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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
In this theorem we identify that for a specified differential equation a set of fundamental solutions will exist. Consider the differential equation y′′ + p (t ) y′ + q (t
Application Interpolation and extrapolation are widely used by businessmen, administrators, sociologists, economists and financial analysts. While interpolation hel
If tanx+secx=sqr rt 3, 0 Ans) sec 2 x=(√3-tanx) 2 1+tan 2 x=3+tan 2 x-2√3tanx 2√3tanx=2 tanx=1/√3 x=30degree
70 multiply 67
on which date of the week does 4th december 2001 falls?
((1-x)/(1+x))^0.5
Find the solution to the following system of equations using substitution:
compare: 643,251: 633,512: 633,893. The answer is 633,512.
whats the best way to solpve
|a.x|=1 where x = i-2j+2k then calculate a
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