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Saffjdaf's questions at Yahoo! Answers regarding difference equations
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Hello Saffjdaf,
1.) We are given the homogeneous recursion:
\(\displaystyle a_{n}=-a_{n-1}\) where \(\displaystyle a_1=1\)
The idea behind solving homogeneous difference equations is to assume a solution of the form:
\(\displaystyle a_n=c_1r^n\) where \(\displaystyle c_1,r\ne0\)
And so, we may write the given difference equation as:
\(\displaystyle c_1r^n+c_1r^{n-1}=0\)
Dividing through by \(\displaystyle c_1r^{n-1}\) we obtain the characteristic equation:
\(\displaystyle r+1=0\implies r=-1\)
And so our solution must be:
\(\displaystyle a_n=c_1(-1)^n\)
Now, we may use the given initial value to determine the parameter $c_1$:
\(\displaystyle a_1=c_1(-1)^1=-c_1=1\implies c_1=-1\)
And so we find the closed form for the solution is:
\(\displaystyle a_{n}=(-1)^{n+1}\)
Thus, the sequence oscillates from 1 to -1 back to 1 indefinitely. Terms with an odd subscript are 1 and terms with an even subscript are -1.
2.) The only difference between this problem as the first is the initial value, we we know:
\(\displaystyle a_n=c_1(-1)^n\)
Next, using the initial value, we find:
\(\displaystyle a_1=c_1(-1)^1=-c_1=2\implies c_1=-2\)
And so we have:
\(\displaystyle a_{n}=2(-1)^{n+1}\)
In fact, if we simply use $a_1$ as the value of the first term, we find:
\(\displaystyle a_1=c_1(-1)^1=-c_1\implies c_1=-a_1\)
and we have:
\(\displaystyle a_{n}=a_1(-1)^{n+1}\)
Thus the sequence oscillate from $a_1$ to $-a_1$ and back again to $a_1$ indefinitely.
3.) This time we have the inhomogeneous recurrence:
\(\displaystyle a_{n}=n+3a_{n-1}\) where \(\displaystyle a_1=-2\)
Now, we want first to find the solution to the corresponding homogenous difference equation, which has an associated characteristic equation of:
\(\displaystyle r-3=0\implies r=3\)
and so the homogeneous solution is:
\(\displaystyle h_n=c_1(3)^n\)
Next, we assume a particular solution of the form:
\(\displaystyle p_n=An+B\) where $A$ and $B$ are constants to be determined.
We assume a linear form for the particular solution because the inhomogeneous term is linear and no part of the homogeneous solution is part of this assume linear solution, that is, they are linearly independent.
So, substituting the particular solution into the difference equation, we find:
\(\displaystyle (An+B)-3(A(n-1)+B)=n\)
\(\displaystyle (-2A)n+(3A-2B)=(1)n+(0)\)
Equating coefficients, we find:
\(\displaystyle -2A=1\implies A=-\frac{1}{2}\)
\(\displaystyle 3A-2B=0\implies B=\frac{3}{2}A=-\frac{3}{4}\)
And so our particular solution is:
\(\displaystyle p_n=-\frac{1}{n}n-\frac{3}{4}=-\frac{2n+3}{4}\)
Hence, by superposition, we find:
\(\displaystyle a_n=h_n+p_n=c_1(3)^n-\frac{2n+3}{4}\)
Next, we mat use the given initial value to determine $c_1$:
\(\displaystyle a_1=c_1(3)^1-\frac{2(1)+3}{4}=3c_1-\frac{5}{4}=-2\,\therefore\,c_1=-\frac{1}{4}\)
And so, the closed form satisfying all given conditions is:
\(\displaystyle a_n=-\frac{3^n+2n+3}{4}\)
And so we have the next 5 terms:
1.) We are given the homogeneous recursion:
\(\displaystyle a_{n}=-a_{n-1}\) where \(\displaystyle a_1=1\)
The idea behind solving homogeneous difference equations is to assume a solution of the form:
\(\displaystyle a_n=c_1r^n\) where \(\displaystyle c_1,r\ne0\)
And so, we may write the given difference equation as:
\(\displaystyle c_1r^n+c_1r^{n-1}=0\)
Dividing through by \(\displaystyle c_1r^{n-1}\) we obtain the characteristic equation:
\(\displaystyle r+1=0\implies r=-1\)
And so our solution must be:
\(\displaystyle a_n=c_1(-1)^n\)
Now, we may use the given initial value to determine the parameter $c_1$:
\(\displaystyle a_1=c_1(-1)^1=-c_1=1\implies c_1=-1\)
And so we find the closed form for the solution is:
\(\displaystyle a_{n}=(-1)^{n+1}\)
Thus, the sequence oscillates from 1 to -1 back to 1 indefinitely. Terms with an odd subscript are 1 and terms with an even subscript are -1.
2.) The only difference between this problem as the first is the initial value, we we know:
\(\displaystyle a_n=c_1(-1)^n\)
Next, using the initial value, we find:
\(\displaystyle a_1=c_1(-1)^1=-c_1=2\implies c_1=-2\)
And so we have:
\(\displaystyle a_{n}=2(-1)^{n+1}\)
In fact, if we simply use $a_1$ as the value of the first term, we find:
\(\displaystyle a_1=c_1(-1)^1=-c_1\implies c_1=-a_1\)
and we have:
\(\displaystyle a_{n}=a_1(-1)^{n+1}\)
Thus the sequence oscillate from $a_1$ to $-a_1$ and back again to $a_1$ indefinitely.
3.) This time we have the inhomogeneous recurrence:
\(\displaystyle a_{n}=n+3a_{n-1}\) where \(\displaystyle a_1=-2\)
Now, we want first to find the solution to the corresponding homogenous difference equation, which has an associated characteristic equation of:
\(\displaystyle r-3=0\implies r=3\)
and so the homogeneous solution is:
\(\displaystyle h_n=c_1(3)^n\)
Next, we assume a particular solution of the form:
\(\displaystyle p_n=An+B\) where $A$ and $B$ are constants to be determined.
We assume a linear form for the particular solution because the inhomogeneous term is linear and no part of the homogeneous solution is part of this assume linear solution, that is, they are linearly independent.
So, substituting the particular solution into the difference equation, we find:
\(\displaystyle (An+B)-3(A(n-1)+B)=n\)
\(\displaystyle (-2A)n+(3A-2B)=(1)n+(0)\)
Equating coefficients, we find:
\(\displaystyle -2A=1\implies A=-\frac{1}{2}\)
\(\displaystyle 3A-2B=0\implies B=\frac{3}{2}A=-\frac{3}{4}\)
And so our particular solution is:
\(\displaystyle p_n=-\frac{1}{n}n-\frac{3}{4}=-\frac{2n+3}{4}\)
Hence, by superposition, we find:
\(\displaystyle a_n=h_n+p_n=c_1(3)^n-\frac{2n+3}{4}\)
Next, we mat use the given initial value to determine $c_1$:
\(\displaystyle a_1=c_1(3)^1-\frac{2(1)+3}{4}=3c_1-\frac{5}{4}=-2\,\therefore\,c_1=-\frac{1}{4}\)
And so, the closed form satisfying all given conditions is:
\(\displaystyle a_n=-\frac{3^n+2n+3}{4}\)
And so we have the next 5 terms:
| $n$ | $a_n$ |
| 1 | -2 |
| 2 | -4 |
| 3 | -9 |
| 4 | -23 |
| 5 | -64 |
| 6 | -186 |