Below is an answer I got for trying to solve a recurrence relation by substitution, however, I am confused on step 3 sum the constants. Can you explain the geometric series here, as well as the rest of step 3 and step 4 solve for the base case? Solving recurrence relations by substitution involves finding a pattern and then using mathematical induction to prove it. Let's walk through it step-by-step. The recurrence relation you provided is: T(n)=2T(n−3)+7 Step 1: Write out the terms by substitution Let's start by expanding the recurrence relation to see if a pattern emerges. 1: Substitute once: T(n)=2T(n−3)+7 2: Substitute T(n−3) in terms of T(n−6): T(n−3)=2T(n−6)+7 Substituting this back into the original recurrence: T(n)=2(2T(n−6)+7)+7 =4T(n−6)+2⋅7+7 =4T(n−6)+14+7 =4T(n−6)+21 3: Substitute again to express T(n−6) in terms of T(n−9): T(n−6)=2T(n−9)+7 Substituting back, we get: T(n)=4(2T(n−9)+7)+21 =8T(n−9)+4⋅7+21 =8T(n−9)+49 After a few substitutions, we see that each step introduces a factor of 2 for T and adds a multiple of 7 to the constant. Step 2: Generalize the pattern From our substitutions, a pattern seems to emerge: T(n)=2kT(n−3k)+7⋅(2k−1+2k−2+⋯+1) Step 3: Sum the constants The constants form a geometric series. The sum of a geometric series 1+r+r2+⋯+rk−1 is r−1rk−1. Here, r=2, so we have: 7⋅(2k−1+2k−2+⋯+1)=7⋅2−12k−1=7(2k−1) Thus, we can rewrite T(n) as: T(n)=2kT(n−3k)+7⋅(2k−1) Step 4: Solve for the base case To get rid of T(n−3k), choose k such that n−3k hits the base case. Suppose the base case is T(0), then let k=3n, so n−3k=0. Then: T(n)=23nT(0)+7⋅(23n−1) If T(0) is given, we can substitute it in to get a closed-form solution. If not, the solution is in terms of T(0): T(n)=23nT(0)+7⋅(23n−1) that's the solution by substitution!
Below is an answer I got for trying to solve a recurrence relation by substitution, however, I am confused on step 3 sum the constants. Can you explain the geometric series here, as well as the rest of step 3 and step 4 solve for the base case?
Solving recurrence relations by substitution involves finding a pattern and then using mathematical induction to prove it. Let's walk through it step-by-step.
The recurrence relation you provided is:
T(n)=2T(n−3)+7
Step 1: Write out the terms by substitution
Let's start by expanding the recurrence relation to see if a pattern emerges.
1: Substitute once:
T(n)=2T(n−3)+7
2: Substitute T(n−3) in terms of T(n−6):
T(n−3)=2T(n−6)+7
Substituting this back into the original recurrence:
T(n)=2(2T(n−6)+7)+7
=4T(n−6)+2⋅7+7
=4T(n−6)+14+7
=4T(n−6)+21
3: Substitute again to express T(n−6) in terms of T(n−9):
T(n−6)=2T(n−9)+7
Substituting back, we get:
T(n)=4(2T(n−9)+7)+21
=8T(n−9)+4⋅7+21
=8T(n−9)+49
After a few substitutions, we see that each step introduces a factor of 2 for T and adds a multiple of 7 to the constant.
Step 2: Generalize the pattern
From our substitutions, a pattern seems to emerge:
T(n)=2kT(n−3k)+7⋅(2k−1+2k−2+⋯+1)
Step 3: Sum the constants
The constants form a geometric series. The sum of a geometric series 1+r+r2+⋯+rk−1 is r−1rk−1. Here, r=2, so we have:
7⋅(2k−1+2k−2+⋯+1)=7⋅2−12k−1=7(2k−1)
Thus, we can rewrite T(n) as:
T(n)=2kT(n−3k)+7⋅(2k−1)
Step 4: Solve for the base case
To get rid of T(n−3k), choose k such that n−3k hits the base case. Suppose the base case is T(0), then let k=3n, so n−3k=0.
Then:
T(n)=23nT(0)+7⋅(23n−1)
If T(0) is given, we can substitute it in to get a closed-form solution. If not, the solution is in terms of T(0):
T(n)=23nT(0)+7⋅(23n−1)
that's the solution by substitution!
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