\frac12,\frac34\} $ so that both $\displaystyle a_n = A\left(-\frac12\right)^n $ and $\displaystyle a_n=B\left(\frac34\right)^n$ are solutions of the original difference equation (where $\displaystyle B $ is another constant). In fact, by trying it in the equation we see that $\displaystyle a_n=A\left(-\frac12\right)^n + B\left(\frac34\right)^n $ is also a solution. It is called the general solution because all possible solutions are special cases of it, generated by selecting specific values of the constants $\displaystyle A $ and $\displaystyle B$.

We use our initial values to find $\displaystyle A $ and $\displaystyle B $. When $\displaystyle n=0$ we have $\displaystyle A+B=0 $ and when $\displaystyle n=1 $ we get $\displaystyle -\frac12 A + \frac34 B = 1 $. Solving this pair of equations gives us $\displaystyle B=- A=\frac45$. So our final solution is
$\displaystyle a_n=\frac45\left(\frac34\right)^n -\frac45\left(-\frac12\right)^n $
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September 23rd, 2015, 07:22 AM   #6
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Originally Posted by fysmat View Post
For $\displaystyle n = 3:$
$\displaystyle L_3 = 2L_2 - \tfrac{3}{2}L_1 = 2 \cdot \tfrac{1}{5} - \tfrac{3}{2} \cdot \tfrac{1}{10} = -\tfrac{1}{10}$

Repeat until $\displaystyle n = 43$.
Sorry. Can explain more
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September 23rd, 2015, 07:23 AM   #7
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How about if I solve a different but similar problem and you apply the method to this one?
Let $\displaystyle 8a_n = 2a_{n-1} + 3a_{n-2} \qquad a_0 = 0, \; a_1 = 1$

Letcus suppose (without justification) that $\displaystyle a_n = Aa^n$ for constants $\displaystyle a$ and $\displaystyle A$. We can put this formula into the given difference equation to get $\displaystyle 8Aa^n = 2Aa^{n-1} + 3Aa^{n-2}$ which can be rearranged to get $\displaystyle Aa^{n-2}\left(8a^2-2a-3\right)=0$.

This is satisfied if $\displaystyle A=0$ or $\displaystyle a=0$ but that is not interesting, because then every $\displaystyle a_n=0$. So the interesting solution comes when the quadratic factor is equal to zero. The quadratic is called the characteristic polynomial of the equation, and we can solve it for $\displaystyle a$ by factorising or by using the quadratic formula. We always find two possible values for $\displaystyle a$, in this case $\displaystyle a \in \{-\frac12,\frac34\} $ so that both $\displaystyle a_n = A\left(-\frac12\right)^n $ and $\displaystyle a_n=B\left(\frac34\right)^n$ are solutions of the original difference equation (where $\displaystyle B $ is another constant). In fact, by trying it in the equation we see that $\displaystyle a_n=A\left(-\frac12\right)^n + B\left(\frac34\right)^n $ is also a solution. It is called the general solution because all possible solutions are special cases of it, generated by selecting specific values of the constants $\displaystyle A $ and $\displaystyle B$.

We use our initial values to find $\displaystyle A $ and $\displaystyle B $. When $\displaystyle n=0$ we have $\displaystyle A+B=0 $ and when $\displaystyle n=1 $ we get $\displaystyle -\frac12 A + \frac34 B = 1 $. Solving this pair of equations gives us $\displaystyle B=- A=\frac45$. So our final solution is
$\displaystyle a_n=\frac45\left(\frac34\right)^n -\frac45\left(-\frac12\right)^n $
Sorry, can give simple explanation? So the answer have to be 2 decimal point
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September 23rd, 2015, 07:49 AM   #8
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What do you not understand in the example I gave? This site is not intended as a place where you come to get your homework done for you.

In the example I gave, we'd put $\displaystyle n = 43$ into the final equation to get $\displaystyle a_43$.
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September 23rd, 2015, 10:24 PM   #9
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Originally Posted by v8archie View Post
How about if I solve a different but similar problem and you apply the method to this one?
Let $\displaystyle 8a_n = 2a_{n-1} + 3a_{n-2} \qquad a_0 = 0, \; a_1 = 1$

Letcus suppose (without justification) that $\displaystyle a_n = Aa^n$ for constants $\displaystyle a$ and $\displaystyle A$. We can put this formula into the given difference equation to get $\displaystyle 8Aa^n = 2Aa^{n-1} + 3Aa^{n-2}$ which can be rearranged to get $\displaystyle Aa^{n-2}\left(8a^2-2a-3\right)=0$.

This is satisfied if $\displaystyle A=0$ or $\displaystyle a=0$ but that is not interesting, because then every $\displaystyle a_n=0$. So the interesting solution comes when the quadratic factor is equal to zero. The quadratic is called the characteristic polynomial of the equation, and we can solve it for $\displaystyle a$ by factorising or by using the quadratic formula. We always find two possible values for $\displaystyle a$, in this case $\displaystyle a \in \{-\frac12,\frac34\} $ so that both $\displaystyle a_n = A\left(-\frac12\right)^n $ and $\displaystyle a_n=B\left(\frac34\right)^n$ are solutions of the original difference equation (where $\displaystyle B $ is another constant). In fact, by trying it in the equation we see that $\displaystyle a_n=A\left(-\frac12\right)^n + B\left(\frac34\right)^n $ is also a solution. It is called the general solution because all possible solutions are special cases of it, generated by selecting specific values of the constants $\displaystyle A $ and $\displaystyle B$.

We use our initial values to find $\displaystyle A $ and $\displaystyle B $. When $\displaystyle n=0$ we have $\displaystyle A+B=0 $ and when $\displaystyle n=1 $ we get $\displaystyle -\frac12 A + \frac34 B = 1 $. Solving this pair of equations gives us $\displaystyle B=- A=\frac45$. So our final solution is
$\displaystyle a_n=\frac45\left(\frac34\right)^n -\frac45\left(-\frac12\right)^n $
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Originally Posted by v8archie View Post
What do you not understand in the example I gave? This site is not intended as a place where you come to get your homework done for you.

In the example I gave, we'd put $\displaystyle n = 43$ into the final equation to get $\displaystyle a_43$.
So I have to get the whole series?
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September 24th, 2015, 12:44 AM   #10
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Sorry, can give simple explanation? So the answer have to be 2 decimal point
Is it correct to do in this way?
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