pi and e. (1 Viewer)

[wanton-wonton]

(pi^4 + pi^5)^1/6 is e or is approximately e. ?

 

[Jago]

approximately

 

[wanton-wonton]

approximately

How do you know? Or did you just use a calculator.

 

[Jago]

((pi^4) + (pi^5))^(1 / 6) = 2.7182818109

e^1 = 2.718281828


with a calculator

 

[wanton-wonton]

((pi^4) + (pi^5))^(1 / 6) = 2.7182818109

e^1 = 2.718281828


with a calculator

But your calculator approximates pi. How do you know it wouldn't work for the real pi.

 

[velox]

The only reason you're saying that the calc approximates it is that it has a limited number of decimal places.

 

[acmilan]

Using taylor polynomials about x=0:

e = 1/0! + 1/1! + 1/2! + 1/3! + ... + 1/n!

From memory theres two other definitions involving a limit and integration

 

[wanton-wonton]

The only reason you're saying that the calc approximates it is that it has a limited number of decimal places.

Well, yeah, what's your argument? Isn't that obvious?

 

[velox]

Both of those numbers would fit on a calculator. They're different. Try on a calc on the computer, they give you much more range.

 

[David_O]

And calculators don't have an error of 3% lol.
The statement implies (e^3-pi^2)/pi^2 = 1; a calculator gives that number to be
1.03509038.

 

[KFunk]

From memory theres two other definitions involving a limit and integration

I remember one where you take the limit of (1 + 1/n)ⁿ as n --> ∞. I think the proof of it requires a bit of integration (one method at least).

 

[acmilan]

Yup thats the limit definition:

y = lim(n --> ∞) (1 + 1/n)ⁿ
lny = ln(lim(n --> ∞) (1 + 1/n)ⁿ)
lny = lim(n --> ∞) ln[(1 + 1/n)ⁿ]
lny = lim(n --> ∞) nln[1 + 1/n]
lny = lim(n --> ∞) ln[1 + 1/n]/(1/n)

You can use l'hopitals rule since that is 0/0

lny = lim(n --> ∞) [1 + 1/n]
lny = 1
y = e (by log definitions)

So e = lim(n --> ∞) (1 + 1/n)ⁿ

edit: e is also a unique number x > 0 such that int(1 to x) (1/t dt) = 1

 

[withoutaface]

That result is hardly significant. The following however, is supercool:

e^i*pi+1=0

 

[acmilan]

That result is hardly significant. The following however, is supercool:

e^i*pi+1=0

haha

 

[Xayma]

That result is hardly significant. The following however, is supercool:

e^i*pi+1=0

As if be lazy

e^iπ+1=0

I like lim (x--->0) (e^x-1)/x=x (which of course you can also see in the taylor polynomial that e^x-1--->sin x as x-->0 or done via a limit using e^ix )

 

[brett86]

 

[brett86]

 

[MAICHI]

(pi^4 + pi^5)^1/6 is e or is approximately e. ?

Where did you get that from? Does anyone know what method is used to come up with that approximation?

 

[brett86]

there seem to be an almost infinite number of ways to approximate pi and euler

2 more:

 

[MAICHI]

there seem to be an almost infinite number of ways to approximate pi and euler

It's Euler's number, not Euler.

Yeah there are many ways to approximate them. I can get pi^4 with the Riemann Zeta Function that you've posted by changing r^2 to r^4. But how do you get a pi^5 approximation?