[math-fun] Identity transformation
Hi, Pi/2==sum((2*n)!*(1+x^(4*n+2))/(2^(2*n)*(n!)^2*(1+x^4)^(n+1/2)*(2*n+1)) ,n=0..inf); an identity transformation gives this result for any real number. I wonder what other process can lead to this formula Best regards.
Hi François, Mathematica immediately gives In[152]:= Assuming[x \[Element] Reals, FullSimplify[Pi/2 == Sum[(2*n)!*(1 + x^(4*n + 2))/(2^(2*n)*(n!)^2*(1 + x^4)^(n + 1/2)*(2*n + 1)), {n, 0, \[Infinity]}]]] Out[152]= 2 (ArcCsc[Sqrt[1 + x^4]] + ArcSin[x^2/Sqrt[1 + x^4]]) == \[Pi] but is unable to finish the proof, nor even plot the constant. You have apparently exposed another headache for the developers. —rwg On Mon, Dec 16, 2019 at 6:29 AM françois mendzina essomba2 <m_essob@yahoo.fr> wrote:
Hi,
Pi/2==sum((2*n)!*(1+x^(4*n+2))/(2^(2*n)*(n!)^2*(1+x^4)^(n+1/2)*(2*n+1)) ,n=0..inf);
an identity transformation gives this result for any real number.
I wonder what other process can lead to this formula
Best regards.
Interestingly, Maple gives the equivalent, after assuming x is real, of arcsin(1/sqrt(x^4+1)) + arcsin(x^2/sqrt(x^4+1)). It also can’t simplify this to Pi/2. By hand it is of course straightforward to use arcsin(x)=arctan(x/sqrt(1-x^2)) and arctan(u)+arctan(v) = arctan((u+v)/(1-uv)) and simplify to arctan(Infinity)=Pi/2. Looks like both Maple and Mathematica are unable to simplify a sum of arcsines like this one. Steve On Dec 16, 2019, at 11:59 AM, Bill Gosper <billgosper@gmail.com<mailto:billgosper@gmail.com>> wrote: Hi François, Mathematica immediately gives In[152]:= Assuming[x \[Element] Reals, FullSimplify[Pi/2 == Sum[(2*n)!*(1 + x^(4*n + 2))/(2^(2*n)*(n!)^2*(1 + x^4)^(n + 1/2)*(2*n + 1)), {n, 0, \[Infinity]}]]] Out[152]= 2 (ArcCsc[Sqrt[1 + x^4]] + ArcSin[x^2/Sqrt[1 + x^4]]) == \[Pi] but is unable to finish the proof, nor even plot the constant. You have apparently exposed another headache for the developers. —rwg On Mon, Dec 16, 2019 at 6:29 AM françois mendzina essomba2 <m_essob@yahoo.fr<mailto:m_essob@yahoo.fr>> wrote: Hi, Pi/2==sum((2*n)!*(1+x^(4*n+2))/(2^(2*n)*(n!)^2*(1+x^4)^(n+1/2)*(2*n+1)) ,n=0..inf); an identity transformation gives this result for any real number. I wonder what other process can lead to this formula Best regards. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com<mailto:math-fun@mailman.xmission.com> https://urldefense.proofpoint.com/v2/url?u=https-3A__mailman.xmission.com_cg...
Agreed maple falls short on this. Like Steve I discovered that, if you assume x is real, maple simplifies the infinite sum to: arcsin(1/sqrt(x^4+1)) + arcsin(x^2/sqrt(x^4+1)) Now convert that to arctan with the command convert(%,arctan); This results in two arctan terms and finally combine and simplify via: simplify(combine(%)); gives the desired result, namely: Pi/2. But, of course, you shouldn't have to do all that. On Mon, Dec 16, 2019 at 11:58 AM Lucas, Stephen K - lucassk <lucassk@jmu.edu> wrote:
Interestingly, Maple gives the equivalent, after assuming x is real, of arcsin(1/sqrt(x^4+1)) + arcsin(x^2/sqrt(x^4+1)). It also can’t simplify this to Pi/2. By hand it is of course straightforward to use arcsin(x)=arctan(x/sqrt(1-x^2)) and arctan(u)+arctan(v) = arctan((u+v)/(1-uv)) and simplify to arctan(Infinity)=Pi/2.
Looks like both Maple and Mathematica are unable to simplify a sum of arcsines like this one.
Steve
On Dec 16, 2019, at 11:59 AM, Bill Gosper <billgosper@gmail.com<mailto: billgosper@gmail.com>> wrote:
Hi François, Mathematica immediately gives In[152]:= Assuming[x \[Element] Reals, FullSimplify[Pi/2 == Sum[(2*n)!*(1 + x^(4*n + 2))/(2^(2*n)*(n!)^2*(1 + x^4)^(n + 1/2)*(2*n + 1)), {n, 0, \[Infinity]}]]]
Out[152]= 2 (ArcCsc[Sqrt[1 + x^4]] + ArcSin[x^2/Sqrt[1 + x^4]]) == \[Pi]
but is unable to finish the proof, nor even plot the constant. You have apparently exposed another headache for the developers. —rwg
On Mon, Dec 16, 2019 at 6:29 AM françois mendzina essomba2 < m_essob@yahoo.fr<mailto:m_essob@yahoo.fr>> wrote:
Hi,
Pi/2==sum((2*n)!*(1+x^(4*n+2))/(2^(2*n)*(n!)^2*(1+x^4)^(n+1/2)*(2*n+1)) ,n=0..inf);
an identity transformation gives this result for any real number.
I wonder what other process can lead to this formula
Best regards.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com<mailto:math-fun@mailman.xmission.com>
https://urldefense.proofpoint.com/v2/url?u=https-3A__mailman.xmission.com_cg...
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (4)
-
Bill Gosper -
françois mendzina essomba2 -
James Buddenhagen -
Lucas, Stephen K - lucassk