# Hi, could somebody assist me please?

• 06-12-2002
Drworthless
Hi, could somebody assist me please?
Hey there, I am a high school student in a C++ 2 programming class. Next week we have our final exam, but yesterday our teacher gave us the 6 programs that we mite have to do.

The longest program is probably 30 lines, and just consists of many for/do while loops.

I have completed three of them, but the other 3 I do not really understand.

If someone could just give me some tips or explain them to me, I would be very appreciative.

Here they are:

Quote:

Number one: code and execute a program that will approximiate the value of (pi)/4 using this formula:
Quote:

pi/4=1-1/3+1/5-1/7+1/9 . . . .
Quote:

First approximation is pi/4=1
Quote:

Second approximation is pi/4=1-1/3
Quote:

third: pi/4=1-1/3+1/5
Required outputs: approximation 1= 1.0
aprox 2: =.0667
3:= .867

Please stop when the absolute difference is between two succesive approximations are less than .01.

//I think that for this one I nest for-loops, but I am not clear on the forumla.

Question 2:
The number e can be approximated by the forumla:
e1= 1+1/1
e2= 1+1/1*(1+1/2)
e3= 1+1/1*(1+1/2*(1+1/3*(1+1/4)))

However, a better approximation would be:
e100= 1+1/1*(1+1/2*(1+1/3*(1+1/4*(1+1/5*(1+1/6( . . .1/99*(...1+1/100))))).

The teacher asks to code and execute the program which will calculate and print out e1, e2, through e10 and also e100.

Outputs:
approximation 1 of e is 2.0
aprox 2 of e is 2.5
aprox of 3 of e is 2.66666675

This one confuses me. It seems like I will be nesting loops, but I am most confused about this program.

Question 3:
It can be shown that the irration number e= 2.71828. . . can be approximatd by taking as many terms as desired in the relation

e=1+1/1!+1/2!+1/3!+1/4!. . . .

1st approximation ois 1
2nd is 1+1/1!=2.
3rd is 1+1/1!+1/2!=2.5
4th is 1+1/1!+1/2!+1/3!=2.666

Please print out the approximation, stop when the absolute difference in 2 successive approximations is less than .001.

output:
approx 1 of e is 1.
aprrox od 2 of e 2.0.
approx of 3 of e is 2.5
the x approximtion of e is x.xxxxxxxxxx

//Do while and while loops? Could someone assist me with the formulas.

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I am not asking anyone here to do it for me, I am just looking for help. If you can, please pm or email me, reply to this, or im me at Worthlessgaming on AIM or Drworthless on AIM.

Drworthless
• 06-12-2002
Mario
Re: Hi, could somebody assist me please?
Quote:

Originally posted by Drworthless
//I think that for this one I nest for-loops, but I am not clear on the forumla.
Not really. I may have missed something but to me a While loop will do the trick. You just need to manage a few counters in order to do the calculation...

Notice the patern in pi/4=1-1/3+1/5-1/7+1/9... that's 1/1 followed by 1/3, followed by 1/5... so you have to manage a counter that increments 2 each iteration... also, it's - then +, then -, then plus... so a boolean that switches values each iteration will be all you need to check on which operator to use...

You will finally have to check on each iteration if the previous 1/x-2 minus the current 1/x is equal or less to .01. If so, exit the loop.

So basically one single loop will do it.

I can get back to you later to check on the other questions
• 06-12-2002
Drworthless
mind if i instant message you?
• 06-12-2002
Mario
err... nope. That's what that thing is there for. :)

Anyways, It's a 1:45am here. I'm just a few keystrokes from going to bed