sum of the first n odd positive integers

[mathon]

hi,

i have the following recursive function:

unsigned int sum_odds(unsigned int n)
{
       if(n==1)
         return 1;
       else
         return sum_odds(n-1)+(2*n -1);
}

Does anybody know what the result is for the input of 10? - the problem
is i do not exactly know how i should handle the +(2*n-1)...

matti

 

[Kai-Uwe Bux]

hi,

i have the following recursive function:

unsigned int sum_odds(unsigned int n)
{
if(n==1)
return 1;
else
return sum_odds(n-1)+(2*n -1);
}

Does anybody know what the result is for the input of 10?

Should be 100:

a ab abc abcd
bb bbc bbcd
ccc cccd
dddd
1 = 1
4 = 1+3
9 = 1+3+5
16 = 1+3+5+7
25 = 1+3+5+7+9
...

- the problem is i do not exactly know how i should handle the

+(2*n-1)...

What about it?


Best

Kai-Uwe Bux

 

[mathon]

the problem is i do not understand when i should add the + (2*n -1) to
the return values...?

hi,

i have the following recursive function:

unsigned int sum_odds(unsigned int n)
{
if(n==1)
return 1;
else
return sum_odds(n-1)+(2*n -1);
}

Does anybody know what the result is for the input of 10?

Should be 100:

a ab abc abcd
bb bbc bbcd
ccc cccd
dddd
1 = 1
4 = 1+3
9 = 1+3+5
16 = 1+3+5+7
25 = 1+3+5+7+9
...

- the problem is i do not exactly know how i should handle the

+(2*n-1)...

What about it?


Best

Kai-Uwe Bux

 

[arnuld]

hi,

i have the following recursive function:

unsigned int sum_odds(unsigned int n)
{
if(n==1)
return 1;
else
return sum_odds(n-1)+(2*n -1);
}

it works fine without any trouble.

Does anybody know what the result is for the input of 10?

(sorry, but i have written /sum/ in Lisp style to make it concise)

the sum of first:

2 +ve integers = 4 (+ 1 3)
3 +ve integers = 9 (+ 1 3 5)
4 +ve integers = 16 (+ 1 3 5 7)
5 +ve integers = 25 (+ 1 3 5 7 9)
......................................................
.......................................................
10 +ve integers = 100 (+ 1 3 5 7 9 11 13 15 17 19)

- the problem
is i do not exactly know how i should handle the +(2*n-1)...

when unsure always use parenthesis. in your case, i find it better like
this:

return (sum_odds(n-1) + (2*n -1))

it makes clearer presentation to me.

thanks

 

[mathon]

hmm...but when i go through the recursion with for example sum_odds(4)

unsigned int sum_odds(unsigned int n)
{
if(n==1)
return 1;
else
return (sum_odds(n-1) + (2*n -1))
}

i got the following:

sum_odds(4)

return (sum_odds(3) + (7)); return 9+7 = 16;

return (sum_odds(2) + (5)); -> return 4+5 = 9;

return (sum_odds(1) + (3)); -> return 1+3 = 4;

return 1

so when i use 4 i got as result 16...??

 

[Greg]

hi,

i have the following recursive function:

unsigned int sum_odds(unsigned int n)
{
if(n==1)
return 1;
else
return sum_odds(n-1)+(2*n -1);
}

Does anybody know what the result is for the input of 10? - the problem
is i do not exactly know how i should handle the +(2*n-1)...

How about a non-recursive implementation:

unsigned sum_odds(unsigned n)
{
return n * n;
}

Should be correct as long as n * n does not overflow.

Greg

 

[Mark P]

hi,

i have the following recursive function:

unsigned int sum_odds(unsigned int n)
{
if(n==1)
return 1;
else
return sum_odds(n-1)+(2*n -1);
}

Even though I rather doubt that this code is being used for any
production software, you should nonetheless get in the habit of looking
for potentially bad inputs. What happens if I call your function on 0?

Does anybody know what the result is for the input of 10? - the problem
is i do not exactly know how i should handle the +(2*n-1)...

What is it that you think you need to "handle"? Better yet, what is it
that this function is supposed to actually do?

 

[arnuld]

How about a non-recursive implementation:

unsigned sum_odds(unsigned n)
{
return n * n;
}

WOW! this is great