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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
Attached Images image.jpg (12.1 KB, 13 views) September 22nd, 2015, 11:06 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra 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. September 23rd, 2015, 01:39 AM   #3
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 Originally Posted by v8archie 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 September 23rd, 2015, 04:25 AM #4 Senior Member   Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics 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 $\displaystyle n = 43$. Thanks from Benit13 September 23rd, 2015, 06:03 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra 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$\displaystyle a_n=\frac45\left(\frac34\right)^n -\frac45\left(-\frac12\right)^n$ Thanks from Benit13 September 23rd, 2015, 07:22 AM   #6
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 Originally Posted by fysmat 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 September 23rd, 2015, 07:23 AM   #7
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 Originally Posted by v8archie 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 September 23rd, 2015, 07:49 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,696 Thanks: 2681 Math Focus: Mainly analysis and algebra 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 $\displaystyle a_43$. September 23rd, 2015, 10:24 PM   #9
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 Originally Posted by v8archie 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$\displaystyle a_n=\frac45\left(\frac34\right)^n -\frac45\left(-\frac12\right)^n$
Quote:
 Originally Posted by v8archie 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? September 24th, 2015, 12:44 AM   #10
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 Originally Posted by babyjo 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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