P1. Write down a computer program to compute the values of the following group of functions:

a) f(n)=lg100 (n) b) f(n)=20*n3 c)f(n)=lglglg(n)=lg(3) (n)

g(n)=n0.05 g(n)=(1.01) n g(n)=lg* (n))

Note that, the domain of f(n) and g(n) are positive natural numbers; n=1,2,3,… . You will compute and display the values f(n) and g(n); and your program will terminate when one of the functions becomes greater than the other for five consecutive n values.

i dont understand what he means at the last sentence. " when one of the functions becomes greater than the other for five consecutive n values."

can anybody help me to understand the problem ?

a) f(n)=lg100 (n) b) f(n)=20*n3 c)f(n)=lglglg(n)=lg(3) (n)

g(n)=n0.05 g(n)=(1.01) n g(n)=lg* (n))

Note that, the domain of f(n) and g(n) are positive natural numbers; n=1,2,3,… . You will compute and display the values f(n) and g(n); and your program will terminate when one of the functions becomes greater than the other for five consecutive n values.

i dont understand what he means at the last sentence. " when one of the functions becomes greater than the other for five consecutive n values."

can anybody help me to understand the problem ?

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You start at n = 1, Then you go n = n + 1. Sooner or later you'll have 5 n's in a row (e.g. 6,7,8,9,10 or 40,41,42,43,44) where f(n) is greater than g(n). Or 5 n's where g(n) is greater than f(n).

Good luck writing that program. Is it homework?

i write small program,

int main( int argc, char *argv[]){

int i = 1;

/* Preliminary 1 - a */

printf("Preliminary 1\n");

while( i <= 20 ){

printf("%d\n",i);

printf("a-) %.2f %.2f\n",pow(log10(i),100),

printf("b-) %.2f %.2f\n",20*pow(i,3),pow(1.

printf("c-) %.2f %.2f\n",pow(log10(i),3),lo

printf("\n");

i+=1;

}

return 0;

}

from the output can you explain which value is satisfy

Preliminary 1

a-) 0.00 1.00

b-) 20.00 1.01

c-) 0.00 0.00

a-) 0.00 1.04

b-) 160.00 1.02

c-) 0.03 0.69

a-) 0.00 1.06

b-) 540.00 1.03

c-) 0.11 1.10

a-) 0.00 1.07

b-) 1280.00 1.04

c-) 0.22 1.39

a-) 0.00 1.08

b-) 2500.00 1.05

c-) 0.34 1.61

a-) 0.00 1.09

b-) 4320.00 1.06

c-) 0.47 1.79

a-) 0.00 1.10

b-) 6860.00 1.07

c-) 0.60 1.95

a-) 0.00 1.11

b-) 10240.00 1.08

c-) 0.74 2.08

a-) 0.01 1.12

b-) 14580.00 1.09

c-) 0.87 2.20

a-) 1.00 1.12

b-) 20000.00 1.10

c-) 1.00 2.30

a-) 57.74 1.13

b-) 26620.00 1.12

c-) 1.13 2.40

a-) 2039.10 1.13

b-) 34560.00 1.13

c-) 1.26 2.48

a-) 48563.57 1.14

b-) 43940.00 1.14

c-) 1.38 2.56

a-) 838134.28 1.14

b-) 54880.00 1.15

c-) 1.51 2.64

a-) 11068842.87 1.14

b-) 67500.00 1.16

c-) 1.63 2.71

a-) 116674612.55 1.15

b-) 81920.00 1.17

c-) 1.75 2.77

a-) 1014749836.62 1.15

b-) 98260.00 1.18

c-) 1.86 2.83

a-) 7478278499.90 1.16

b-) 116640.00 1.20

c-) 1.98 2.89

a-) 47718520380.19 1.16

b-) 137180.00 1.21

c-) 2.09 2.94

a-) 268368134667.23 1.16

b-) 160000.00 1.22

c-) 2.20 3.00

(20 iteration)

Can you?

If you cannot tell whether the output is correct or not, my guess is you don't fully understand the question yet.

Just say so. I'll be happy to try another explanation.

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one of those functions returns a value that is greater than the other five for a certain n. When that same function returns a value that is greater than the other five for n+1, n+2, n+3 and n+4, the program should terminate.

HTH,

Ruud