Come up with a new linear function that has a slope that falls in the range −1 < m < 0. Choose two different initial values. For this new linear function, what happens to the function’s values after many iterations? Are the function’s values getting close to a particular number in each case? I don't really understand what it's asking.
So nobody is smart enough to answer my simple questions
┐(´(エ)｀)┌
Answers
#6 was skipped for my teacher.
Question 7 follows a pattern with answers:
3.4
1.2
2.4
3.2
1.6
3.2
1.6 which is the loop
8. Question 8 answers
3.44
1.12
2.24
3.52
0.96
1.92
3.84*
0.32
0.64
1.28
2.56
2.88
2.24
-2.48
3.04
1.92
3.84*
0.32*
Which is the loop.
Problems with answers like this make me wonder if something is wrong with my Algebra 2 teacher, or if he is just mean. One little mistake and you would never find the answer.
9-11 were skipped.
Let's assume that we have a function :
f(n)=mf(n-1) or something like it, where -1<m<0.
For example,
If m=-½, and f₁=1, f₂= -½f(1)= -½.
f₃= -½f₂= ¼, f₄= -⅛ etc.
Explanation :
If f(1)=a, in the general case, f₂= -a/2, f₃= a/4, f₄= -a/8
So f(n)= (-1)ⁿ⁻¹a/2ⁿ⁻¹.
As n gets bigger, f(n)→0 no matter what the initial value of a was, provided a ≠ 0.
For a general linear function f(n) = mf(n-1)+b where b is any non-zero number and f(1) = a ≠ 0, we have:
f₂=ma+b, f₃=m(ma+b)+b=m²a+mb+b=m²a+b(1+m), f₄=m³a+b(1+m+m²), ...
f(n)=mⁿ⁻¹a+b(1+m+m²+...+mⁿ⁻²)=mⁿ⁻¹a+b(mⁿ⁻¹-1)/(m-1).
So, since -1<m<0, the first term converges to 0 but the second term converges to a finite non-zero value for b≠0, because we can write this as b/(1-m). The denominator is positive because m is negative and mⁿ⁻¹→0. If b is positive the convergence is to a positive value, otherwise it’s negative.
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