Determine whetherfis a function from the set of all bitstrings to the set of integers if

Mathematics

1 answer:

ipn [44]3 years ago

8 0

Answer:

a) Not a function

b) It is a function

c) It is a function

Step-by-step explanation:

a) f(S) is not completly well defined, becuase if the bistring has multiple zeros, you need to pick one and there is not specified method in how you would do so. Another reason for which f is not well defined as a function is that for a bistring with only 1's, f is not defined because there is no zero.

b) This is a function. f(S) returns a concrete and unique integer number for each bistring S. If the bistring has only zeros, the f(S) = 0.

c) This is well defined as a function. It compensates for the failures of the first definition:

- If multiple zeros appears in S, you pick the first one, so there is one possible value for f(S)

-If no zeros appear, f is still well defined because it takes the value 0, as specified in the definition.

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Can anyone explain those marked steps in case (iii)

Tema [17]

$\cos\dfrac{3x}2=\cos\left(\dfrac\pi2+\dfrac x2\right)$
$\implies \dfrac{3x}2=2n\pi+\dfrac\pi2+\dfrac x2$

follows from the fact that the cosine function is $2\pi$ -periodic, which means $\cos x=\cos(2\pi+x)$ . Roughly speaking, this is the same as saying that a point on a circle is the same as the point you get by completing a full revolution around the circle (i.e. add $2\pi$ to the original point's angle with respect to the horizontal axis).

If you make another complete revolution (so we're effectively adding $4\pi$ ) we get the same result: $\cos x=\cos(4\pi+x)$ . This is true for any number of complete revolutions, so that this pattern holds for any even multiple of $\pi$ added to the argument. Therefore $\cos x=\cos(2n\pi+x)$ for any integer $n$ .

Next, because $\cos(-x)=\cos x$ , it follows that $\cos x=\cos(2n\pi-x)$ is also true for any integer $n$ . So we have

$\implies \dfrac{3x}2=2n\pi\pm\left(\dfrac\pi2+\dfrac x2\right)$

The rest follows from considering either case and solving for $x$ .