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Let $f(x)$ mean that function $f$ ,applied to $x$,and $f^{n}(x)$ mean $f(f(........f(x)))$,that is $f$ applied to $x$ ,$n$ times.Let $g(x) = x+1$ and $h_{n}(x)=g^{n}(x).$Then what is $h_{9}^{8}(72)?$
asked in Set Theory & Algebra by Boss (41.5k points) | 52 views
0
144?
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@minipanda, $h_{9}$ is fine, but what is $h_{9}^{8}$
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@MiNiPanda

Yes 144 is the right answer

can you explain?

+4

First compute h9(x)

Given: hn(x)=gn(x) , g(x)= x+1

g2(x)= g(g(x)) = g(x+1)=(x+1)+1 =x + 2*1

g3(x)= g(g(g(x))) = g(g2(x))=g(x+2)= (x+2)+1 = x+3 = x+ 3*1

Seeing this pattern we can infer that g9(x) = x+ 9*1=x +9

So, h9(x)=g9(x)=x+9

Now we have to compute h98(x)  where x=72

h91(x) = x+9

h92(x) = h9(h9(x)) = h9(x+9) = (x+9) +9 = x+2*9

h93(x) = h9(h9(h9(x))) = h9(h92(x))= h9(x+2*9) = (x+2*9) +9 = x+3*9

Again following the same pattern we can conclude that

h98(x) = x +8*9 =x+72

Put x=72 , it becomes 72+72=144

 

0
Why dont you put that as an answer.
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@MiNiPanda

This is the nice method

I think this you add to the answer

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