Show that if M = (S, I, f, s0,F)is a deterministic finitestate automaton and f (s, x) = s for the state s ∈ S and the input string x ∈ I ∗, then f (s, xn) = s for every nonnegative integer n. (Here xn is the concatenation of n copies of the string x, defined recursively in Exercise 37 in Section 5.3.)

we were able to show that f(s,x^n)=s holds for n>0

Step-by-step explanation:

Given:

M=(S,I,f,so,F) a deterministic finite automaton

f(s,x)=s state s in S

proven f(s,x^n)=s for n>0

if n=0

f(s,x^0)=f(s,o)=s

if assume that f(s,x^n)=s prove that f(s,x^n+1)=s is true

the string of x^n+1 is:

x^n+1=(x^n)*x (eq. 1)

This can be written as:

f(s,x^n+1)=f(s,(x^n)*x) (eq. 2)

The states s of S we have:

f(s,xy)=f(f(s,x),y) (eq. 3)

equation 3 in equation 2:

f(s,(x^n)*x)=f(f(s,x^n),x) (eq. 4)

if f(s,x^n)=s is true, equation 4 will be equal to:

f(s,(x^n)*x)=f(f(s,x^n),x)=f(s,a)=s

we were able to show that f(s,x^n)=s holds for n>0

a, b, c, d are correct.

step-by-step explanation:

we have g(x) = 3 and are required to find which option is a translation of g(x).

now, 'translation' is a transformation which means to shift the function in any direction and the general form of translation is g(x) + k, where k is any real number.

so, according to the options we see that all are correct as they satisfy the general form of translation.

also, it can be seen from the graph below that the function g(x) = 3 is shifted from its position in all the four options.