Mixing
How to prove that $pi=pi$?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
I am trying to prove that indeed $pi=pi$. More precisely, that:
$$6sum_n=0^infty frac(2n)!4^n (n!)^2 (2n+1) 2^2n+1=pi$$
Where the definition of $pi:$
$$pi=4sum_n=1^inftyfrac(-1)^(n+1)(2n-1)$$
Using the epsilon delta definition, we should prove:
$$forall epsilon_+existsdeltaforall k(k>deltarightarrow|3+6sum_n=1^k frac(2n)!4^n (n!)^2 (2n+1) 2^2n+1-4sum_n=1^kfrac(-1)^(n+1)(2n-1)|<epsilon)$$
Re-arranging the sum, it can be written as:
$$sum_n=1^kfrac(-1)^(n+1) 4^n+1 (n!)^2 (2n+1) 2^(2n+1)-6(2n)!(2n-1)(4n^2-1) 4^n (n!)^2 2^2n+1 -3$$
But from this point it's difficult to go on and find delta as a function of epsilon. What techniques should be used to prove the identity? Looking at a graph I think it's clear that $delta=textceil(frac1 epsilon)$ would satisfy the criteria, but proving it is a other matter.
real-analysis
asked Jul 26 at 19:46
Dole
792514
 |Â
show 4 more comments
up vote
3
down vote
favorite
I am trying to prove that indeed $pi=pi$. More precisely, that:
$$6sum_n=0^infty frac(2n)!4^n (n!)^2 (2n+1) 2^2n+1=pi$$
Where the definition of $pi:$
$$pi=4sum_n=1^inftyfrac(-1)^(n+1)(2n-1)$$
Using the epsilon delta definition, we should prove:
$$forall epsilon_+existsdeltaforall k(k>deltarightarrow|3+6sum_n=1^k frac(2n)!4^n (n!)^2 (2n+1) 2^2n+1-4sum_n=1^kfrac(-1)^(n+1)(2n-1)|<epsilon)$$
Re-arranging the sum, it can be written as:
$$sum_n=1^kfrac(-1)^(n+1) 4^n+1 (n!)^2 (2n+1) 2^(2n+1)-6(2n)!(2n-1)(4n^2-1) 4^n (n!)^2 2^2n+1 -3$$
But from this point it's difficult to go on and find delta as a function of epsilon. What techniques should be used to prove the identity? Looking at a graph I think it's clear that $delta=textceil(frac1 epsilon)$ would satisfy the criteria, but proving it is a other matter.
real-analysis
asked Jul 26 at 19:46
Dole
792514
5
What you're trying to prove and your $epsilon$-$delta$ statement are not the same thing. The two series expressions for $pi$ need not converge at the same speed. In fact, there are techniques known as series acceleration used to, well, accelerate speed of convergence of a given series.
– Fimpellizieri
Jul 26 at 19:51
1
The way you are approaching this problem is very unusual. It is customary to use the geometric/trigonometric definition (area of a unit circle, or one half the circumference) of $pi$ as the base definition and then show that each of these series converge to that value of $pi$.
– Hamed
Jul 26 at 19:54
2
The sum $ sum_n=0^infty binom2nn8^-n/(2n+1)=pi/(2sqrt2) $ according to Mathematica and not the value $pi /3$ as implied in the statement of the problem.
– skbmoore
Jul 26 at 20:15
You could try $2 sum _n=0^infty frac1 4^n (2 n+1)binom2 nn=pi$ instead perhaps.
– James Arathoon
Jul 26 at 20:22
@skbmoore More specifically, $$sum_n=0^infty binom2nn fracx^n2n+1 = fracarcsin(2sqrtx)2sqrtx$$ for $lvert xrvert < 1/4$.
– Clement C.
Jul 26 at 20:23
 |Â
show 4 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am trying to prove that indeed $pi=pi$. More precisely, that:
$$6sum_n=0^infty frac(2n)!4^n (n!)^2 (2n+1) 2^2n+1=pi$$
Where the definition of $pi:$
$$pi=4sum_n=1^inftyfrac(-1)^(n+1)(2n-1)$$
Using the epsilon delta definition, we should prove:
$$forall epsilon_+existsdeltaforall k(k>deltarightarrow|3+6sum_n=1^k frac(2n)!4^n (n!)^2 (2n+1) 2^2n+1-4sum_n=1^kfrac(-1)^(n+1)(2n-1)|<epsilon)$$
Re-arranging the sum, it can be written as:
$$sum_n=1^kfrac(-1)^(n+1) 4^n+1 (n!)^2 (2n+1) 2^(2n+1)-6(2n)!(2n-1)(4n^2-1) 4^n (n!)^2 2^2n+1 -3$$
But from this point it's difficult to go on and find delta as a function of epsilon. What techniques should be used to prove the identity? Looking at a graph I think it's clear that $delta=textceil(frac1 epsilon)$ would satisfy the criteria, but proving it is a other matter.
real-analysis
asked Jul 26 at 19:46
Dole
792514
I am trying to prove that indeed $pi=pi$. More precisely, that:
$$6sum_n=0^infty frac(2n)!4^n (n!)^2 (2n+1) 2^2n+1=pi$$
Where the definition of $pi:$
$$pi=4sum_n=1^inftyfrac(-1)^(n+1)(2n-1)$$
Using the epsilon delta definition, we should prove:
$$forall epsilon_+existsdeltaforall k(k>deltarightarrow|3+6sum_n=1^k frac(2n)!4^n (n!)^2 (2n+1) 2^2n+1-4sum_n=1^kfrac(-1)^(n+1)(2n-1)|<epsilon)$$
Re-arranging the sum, it can be written as:
$$sum_n=1^kfrac(-1)^(n+1) 4^n+1 (n!)^2 (2n+1) 2^(2n+1)-6(2n)!(2n-1)(4n^2-1) 4^n (n!)^2 2^2n+1 -3$$
But from this point it's difficult to go on and find delta as a function of epsilon. What techniques should be used to prove the identity? Looking at a graph I think it's clear that $delta=textceil(frac1 epsilon)$ would satisfy the criteria, but proving it is a other matter.
real-analysis
asked Jul 26 at 19:46
Dole
792514
edited Jul 26 at 20:24
asked Jul 26 at 19:46
Dole
792514
asked Jul 26 at 19:46
Dole
792514
asked Jul 26 at 19:46
Dole
792514
792514
5
What you're trying to prove and your $epsilon$-$delta$ statement are not the same thing. The two series expressions for $pi$ need not converge at the same speed. In fact, there are techniques known as series acceleration used to, well, accelerate speed of convergence of a given series.
– Fimpellizieri
Jul 26 at 19:51
1
The way you are approaching this problem is very unusual. It is customary to use the geometric/trigonometric definition (area of a unit circle, or one half the circumference) of $pi$ as the base definition and then show that each of these series converge to that value of $pi$.
– Hamed
Jul 26 at 19:54
2
The sum $ sum_n=0^infty binom2nn8^-n/(2n+1)=pi/(2sqrt2) $ according to Mathematica and not the value $pi /3$ as implied in the statement of the problem.
– skbmoore
Jul 26 at 20:15
You could try $2 sum _n=0^infty frac1 4^n (2 n+1)binom2 nn=pi$ instead perhaps.
– James Arathoon
Jul 26 at 20:22
@skbmoore More specifically, $$sum_n=0^infty binom2nn fracx^n2n+1 = fracarcsin(2sqrtx)2sqrtx$$ for $lvert xrvert < 1/4$.
– Clement C.
Jul 26 at 20:23
 |Â
show 4 more comments
5
What you're trying to prove and your $epsilon$-$delta$ statement are not the same thing. The two series expressions for $pi$ need not converge at the same speed. In fact, there are techniques known as series acceleration used to, well, accelerate speed of convergence of a given series.
– Fimpellizieri
Jul 26 at 19:51
1
The way you are approaching this problem is very unusual. It is customary to use the geometric/trigonometric definition (area of a unit circle, or one half the circumference) of $pi$ as the base definition and then show that each of these series converge to that value of $pi$.
– Hamed
Jul 26 at 19:54
2
The sum $ sum_n=0^infty binom2nn8^-n/(2n+1)=pi/(2sqrt2) $ according to Mathematica and not the value $pi /3$ as implied in the statement of the problem.
– skbmoore
Jul 26 at 20:15
You could try $2 sum _n=0^infty frac1 4^n (2 n+1)binom2 nn=pi$ instead perhaps.
– James Arathoon
Jul 26 at 20:22
@skbmoore More specifically, $$sum_n=0^infty binom2nn fracx^n2n+1 = fracarcsin(2sqrtx)2sqrtx$$ for $lvert xrvert < 1/4$.
– Clement C.
Jul 26 at 20:23
5
5
What you're trying to prove and your $epsilon$-$delta$ statement are not the same thing. The two series expressions for $pi$ need not converge at the same speed. In fact, there are techniques known as series acceleration used to, well, accelerate speed of convergence of a given series.
– Fimpellizieri
Jul 26 at 19:51
What you're trying to prove and your $epsilon$-$delta$ statement are not the same thing. The two series expressions for $pi$ need not converge at the same speed. In fact, there are techniques known as series acceleration used to, well, accelerate speed of convergence of a given series.
– Fimpellizieri
Jul 26 at 19:51
1
1
The way you are approaching this problem is very unusual. It is customary to use the geometric/trigonometric definition (area of a unit circle, or one half the circumference) of $pi$ as the base definition and then show that each of these series converge to that value of $pi$.
– Hamed
Jul 26 at 19:54
The way you are approaching this problem is very unusual. It is customary to use the geometric/trigonometric definition (area of a unit circle, or one half the circumference) of $pi$ as the base definition and then show that each of these series converge to that value of $pi$.
– Hamed
Jul 26 at 19:54
2
2
The sum $ sum_n=0^infty binom2nn8^-n/(2n+1)=pi/(2sqrt2) $ according to Mathematica and not the value $pi /3$ as implied in the statement of the problem.
– skbmoore
Jul 26 at 20:15
The sum $ sum_n=0^infty binom2nn8^-n/(2n+1)=pi/(2sqrt2) $ according to Mathematica and not the value $pi /3$ as implied in the statement of the problem.
– skbmoore
Jul 26 at 20:15
You could try $2 sum _n=0^infty frac1 4^n (2 n+1)binom2 nn=pi$ instead perhaps.
– James Arathoon
Jul 26 at 20:22
You could try $2 sum _n=0^infty frac1 4^n (2 n+1)binom2 nn=pi$ instead perhaps.
– James Arathoon
Jul 26 at 20:22
@skbmoore More specifically, $$sum_n=0^infty binom2nn fracx^n2n+1 = fracarcsin(2sqrtx)2sqrtx$$ for $lvert xrvert < 1/4$.
– Clement C.
Jul 26 at 20:23
@skbmoore More specifically, $$sum_n=0^infty binom2nn fracx^n2n+1 = fracarcsin(2sqrtx)2sqrtx$$ for $lvert xrvert < 1/4$.
– Clement C.
Jul 26 at 20:23
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
Consider the expression
begineqnarray*
sum_n=0^infty binom2nn fracx^2n+12n+1.
endeqnarray*
Now use
begineqnarray*
fracx^2n+12n+1= int_0^x x^2n dx
endeqnarray*
and
begineqnarray*
sum_n=0^infty binom2nn x^2n = frac1sqrt1-4x^2.
endeqnarray*
Invert the order of the plum & integral and the expression becomes
begineqnarray*
int_0^x frac1sqrt1-4x^2 dx.
endeqnarray*
This integral can be done by the substition $2x=sin( theta)$ to give
begineqnarray*
frac12sin^-1 (2x).
endeqnarray*
Now substitute $x^2=1/8$ and we have
begineqnarray*
sum_n=0^infty frac12n+1binom2nn frac18^n= fracpi2 sqrt2.
endeqnarray*
answered Jul 26 at 20:34
Donald Splutterwit
21.3k21143
5
Do you have a reference for the theorem on plum and integral swapping? :)
– Clement C.
Jul 26 at 20:45
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Consider the expression
begineqnarray*
sum_n=0^infty binom2nn fracx^2n+12n+1.
endeqnarray*
Now use
begineqnarray*
fracx^2n+12n+1= int_0^x x^2n dx
endeqnarray*
and
begineqnarray*
sum_n=0^infty binom2nn x^2n = frac1sqrt1-4x^2.
endeqnarray*
Invert the order of the plum & integral and the expression becomes
begineqnarray*
int_0^x frac1sqrt1-4x^2 dx.
endeqnarray*
This integral can be done by the substition $2x=sin( theta)$ to give
begineqnarray*
frac12sin^-1 (2x).
endeqnarray*
Now substitute $x^2=1/8$ and we have
begineqnarray*
sum_n=0^infty frac12n+1binom2nn frac18^n= fracpi2 sqrt2.
endeqnarray*
answered Jul 26 at 20:34
Donald Splutterwit
21.3k21143
5
Do you have a reference for the theorem on plum and integral swapping? :)
– Clement C.
Jul 26 at 20:45
add a comment |Â
up vote
1
down vote
Consider the expression
begineqnarray*
sum_n=0^infty binom2nn fracx^2n+12n+1.
endeqnarray*
Now use
begineqnarray*
fracx^2n+12n+1= int_0^x x^2n dx
endeqnarray*
and
begineqnarray*
sum_n=0^infty binom2nn x^2n = frac1sqrt1-4x^2.
endeqnarray*
Invert the order of the plum & integral and the expression becomes
begineqnarray*
int_0^x frac1sqrt1-4x^2 dx.
endeqnarray*
This integral can be done by the substition $2x=sin( theta)$ to give
begineqnarray*
frac12sin^-1 (2x).
endeqnarray*
Now substitute $x^2=1/8$ and we have
begineqnarray*
sum_n=0^infty frac12n+1binom2nn frac18^n= fracpi2 sqrt2.
endeqnarray*
answered Jul 26 at 20:34
Donald Splutterwit
21.3k21143
5
Do you have a reference for the theorem on plum and integral swapping? :)
– Clement C.
Jul 26 at 20:45
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Consider the expression
begineqnarray*
sum_n=0^infty binom2nn fracx^2n+12n+1.
endeqnarray*
Now use
begineqnarray*
fracx^2n+12n+1= int_0^x x^2n dx
endeqnarray*
and
begineqnarray*
sum_n=0^infty binom2nn x^2n = frac1sqrt1-4x^2.
endeqnarray*
Invert the order of the plum & integral and the expression becomes
begineqnarray*
int_0^x frac1sqrt1-4x^2 dx.
endeqnarray*
This integral can be done by the substition $2x=sin( theta)$ to give
begineqnarray*
frac12sin^-1 (2x).
endeqnarray*
Now substitute $x^2=1/8$ and we have
begineqnarray*
sum_n=0^infty frac12n+1binom2nn frac18^n= fracpi2 sqrt2.
endeqnarray*
answered Jul 26 at 20:34
Donald Splutterwit
21.3k21143
Consider the expression
begineqnarray*
sum_n=0^infty binom2nn fracx^2n+12n+1.
endeqnarray*
Now use
begineqnarray*
fracx^2n+12n+1= int_0^x x^2n dx
endeqnarray*
and
begineqnarray*
sum_n=0^infty binom2nn x^2n = frac1sqrt1-4x^2.
endeqnarray*
Invert the order of the plum & integral and the expression becomes
begineqnarray*
int_0^x frac1sqrt1-4x^2 dx.
endeqnarray*
This integral can be done by the substition $2x=sin( theta)$ to give
begineqnarray*
frac12sin^-1 (2x).
endeqnarray*
Now substitute $x^2=1/8$ and we have
begineqnarray*
sum_n=0^infty frac12n+1binom2nn frac18^n= fracpi2 sqrt2.
endeqnarray*
answered Jul 26 at 20:34
Donald Splutterwit
21.3k21143
answered Jul 26 at 20:34
Donald Splutterwit
21.3k21143
answered Jul 26 at 20:34
Donald Splutterwit
21.3k21143
answered Jul 26 at 20:34
Donald Splutterwit
21.3k21143
21.3k21143
5
Do you have a reference for the theorem on plum and integral swapping? :)
– Clement C.
Jul 26 at 20:45
add a comment |Â
5
Do you have a reference for the theorem on plum and integral swapping? :)
– Clement C.
Jul 26 at 20:45
5
5
Do you have a reference for the theorem on plum and integral swapping? :)
– Clement C.
Jul 26 at 20:45
Do you have a reference for the theorem on plum and integral swapping? :)
– Clement C.
Jul 26 at 20:45
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863745%2fhow-to-prove-that-pi-pi%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
5
What you're trying to prove and your $epsilon$-$delta$ statement are not the same thing. The two series expressions for $pi$ need not converge at the same speed. In fact, there are techniques known as series acceleration used to, well, accelerate speed of convergence of a given series.
– Fimpellizieri
Jul 26 at 19:51
1
The way you are approaching this problem is very unusual. It is customary to use the geometric/trigonometric definition (area of a unit circle, or one half the circumference) of $pi$ as the base definition and then show that each of these series converge to that value of $pi$.
– Hamed
Jul 26 at 19:54
2
The sum $ sum_n=0^infty binom2nn8^-n/(2n+1)=pi/(2sqrt2) $ according to Mathematica and not the value $pi /3$ as implied in the statement of the problem.
– skbmoore
Jul 26 at 20:15
You could try $2 sum _n=0^infty frac1 4^n (2 n+1)binom2 nn=pi$ instead perhaps.
– James Arathoon
Jul 26 at 20:22
@skbmoore More specifically, $$sum_n=0^infty binom2nn fracx^n2n+1 = fracarcsin(2sqrtx)2sqrtx$$ for $lvert xrvert < 1/4$.
– Clement C.
Jul 26 at 20:23