Answer:
Step-by-step explanation:
Problem A
t(1) = 2(1) + 5
t(2) = 2*2 + 5 = 9
t(3) = 2*3 + 5 = 11
t(4) = 2*4 + 5 = 13
So this is the explicit result. Now try it recursively.
t_3 = t_2 + 2
t_3 = 9 + 2
t_3 = 11 which is just what it should do.
t_n = t_(n - 1) + 2
Problem B
t(1) = 3 * 1/2
t(1) = 3/2
t(2) = 3*(1/2)^2
t(2) = 3 * 1/4
t(2) = 3/4
t(3) = 3*(1/2)^3
t(3) = 3 * 1/8
t(3) = 3/8
t(4) = 3 (1/2)^4
t(4) = 3 (1/16)
t(4) = 3/16
So in general
t_n = t_n-1 * 1/2
For example t(5)
t_5 = t_4 * 1/2
t_5 = 3 /16 * 1/2 = 3/32
By applying the variational approach and then comparing the result to the exact solution, we can calculate the error in the approximation. That is the main and major use of Terminal notation of pi.
π/2 = [tex] \lim_{n \to \infty} π (2j)(2j) / (2j-1)(2j+1)
Here, by this expression, we set the limits, and get the approximate error in the experiment.
Hope this helps!
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Answer: 0.5
Step-by-step explanation:
Answer:
Decimal
Step-by-step explanation:
A DECIMAL is a symbol used to separate the ones place from the tenths place in decimal numbers.
