Integral with two different answers using real and complex analysis












9












$begingroup$


The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
$$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






complex-analysis definite-integrals cauchy-integral-formula







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
    $endgroup$
    – Ben Millwood
    Jan 2 at 12:56
















9












$begingroup$


The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
$$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






complex-analysis definite-integrals cauchy-integral-formula







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
    $endgroup$
    – Ben Millwood
    Jan 2 at 12:56














9












9








9


2



$begingroup$


The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
$$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






complex-analysis definite-integrals cauchy-integral-formula







share|cite|improve this question











$endgroup$




The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
$$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






complex-analysis definite-integrals cauchy-integral-formula




complex-analysis definite-integrals cauchy-integral-formula






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 1:22









Saad

19.7k92352




19.7k92352










asked Jan 2 at 0:17









khaled014zkhaled014z

1769




1769








  • 2




    $begingroup$
    OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
    $endgroup$
    – Ben Millwood
    Jan 2 at 12:56














  • 2




    $begingroup$
    OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
    $endgroup$
    – Ben Millwood
    Jan 2 at 12:56








2




2




$begingroup$
OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
$endgroup$
– Ben Millwood
Jan 2 at 12:56




$begingroup$
OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
$endgroup$
– Ben Millwood
Jan 2 at 12:56










2 Answers
2






active

oldest

votes


















13












$begingroup$

The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



$$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
= lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
    $endgroup$
    – khaled014z
    Jan 2 at 0:26






  • 1




    $begingroup$
    @khaled014z See the edit. Let me know if you want more details.
    $endgroup$
    – N. S.
    Jan 2 at 0:27












  • $begingroup$
    Brilliant, that was kind of a tricky substitution, thank you
    $endgroup$
    – khaled014z
    Jan 2 at 0:31










  • $begingroup$
    When this is next edited, you want tan^{-1} twice in the last line.
    $endgroup$
    – Teepeemm
    Jan 2 at 14:12



















3












$begingroup$

Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



Instead, we have



$$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Don't you mean $x = frac{pi}{2} + npi$?
    $endgroup$
    – DavidG
    Jan 2 at 3:18










  • $begingroup$
    Hi Mark ! Happy New Year !
    $endgroup$
    – Claude Leibovici
    Jan 2 at 3:33










  • $begingroup$
    @DavidG Yes, of course. Thank you for the comment.
    $endgroup$
    – Mark Viola
    Jan 2 at 14:36










  • $begingroup$
    @ClaudeLeibovici Hi Claude! Happy New Year to you too!
    $endgroup$
    – Mark Viola
    Jan 2 at 14:36











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









13












$begingroup$

The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



$$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
= lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
    $endgroup$
    – khaled014z
    Jan 2 at 0:26






  • 1




    $begingroup$
    @khaled014z See the edit. Let me know if you want more details.
    $endgroup$
    – N. S.
    Jan 2 at 0:27












  • $begingroup$
    Brilliant, that was kind of a tricky substitution, thank you
    $endgroup$
    – khaled014z
    Jan 2 at 0:31










  • $begingroup$
    When this is next edited, you want tan^{-1} twice in the last line.
    $endgroup$
    – Teepeemm
    Jan 2 at 14:12
















13












$begingroup$

The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



$$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
= lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
    $endgroup$
    – khaled014z
    Jan 2 at 0:26






  • 1




    $begingroup$
    @khaled014z See the edit. Let me know if you want more details.
    $endgroup$
    – N. S.
    Jan 2 at 0:27












  • $begingroup$
    Brilliant, that was kind of a tricky substitution, thank you
    $endgroup$
    – khaled014z
    Jan 2 at 0:31










  • $begingroup$
    When this is next edited, you want tan^{-1} twice in the last line.
    $endgroup$
    – Teepeemm
    Jan 2 at 14:12














13












13








13





$begingroup$

The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



$$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
= lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






share|cite|improve this answer











$endgroup$



The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



$$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
= lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 0:55









Saad

19.7k92352




19.7k92352










answered Jan 2 at 0:25









N. S.N. S.

104k7114209




104k7114209












  • $begingroup$
    Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
    $endgroup$
    – khaled014z
    Jan 2 at 0:26






  • 1




    $begingroup$
    @khaled014z See the edit. Let me know if you want more details.
    $endgroup$
    – N. S.
    Jan 2 at 0:27












  • $begingroup$
    Brilliant, that was kind of a tricky substitution, thank you
    $endgroup$
    – khaled014z
    Jan 2 at 0:31










  • $begingroup$
    When this is next edited, you want tan^{-1} twice in the last line.
    $endgroup$
    – Teepeemm
    Jan 2 at 14:12


















  • $begingroup$
    Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
    $endgroup$
    – khaled014z
    Jan 2 at 0:26






  • 1




    $begingroup$
    @khaled014z See the edit. Let me know if you want more details.
    $endgroup$
    – N. S.
    Jan 2 at 0:27












  • $begingroup$
    Brilliant, that was kind of a tricky substitution, thank you
    $endgroup$
    – khaled014z
    Jan 2 at 0:31










  • $begingroup$
    When this is next edited, you want tan^{-1} twice in the last line.
    $endgroup$
    – Teepeemm
    Jan 2 at 14:12
















$begingroup$
Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
$endgroup$
– khaled014z
Jan 2 at 0:26




$begingroup$
Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
$endgroup$
– khaled014z
Jan 2 at 0:26




1




1




$begingroup$
@khaled014z See the edit. Let me know if you want more details.
$endgroup$
– N. S.
Jan 2 at 0:27






$begingroup$
@khaled014z See the edit. Let me know if you want more details.
$endgroup$
– N. S.
Jan 2 at 0:27














$begingroup$
Brilliant, that was kind of a tricky substitution, thank you
$endgroup$
– khaled014z
Jan 2 at 0:31




$begingroup$
Brilliant, that was kind of a tricky substitution, thank you
$endgroup$
– khaled014z
Jan 2 at 0:31












$begingroup$
When this is next edited, you want tan^{-1} twice in the last line.
$endgroup$
– Teepeemm
Jan 2 at 14:12




$begingroup$
When this is next edited, you want tan^{-1} twice in the last line.
$endgroup$
– Teepeemm
Jan 2 at 14:12











3












$begingroup$

Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



Instead, we have



$$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Don't you mean $x = frac{pi}{2} + npi$?
    $endgroup$
    – DavidG
    Jan 2 at 3:18










  • $begingroup$
    Hi Mark ! Happy New Year !
    $endgroup$
    – Claude Leibovici
    Jan 2 at 3:33










  • $begingroup$
    @DavidG Yes, of course. Thank you for the comment.
    $endgroup$
    – Mark Viola
    Jan 2 at 14:36










  • $begingroup$
    @ClaudeLeibovici Hi Claude! Happy New Year to you too!
    $endgroup$
    – Mark Viola
    Jan 2 at 14:36
















3












$begingroup$

Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



Instead, we have



$$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Don't you mean $x = frac{pi}{2} + npi$?
    $endgroup$
    – DavidG
    Jan 2 at 3:18










  • $begingroup$
    Hi Mark ! Happy New Year !
    $endgroup$
    – Claude Leibovici
    Jan 2 at 3:33










  • $begingroup$
    @DavidG Yes, of course. Thank you for the comment.
    $endgroup$
    – Mark Viola
    Jan 2 at 14:36










  • $begingroup$
    @ClaudeLeibovici Hi Claude! Happy New Year to you too!
    $endgroup$
    – Mark Viola
    Jan 2 at 14:36














3












3








3





$begingroup$

Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



Instead, we have



$$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






share|cite|improve this answer











$endgroup$



Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



Instead, we have



$$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 14:35

























answered Jan 2 at 0:28









Mark ViolaMark Viola

133k1277175




133k1277175












  • $begingroup$
    Don't you mean $x = frac{pi}{2} + npi$?
    $endgroup$
    – DavidG
    Jan 2 at 3:18










  • $begingroup$
    Hi Mark ! Happy New Year !
    $endgroup$
    – Claude Leibovici
    Jan 2 at 3:33










  • $begingroup$
    @DavidG Yes, of course. Thank you for the comment.
    $endgroup$
    – Mark Viola
    Jan 2 at 14:36










  • $begingroup$
    @ClaudeLeibovici Hi Claude! Happy New Year to you too!
    $endgroup$
    – Mark Viola
    Jan 2 at 14:36


















  • $begingroup$
    Don't you mean $x = frac{pi}{2} + npi$?
    $endgroup$
    – DavidG
    Jan 2 at 3:18










  • $begingroup$
    Hi Mark ! Happy New Year !
    $endgroup$
    – Claude Leibovici
    Jan 2 at 3:33










  • $begingroup$
    @DavidG Yes, of course. Thank you for the comment.
    $endgroup$
    – Mark Viola
    Jan 2 at 14:36










  • $begingroup$
    @ClaudeLeibovici Hi Claude! Happy New Year to you too!
    $endgroup$
    – Mark Viola
    Jan 2 at 14:36
















$begingroup$
Don't you mean $x = frac{pi}{2} + npi$?
$endgroup$
– DavidG
Jan 2 at 3:18




$begingroup$
Don't you mean $x = frac{pi}{2} + npi$?
$endgroup$
– DavidG
Jan 2 at 3:18












$begingroup$
Hi Mark ! Happy New Year !
$endgroup$
– Claude Leibovici
Jan 2 at 3:33




$begingroup$
Hi Mark ! Happy New Year !
$endgroup$
– Claude Leibovici
Jan 2 at 3:33












$begingroup$
@DavidG Yes, of course. Thank you for the comment.
$endgroup$
– Mark Viola
Jan 2 at 14:36




$begingroup$
@DavidG Yes, of course. Thank you for the comment.
$endgroup$
– Mark Viola
Jan 2 at 14:36












$begingroup$
@ClaudeLeibovici Hi Claude! Happy New Year to you too!
$endgroup$
– Mark Viola
Jan 2 at 14:36




$begingroup$
@ClaudeLeibovici Hi Claude! Happy New Year to you too!
$endgroup$
– Mark Viola
Jan 2 at 14:36


















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