Question

(c) Since π/4 = arctan, you can find the value of π/4 using your expansion. How many terms of the expansion would you have to add in order to obtain π correct to four decimal places?

Answer 1

I got the Maclaurin series for arctan(x) = \[\sum_{n=0}^{\inf} \dfrac{ (-1^n)x^{2n+1} }{ 2n+1 }\]

Since it's alternating, when should the approximation end?

Answer 2

\[x-\frac{ x^3 }{3 } + \frac{ x^5 }{5 } -\frac{ x^7 }{7 } + ...\]

how far does this go?

Answer 3

lol, no. I mean if it's alternating you cut off right before the one you don't want to include, right?
would that be x^1001/1001?

Answer 4

What do you mean by how far does this go?

Answer 5

the expansion. For approximation

Answer 6

.0001 whoops. Just how much terms I'd have to add up to get an accuracy of .0001

Answer 7

Couldn't we find out what the decimal approximation of pi over four is and then add enough terms of the series to become within .0001 of that value.

Answer 8

oooh I mean arctan 1

Answer 9

Right, so arctan(1) = pi/4

Answer 10

So we can write out a few terms for arctan(x) using the series you noted above and plug in 1 for x.

Answer 11

exactly.
\[1 - \frac{ 1 }{ 3 } + \frac{ 1 }{ 5 } - \frac{ 1 }{ 7 } +\frac{ 1 }{ 9 } - \frac{ 1 }{ 11 }+...\] and then what?

Answer 12

We know \[\pi/4 = 0.78539...\]

Find the value of n where the absolute value of the difference between the nth partial sum of the series and pi/4 is less than or equal to .0001

Answer 13

I'm admittedly struggling with taking the next step...

Answer 14

It's okay. What we want to do is start calculating each partial sum, like so,\[1-\frac{ 1 }{ 3 } = \frac{ 2 }{ 3}\approx.66666\]\[1-\frac{ 1 }{ 3 } + \frac{ 1 }{ 5 } = \frac{ 2 }{ 3}\ + \frac{ 1 }{ 5}\approx.66666 + .2 = .86666\] \[1-\frac{ 1 }{ 3 } + \frac{ 1 }{ 5 } = \frac{ 13 }{ 15}\ - \frac{ 1 }{ 7}\approx.86666 - .14285 = .72389\] Observe that the more terms we add, the closer we get to the actual value of pi/4. Once we calculate enough terms, some absolute difference of that nth partial sum and the actual value of pi/4 will be less than or equal to .0001

Answer 15

Could I create a program to automate this and see how close I get?

Answer 16

Wow. That's a great question. I'm sure we could. Do you know how to code in c?

Answer 17

I'll use c++ gimme a few minutes. Terribly sorry I'm so slow lol

Answer 18

Great. I'm interested in seeing how accurate you can get it.

Answer 19

I immediately regret this suggestion
...
double sum = 0;

for(int i=1;i<8;i+=2)
{
sum += ((pow((-1),i))/(2*i+1));
}
...

Answer 20

is not working

Answer 21

I was going to use some if statements to dictate when to add or subtract.

Answer 22

I was trying something like using a counter "if(counter%2!=0)"
but that would not work at all lol

Answer 23

Right that is what i'm trying lol

Answer 24

Mine works
Wanna see?

Answer 25

It's in c though. Not too different.

Answer 26

I'll manage with the syntax and whatnot

Answer 27

//pi over 4 approx


#include<stdio.h>
#include<stdlib.h>
#include<math.h.>


#define PI_OVER_FOUR 0.78539816339744830961566084581988

int main(){

int ans, i;
double sum = 0, error = 0;

printf("How many sums would you like to calculate?\n");
scanf("%d",&ans);

for (i = 0; i < ans; i++){

if (i%2 == 0)

sum += 1/(double)(2*i + 1);

else

sum -= 1/(double)(2*i + 1);

}

printf("We calculated %lf\n",sum);
printf("Our error is %lf\n", fabs(sum - PI_OVER_FOUR));

return 0;
}

Answer 28

I think I've stolen your code well enough. This is what I get when I input 8.

0.754268
Our error is 0.0311302

Answer 29

Yes. Apparently, we have to sum a LOT of numbers to get close the error bound.

Answer 30

There's an identity we can use that makes it a lot easier to approximate
\[π/4=4\arctan \frac{ 1 }{ 5 }−\arctan \frac{ 1 }{ 239 }\]

Answer 31

Hmmm. What identity is this?

Answer 32

The euler-machin one IIRC

Answer 33

That's pretty interesting. I have to go. Do you understand the ideas here?

Answer 34

Yep. Sorry for holding onto you for so damn long. I feel terrible. I'll get some buddies to medal you appropriately.

thanks for stepping me through this!

Answer 35

It's not a problem. This was really fun. We should talk again. Good luck on any exams you may have.

Answer 36

PM me if you'd like. The feeling's mutual :)

Answer 37

Great, see ya!