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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
Differences of Squares (and other even powers) ? A square monomial is a monomial which is the square of another monomial. Here are some examples: 25 is the square of 5 x 2 i
Assume that Y 1 (t) and Y 2 (t) are two solutions to (1) and y 1 (t) and y 2 (t) are a fundamental set of solutions to the associated homogeneous differential equation (2) so, Y
As1212uestion #Minimum 100 words accepted#
Let a 0 , a 1 ::: be the series recursively defined by a 0 = 1, and an = 3 + a n-1 for n ≥ 1. (a) Compute a 1 , a 2 , a 3 and a 4 . (b) Compute a formula for an, n ≥ 0.
lbl 2 lcl 2 sin 2 θ
how does sin of x equal negative 1/3
2+4
A,B,C are natural numbers and are in arithmetic progressions and a+b+c=21.then find the possible values for a,b,c Solution) a+b+c=21 a+c=2b 3b=21 b=7 a can be 1,2,3,4,5,6 c c
round to the nearest hundreths 1677.76
Standard form of the line Let's begin this section off along a quick mathematical definition of a line. Any equation that can be written in the following form,
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