K
kathy
if:
1. s[n]=2*S[n-1]
2. s[0]=2
then:
s[n]=?
1. s[n]=2*S[n-1]
2. s[0]=2
then:
s[n]=?
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D
Dave Quigley[work]
2 ^ n+1
D
Dave Quigley[work]
if you want that to be recursive for 2^n then make s[0] = 1 otherwise its
2^(n+1)
2^(n+1)
Z
Zack
-----Original Message-----
if:
1. s[n]=2*S[n-1]
2. s[0]=2
then:
s[n]=?
.
s[n] = 2 ^ n+1 (2 raise to the power n+1)
G
GC
s[n]=2^(n+1)
J
Jon Skeet
kathy said:if:
1. s[n]=2*S[n-1]
2. s[0]=2
then:
s[n]=?
s[n]=2^(n+1) (for all n>=0)
However, what's that got to do with C#?
K
kathy
correction:
if:
1. s[n]=2*S[n-1]-1
2. s[0]=2
then:
s[n]=?
if:
1. s[n]=2*S[n-1]-1
2. s[0]=2
then:
s[n]=?
G
GC
s[n]=2^(n+1)-1
come on now...
come on now...
B
Bruno Jouhier [MVP]
s[n] is (2^n) + 1
series looks like: 2, 3, 5, 9, 17, 33, 65, 129, ...
What's the point?
series looks like: 2, 3, 5, 9, 17, 33, 65, 129, ...
What's the point?
J
Jon Skeet
GC said:s[n]=2^(n+1)-1
No it doesn't.
s[0]=2
s[1]=3
s[2]=5
s[3]=9
s[4]=17
s[5]=33
Your sequence would go:
s[0]=1 // Disagreement even on the axioms!
s[1]=3
s[2]=7
s[3]=15
s[4]=31
s[5]=63
G
GC
Ok, I am being dumb today. Tomorrow I will be smart again.
Jon Skeet said:GC said:s[n]=2^(n+1)-1
No it doesn't.
s[0]=2
s[1]=3
s[2]=5
s[3]=9
s[4]=17
s[5]=33
Your sequence would go:
s[0]=1 // Disagreement even on the axioms!
s[1]=3
s[2]=7
s[3]=15
s[4]=31
s[5]=63
F
Fergus Cooney
Hi Kathy, Jon,
You've got me puzzled. My question is: What's this got to do with
<anything> ?
Regards,
Fergus
You've got me puzzled. My question is: What's this got to do with
<anything> ?
Regards,
Fergus
P
Pete
Hi,
s[n] = a big error 'cause when n=0, (n-1) will be an invalid index into s.
Unless of course you reorder your statements and correctly setup your loop.
Is this a programming group or what?
Pete
s[n] = a big error 'cause when n=0, (n-1) will be an invalid index into s.
Unless of course you reorder your statements and correctly setup your loop.
Is this a programming group or what?
Pete
S
Stan
Given you say algorithm maybe your assignment expects an answer something
along the lines of,
function int S(int n)
{
if (n==0)
return 2;
else
return 2 * S(n - 1) - 1;
}
Why don't you just tell us exactly what the questions are and we'll get you
an A+.
along the lines of,
function int S(int n)
{
if (n==0)
return 2;
else
return 2 * S(n - 1) - 1;
}
Why don't you just tell us exactly what the questions are and we'll get you
an A+.
Z
Zack
s[n] = (2^n) + 1;
-----Original Message-----
Given you say algorithm maybe your assignment expects an answer something
along the lines of,
function int S(int n)
{
if (n==0)
return 2;
else
return 2 * S(n - 1) - 1;
}
Why don't you just tell us exactly what the questions are and we'll get you
an A+.
kathy said:correction:
if:
1. s[n]=2*S[n-1]-1
2. s[0]=2
then:
s[n]=?
.