Anyone good with math?

[chipsdeluxe888]

I'm doing my hw and i'm stuck on this problem. I'm completely lost

for what values of the constant c do the following define mass functions on the positive integers: x=1,2,....?

a. Logarithmic : f(x) = c2^(-x) / (x)


I don't even know where to start.

Any help would be great

[rctriplefresh5]

Originally Posted by chipsdeluxe888

I'm doing my hw and i'm stuck on this problem. I'm completely lost

for what values of the constant c do the following define mass functions on the positive integers: x=1,2,....?

a. Logarithmic : f(x) = c2^(-x) / (x)


I don't even know where to start.

Any help would be great

what are you taking precalc? anyways i have no idea.

[max4225]

to be a mass function the sum of all the f(x)'s have to sum to 1:
f(1)+f(2)+f(3)+...+f(infinity)=1

given f(x) = c*2^(-x) / (x)

c2^-1/1+c2^-2/2+c2^-3/3+...c2^-n/n = 1 (lim n->oo)

c*( infinite sum of 1/(n2^n) )=1

c=1/( infinite sum of 1/(n2^n) )=1/B

1/(n2^n)=(1/2)^n /n
Taylor series of the natural log:
ln(1-a)=-infinite sum (a^n)/n (n=1 to infinity) |a|<1
a=1/2 => B=-ln(1-1/2)=-ln(1/2)= ln(2)
________________________
c=1/(ln(2))=1.442695041 ?
________________________

try the numbers out for 20 terms:
x f(x)=(1.442695041*(2^-x))/x
1 0.721347520500000
2 0.180336880125000
3 0.060112293375000
4 0.022542110015625
5 0.009016844006250
6 0.003757018335937
7 0.001610150715402
8 0.000704440937988
9 0.000313084861328
10 0.000140888187598
11 0.000064040085272
12 0.000029351705750
13 0.000013546941115
14 0.000006289651232
15 0.000002935170575
16 0.000001375861207
17 0.000000647464097
18 0.000000305746935
19 0.000000144827495
20 0.000000068793060
+________________
0.999999937306867 = f(1)+f(2)+f(3)+..
yeah it's approaching 1, so c=1/ln(2) seems to work

I'm very tired so I'm not sure if I'm doing that right. Something like that. Good luck

[chipsdeluxe888]

Originally Posted by rctriplefresh5

what are you taking precalc? anyways i have no idea.

No, probability

[chipsdeluxe888]

Originally Posted by max4225

to be a mass function the sum of all the f(x)'s have to sum to 1:
f(1)+f(2)+f(3)+...+f(infinity)=1

given f(x) = c*2^(-x) / (x)

c2^-1/1+c2^-2/2+c2^-3/3+...c2^-n/n = 1 (lim n->oo)

c*( infinite sum of 1/(n2^n) )=1

c=1/( infinite sum of 1/(n2^n) )=1/B

1/(n2^n)=(1/2)^n /n
Taylor series of the natural log:
ln(1-a)=-infinite sum (a^n)/n (n=1 to infinity) |a|<1
a=1/2 => B=-ln(1-1/2)=-ln(1/2)= ln(2)
________________________
c=1/(ln(2))=1.442695041 ?
________________________

try the numbers out for 20 terms:
x f(x)=(1.442695041*(2^-x))/x
1 0.721347520500000
2 0.180336880125000
3 0.060112293375000
4 0.022542110015625
5 0.009016844006250
6 0.003757018335937
7 0.001610150715402
8 0.000704440937988
9 0.000313084861328
10 0.000140888187598
11 0.000064040085272
12 0.000029351705750
13 0.000013546941115
14 0.000006289651232
15 0.000002935170575
16 0.000001375861207
17 0.000000647464097
18 0.000000305746935
19 0.000000144827495
20 0.000000068793060
+________________
0.999999937306867 = f(1)+f(2)+f(3)+..
yeah it's approaching 1, so c=1/ln(2) seems to work

I'm very tired so I'm not sure if I'm doing that right. Something like that. Good luck

thank you!!! that's very helpful

[ecotec83]

Wow i think my brain just melted down trying to understand that equation