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Maths help

Koing

Koing

Elite Member <br> Super Moderator<br> Health and F
Help!

I have this:

class Calc
{
static int fun( int i)
{
if (i==0) return 3;
else
return i*i*i+2*i+3+fun(i-1);
}

public static void main(String[] args)
{
System.out.println( "The answer is "+fun(4));
}
/* The answer is 135 */
}

I am asked

b) What is the program Calc doing?
The series I worked out is;

3, 9, 24, 60, 135

I have tried to work out what it does but I don't get it.

Any help is appreciated.

Koing
 
Thanks a lot.

It is probably correct but I can not get it to work out.

example f(2) = 2^3 * 2(2)+f(1)
= 8 * 4 + 9 = 41 The answer should be 24.

Have you left out a +3 from the equation?

Koing
 
Oh, yes. Well, that's the advantage of C++ functions over standard math functions... they can have decision statements.
f(2) = 2^3 * 2*3 + f(1)
f(1) = 1^3 * 1 * 3 + f(0)
f(0) = 3
therefore
f(1) = 1 * 3 + 3 = 3 + 3 = 6
f(2) = 8 * 6 + 6 = 48 + 6 = 54

I hope I didn't make a silly mistake there somewhere.
 
Originally posted by: GigaCluster
Oh, yes. Well, that's the advantage of C++ functions over standard math functions... they can have decision statements.
f(2) = 2^3 * 2*3 + f(1)
f(1) = 1^3 * 1 * 3 + f(0)
f(0) = 3
therefore
f(1) = 1 * 3 + 3 = 3 + 3 = 6
f(2) = 8 * 6 + 6 = 48 + 6 = 54

I hope I didn't make a silly mistake there somewhere.
Click to expand...

Jesus H. Christ, you're a genius.
 
Originally posted by: GigaCluster
Oh, yes. Well, that's the advantage of C++ functions over standard math functions... they can have decision statements.
f(2) = 2^3 * 2*3 + f(1)
f(1) = 1^3 * 1 * 3 + f(0)
f(0) = 3
therefore
f(1) = 1 * 3 + 3 = 3 + 3 = 6
f(2) = 8 * 6 + 6 = 48 + 6 = 54

I hope I didn't make a silly mistake there somewhere.
Click to expand...

The original code is this basically

if (i==0) return 3;
else
return i*i*i+2*i+3+fun(i-1);

Yours is
f(n)=n^3*2n+f(n-1)

The series I get from the code is this

f(0) = 3
f(1) = 9
f(2) = 24
f(3) = 60
f(4) = 135

How did you get the extra 3 in your working out?

f(2) = 2^3 * 2*3 + f(1)
f(1) = 1^3 * 1 * 3 + f(0)
f(0) = 3
therefore
f(1) = 1 * 3 + 3 = 3 + 3 = 6
f(2) = 8 * 6 + 6 = 48 + 6 = 54
Click to expand...

From this formula?
f(n)=n^3*2n+f(n-1)

I follow the n^3 part.

The "*2n" part you don't seem to have in your examples?

The last part "+f(n-1) you have and I get that.

Koing
 
f(0) = 3
f(1) = 9 = f(0) + (1)^3 + 2(1) + 3
f(2) = 24 = f(1) + (2)^3 + 2(2) + 3
f(3) = 60 = f(2) + (3)^3 + 2(3) + 3
f(4) = 135 = f(3) + (4)^3 + 2(4) + 3

f(n) = f(n-1) + (n)^3 + 2(n) + 3

He messed it up slightly.
 
Pfft... I did make a mistake. I omitted the "+3+".
Instead of: f(n)=n^3*2n+f(n-1)
It should be: f(n)=n^3*2n+3+f(n-1) unless n is zero.

So, my calculations are wrong as well. Let's see... second try:

f(2) = 2^3 * 2*3 + 3 + f(1)
f(1) = 1^3 * 1*3 + 3 + f(0)
f(0) = 3
therefore
f(1) = 1 * 3 + 3 + 3 = 3 + 3 + 3 = 9
f(2) = 8 * 6 + 3 + 9 = 48 + 3 + 9 = 60

Sorry for all the confusion... I will kick myself if this is also wrong.
 
Originally posted by: GigaCluster
Pfft... I did make a mistake. I omitted the "+3+".
Instead of: f(n)=n^3*2n+f(n-1)
It should be: f(n)=n^3*2n+3+f(n-1) unless n is zero.

So, my calculations are wrong as well. Let's see... second try:

f(2) = 2^3 * 2*3 + 3 + f(1)
f(1) = 1^3 * 1*3 + 3 + f(0)
f(0) = 3
therefore
f(1) = 1 * 3 + 3 + 3 = 3 + 3 + 3 = 9
f(2) = 8 * 6 + 3 + 9 = 48 + 3 + 9 = 60

Sorry for all the confusion... I will kick myself if this is also wrong.
Click to expand...

Ahhhhh no biggie at least I see I'm not going mad and not getting it.

Thank you VERY much for the help and everybody else that contributed.

😀

Koing

 
You messed up again...don't kick yourself, just get some sleep.

It is:
f(n) = f(n-1) + (n)^3 + 2(n) + 3
NOT
f(n) = n^3 *2n + 3+ f(n-1)
 
It's OK. I'm currently a math major, and you're probably tired. I am curious as to why Koing is asking for homework help in the early morning hours of Saturday (he lives in Great Britain).
 
Originally posted by: crt1530
It's OK. I'm currently a math major, and you're probably tired. I am curious as to why Koing is asking for homework help in the early morning hours of Saturday (he lives in Great Britain).
Click to expand...

Thanks for ALL your help everybody.

I was doing it about 4am or so or whatever as I was ready to go to sleep (Saturday no training as this beer festival is on and the gym is closed today = wake up late 😀) then a friend messaged me. Had a few problems and such about the work and I decided to have a look. Got interested and wanted to finish this part off.

How do you guys in the states call "Maths" help then? The formal way would be Mathematics. Mathes with an e is not correct in UK.

Was wondering why I needed English help, as my post looked alright as far as things went I thought.

Koing
 
Originally posted by: Koing


How do you guys in the states call "Maths" help then? The formal way would be Mathematics. Mathes with an e is not correct in UK.

Was wondering why I needed English help, as my post looked alright as far as things went I thought.

Koing
Click to expand...


Math help. And it's Stat help for Statistics help.
Click to expand...
 
Originally posted by: OulOat
Originally posted by: Koing


How do you guys in the states call "Maths" help then? The formal way would be Mathematics. Mathes with an e is not correct in UK.

Was wondering why I needed English help, as my post looked alright as far as things went I thought.

Koing
Click to expand...


Math help. And it's Stat help for Statistics help.
Click to expand...
Click to expand...


So you don't say the s then? Hmm doesn't sound right to me.

I guess being in England too long has this affect on a person. Much like your use of "z" for the "ized" types of words and English use "ised" mainly. Same with colour and color etc.

Koing
 
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