Answer:
The linear inequality is represented by y < x _ 2 and y > x + 1
I will be using the language C++. Given the problem specification, there are an large variety of solving the problem, ranging from simple addition, to more complicated bit testing and selection. But since the problem isn't exactly high performance or practical, I'll use simple addition. For a recursive function, you need to create a condition that will prevent further recursion, I'll use the condition of multiplying by 0. Also, you need to define what your recursion is.
To wit, consider the following math expression
f(m,k) = 0 if m = 0, otherwise f(m-1,k) + k
If you calculate f(0,k), you'll get 0 which is exactly what 0 * k is.
If you calculate f(1,k), you'll get 0 + k, which is exactly what 1 * k is.
So here's the function
int product(int m, int k)
{
if (m == 0) return 0;
return product(m-1,k) + k;
}
Answer:
(a) true
(b) true
(c) false; {y = x, t < 1; y = 2x, t ≥ 1}
(d) false; y = 200x for .005 < |x| < 1
Step-by-step explanation:
(a) "s(t) is periodic with period T" means s(t) = s(t+nT) for any integer n. Since values of n may be of the form n = 2m for any integer m, then this also means ...
s(t) = s(t +2mt) = s(t +m(2T)) . . . for any integer m
This equation matches the form of a function periodic with period 2T.
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(b) A system being linear means the output for the sum of two inputs is the sum of the outputs from the separate inputs:
s(a) +s(b) = s(a+b) . . . . definition of linear function
Then if a=b, you have
2s(a) = s(2a)
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(c) The output from a time-shifted input will only be the time-shifted output of the unshifted input if the system is time-invariant. The problem conditions here don't require that. A system can be "linear continuous time" and still be time-varying.
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(d) A restriction on an input magnitude does not mean the same restriction applies to the output magnitude. The system may have gain, for example.
0.0454045404540454045404540454045404540 that is repeating for ever.