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Difference Equations
Solve fn = fn-1 + fn-2 f 0 = 2, f 1 = 1 Step 1: Find Roots fn = fn-1 + fn- fn - fn-1 - fn-2 = 0 λ^2 - λ - 1 = 0 (1±√(1^2 - 41(-1)))/(2*1) = (1±√5)/ ⇒λ 1 = (1+√5)/2, λ 2 = (1-√5)/
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Step 2: Find General Solution fn = a 1 λ 1 n^ + a 2 λ 2 n substitute in initial conditions and solve n = 0: f 0 = 2 = a 1 λ 10 + a 2 λ 20 = a 1 + a 2 2 - a 1 = a 2 n = 1: f 1 = 1 = a 1 λ 11 + a 2 λ 21 1 = a 1 λ 1 + (2 - a 1 ) λ 2 1 = a 1 λ 1 + 2 λ 2 - a 1 λ 2 1 - 2 λ 2 = (λ 1 - λ 2 ) a 1 a 1 = (1 - 2 λ 2 )/ (λ 1 - λ 2 ) a 1 = 1 ⇒ a 2 = 2 - 1 = 1 ⇒ fn = 1λ 1 n^ + 1λ 2 n^ = ((1+√5)/2)n^ + ((1-√5)/2)n
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Solve xn = xn-1 + 2xn-2 x 0 = 1, x 1 = 1 Step 1: Find Roots xn = xn-1 + 2xn- xn - xn-1 - 2xn-2 = 0 λ^2 - λ - 2 = 0 (1±√(1^2 - 41(-2)))/(2*1) = (1±√9)/ ⇒ λ 1 = (1+√9)/2 = 2, λ 2 = (1-√9)/2 = - Step 2: Find General Solution xn = a 1 λ 1 n^ + a 2 λ 2 n substitute in initial conditions and solve
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n = 0: x 0 = 1 = a 1 λ 10 + a 2 λ 20 1 = a 1 + a 2 1 - a 1 = a 2 n = 1: x 1 = 1 = a 1 λ 11 + a 2 λ 21 1 = a 1 λ 1 + (1 - a 1 )λ 2 1 = a 1 λ 1 + λ 2 - a 1 λ 2 1 - λ 2 = (λ 1 - λ 2 ) a 1 a 1 = (1 - λ 2 )/ (λ 1 - λ 2 ) a 1 = (1 - (-1))/(2 - (-1)) = 2/ ⇒ a 2 = 1 - 2/3 = 1/ ⇒ xn = 2/3 λ 1 n^ + 1/3 λ 2 n^ = 2/3(2)n^ + 1/3(-1)n
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Forcing Functions
Equations of the form: xn = ∑ cixn-i + g(n) where g(n) is the forcing function
n i=
Forcing function
Homogeneous part
Output of equation Output will look like either the homogeneous solution or the forcing function, which ever is larger
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Solve: xn = 3xn-1 + 5 x 0 = xn = hn + vn hn = 3hn- vn = 3vn-1 + 5 Step 1: Find Roots hn = 3hn- hn - 3hn-1 = 0 λ - 3 = 0 λ = 3 ⇒ hn = a 13 n
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Step 2: Find Particular Solution vn = 3vn-1 + 5 Guess vn = general equation of same form as forcing function GUESS: vn = c where c = constant Solve for variables by substituting in guess vn = 3vn-1 + 5 c = 3c + 5 -2c = 5 c = -5/ Substitute results into the guess equation ⇒ vn = -5/
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Step 3: Find General Solution hn = a 13 n xn = hn + vn xn = a 13 n^ - 5/ substitute in initial conditions and solve n = 0: x 0 = 1 = a 130 - 5/ 1 = a 1 - 5/ 7/2 = a 1 ⇒ xn = 7/2 * 3n^ - 5/
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c 1 n + c 0 = 3c 1 n - 3c 1 + 3c 0 + 2n - 4 pull out terms by their power n: c 1 n = 3c 1 n + 2n c 1 = 3c 1 + 2 -2 = 2 c 1 c 1 = - 1: c 0 = - 3c 1 + 3c 0 - 4 c 0 = - 3(-1) + 3c 0 - 4 //sub in answer c 0 = 3 + 3c 0 - 4 -2 c 0 = - 1 c 0 = 1/ ⇒ vn = c 1 n + c 0 = -1n + 1/2 = -n + 1/
Equation from last slide:
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if ax^3 +bx^2 + cx + d = ex^3 + fx^2 + gx + h this means that they = 0 for the same values of n. Therefore, 0 = ax^3 + bx^2 + cx + d 0 = ex^3 + fx^2 + gx + h this means that a = e, b = f, c = g, and d = h otherwise the polynomials wouldn’t be equal
Why Can Break Up By Powers
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Step 3: Find General Solution hn = a 13 n xn = hn + vn xn = a 13 n^ - n + 1/ substitute in initial conditions and solve n = 0: x 0 = 1 = a 130 - 0 + 1/ 1 = a 1 + 1/ 1/2 = a 1 ⇒ xn = 1/2 * 3n^ - n + 1/
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Solve xn = 3xn-1 + 2n^ x 0 = 1
xn = hn + vn hn = 3hn- vn = 3vn-1 + 2n
Step 1: Find Roots hn = 3hn- hn - 3hn-1 = 0 λ - 3 = 0 λ = 3 ⇒ hn = a 13 n
