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Solve the recurrence relation
T (K) = 2T (K-1), T (0) = 1
Ans: The following equation can be written in the subsequent form:
tn - 2tn-1 = 0
Here now successively replacing n by (n - 1) and then by (n - 2) and so on we obtain a set of equations.
The method is continued till terminating condition. Add these equations in such type of a way that all intermediate terms get cancelled. The equation can be rearranged as
Multiplying all the equations correspondingly by 20, 21, ..., 2n - 1 and then adding them together, we get
tn - 2nt0 = 0
or, tn = 2n
A business has the opportunity to expand by purchasing a machine at a cost of £80,000. The machine has an estimated life of 5 years and is projected to generate a cashflow of £20,0
40000*1000
Is this given matrices are called equal Matrices?
Solve the inequation: |x|
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If e were rational, then e = n/m for some positive integers m, n. So then 1/e = m/n. But the series expansion for 1/e is 1/e = 1 - 1/1! + 1/2! - 1/3! + ... Call the first n v
Suppose research on three major cell phones companies revealed the following transition matrix for the probability that a person with one cell phone carrier switches to another.
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Ut=Uxx+A exp(-bx) u(x,0)=A/b^2(1-exp(-bx)) u(0,t)=0 u(1,t)=-A/b^2 exp(-b)
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