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Alignment Problem of Mathematical Equation in Beamer Presentation
Problem converting images for beamer presentationText alignment in beamerBeamer equation problem settingtext alignment problem in beamerBeamer presentationBeamer presentation: Problem with captionsUsing printeranswers environment, how to print solutions in the box without the title “Solutions”Alignment problem in BeamerAlignment Problem in Beamer for Maths codeAlignment problem with a mathematical equation in a presentation in beamer
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
3
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[utf8]inputenc
usepackage[compat=1.1.0]tikz-feynman
usepackageamsmath, amsthm, amssymb,amsfonts
usepackagegraphicx
usepackagetikz
usepackagetcolorbox
usepackageamsmath
usepackageamsthm
usepackageamssymb
usepackagetikz-feynman
setcounterMaxMatrixCols20
%usepackagebreqn
usepackagetikz-feynman
usepackageamsmath
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
usecolortheme
date
begin document
begin frame
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
beginequation*
beginsplit
G_1 = frac(-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3Gamma(a_1+a_2+a_3-fracD2)Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)\\
times _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endsplit
endequation*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
end document
beamer math-mode horizontal-alignment
edited Jul 22 at 6:13
Red-Cloud
4,9852 silver badges17 bronze badges
asked Jul 22 at 6:08
Sumit BanikSumit Banik
716 bronze badges
add a comment |
3
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[utf8]inputenc
usepackage[compat=1.1.0]tikz-feynman
usepackageamsmath, amsthm, amssymb,amsfonts
usepackagegraphicx
usepackagetikz
usepackagetcolorbox
usepackageamsmath
usepackageamsthm
usepackageamssymb
usepackagetikz-feynman
setcounterMaxMatrixCols20
%usepackagebreqn
usepackagetikz-feynman
usepackageamsmath
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
usecolortheme
date
begin document
begin frame
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
beginequation*
beginsplit
G_1 = frac(-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3Gamma(a_1+a_2+a_3-fracD2)Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)\\
times _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endsplit
endequation*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
end document
beamer math-mode horizontal-alignment
edited Jul 22 at 6:13
Red-Cloud
4,9852 silver badges17 bronze badges
asked Jul 22 at 6:08
Sumit BanikSumit Banik
716 bronze badges
add a comment |
3
3
3
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[utf8]inputenc
usepackage[compat=1.1.0]tikz-feynman
usepackageamsmath, amsthm, amssymb,amsfonts
usepackagegraphicx
usepackagetikz
usepackagetcolorbox
usepackageamsmath
usepackageamsthm
usepackageamssymb
usepackagetikz-feynman
setcounterMaxMatrixCols20
%usepackagebreqn
usepackagetikz-feynman
usepackageamsmath
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
usecolortheme
date
begin document
begin frame
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
beginequation*
beginsplit
G_1 = frac(-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3Gamma(a_1+a_2+a_3-fracD2)Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)\\
times _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endsplit
endequation*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
end document
beamer math-mode horizontal-alignment
edited Jul 22 at 6:13
Red-Cloud
4,9852 silver badges17 bronze badges
asked Jul 22 at 6:08
Sumit BanikSumit Banik
716 bronze badges
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[utf8]inputenc
usepackage[compat=1.1.0]tikz-feynman
usepackageamsmath, amsthm, amssymb,amsfonts
usepackagegraphicx
usepackagetikz
usepackagetcolorbox
usepackageamsmath
usepackageamsthm
usepackageamssymb
usepackagetikz-feynman
setcounterMaxMatrixCols20
%usepackagebreqn
usepackagetikz-feynman
usepackageamsmath
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
usecolortheme
date
begin document
begin frame
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
beginequation*
beginsplit
G_1 = frac(-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3Gamma(a_1+a_2+a_3-fracD2)Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)\\
times _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endsplit
endequation*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
end document
beamer math-mode horizontal-alignment
beamer math-mode horizontal-alignment
edited Jul 22 at 6:13
Red-Cloud
4,9852 silver badges17 bronze badges
asked Jul 22 at 6:08
Sumit BanikSumit Banik
716 bronze badges
edited Jul 22 at 6:13
Red-Cloud
4,9852 silver badges17 bronze badges
asked Jul 22 at 6:08
Sumit BanikSumit Banik
716 bronze badges
edited Jul 22 at 6:13
Red-Cloud
4,9852 silver badges17 bronze badges
edited Jul 22 at 6:13
Red-Cloud
4,9852 silver badges17 bronze badges
edited Jul 22 at 6:13
Red-Cloud
4,9852 silver badges17 bronze badges
4,9852 silver badges17 bronze badges
asked Jul 22 at 6:08
Sumit BanikSumit Banik
716 bronze badges
asked Jul 22 at 6:08
Sumit BanikSumit Banik
716 bronze badges
asked Jul 22 at 6:08
Sumit BanikSumit Banik
716 bronze badges
716 bronze badges
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
3
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[compat=1.1.0]tikz-feynman
usepackagemathtools
usepackageamsthm, amssymb,amsfonts
usepackagegraphicx
usepackagetcolorbox
setcounterMaxMatrixCols20
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
begindocument
beginframeEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original
Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of
poles to get.
beginmultline*
G_1 = (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3\
timesfracGamma(a_1+a_2+a_3-fracD2)
Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)
Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)\
times _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endmultline*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
enddocument
answered Jul 22 at 6:28
Red-CloudRed-Cloud
4,9852 silver badges17 bronze badges
add a comment |
2
An alignment on 3 lines, and various improvements as to the size of fractions, with the medium-size fractions frm nccmath. Also, I simplified the code, deleting the multiply loaded packages, and replacing amsmath with its extension mathtools, for its matrix* environments, which accept an optional argument for the columns alignment.
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[utf8]inputenc
usepackagemathtools, nccmath
usepackageamsthm, amssymb
usepackagegraphicx
usepackagetcolorbox
usepackagetikz
usepackage[compat=1.1.0]tikz-feynman
setcounterMaxMatrixCols20
%usepackagebreqn
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
usecolortheme
date
begin document
%
begin frame
setlengthleftmargini12pt
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
beginalign*
G_1 =&(-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3times \
& fracGammaBigl(a_1+a_2+a_3-mfracD2Bigr) GammaBigl(mfracD2-a_1-a_2Bigr) GammaBigl(mfracD2-a_1-a_3Bigr)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3) times \
& _2F_1
Biggl(
beginmatrix*[r]
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-mfracD2\[0.5ex]
1+a_1+a_3-mfracD2
endmatrix*
Biggm | mfrac-M^2Q^2
Biggr)
endalign*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
end document
answered Jul 22 at 9:53
BernardBernard
186k7 gold badges84 silver badges220 bronze badges
add a comment |
2
I'd split the thing into three lines.
documentclassbeamer
usethemeAntibes
begindocument
beginframe
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full
solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get
three series for three set of poles to get.
beginequation*
beginsplit
G_1 =& (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3
\
×
frac
Gamma(a_1+a_2+a_3-fracD2)
Gamma(fracD2-a_1-a_2)
Gamma(fracD2-a_1-a_3)
Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)
\
× _2F_1
left(
beginsmallmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\
1+a_1+a_3-fracD2
endsmallmatrix
;middle|; frac-M^2Q^2
right)
endsplit
endequation*
For poles: $z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
enddocument
Note the before the symbol for the hypergeometric function, that's meant to avoid the subscript being attached to times. Note also middle.
answered Jul 22 at 10:12
egregegreg
759k90 gold badges1986 silver badges3332 bronze badges
add a comment |
0
You can divide your equation into three part. Consider the following code:
beginalign
G_1 =& (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3nonumber\
×
fracGamma(a_1+a_2+a_3-fracD2)Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)
nonumber\
× _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, a_1+a_2+a_3-fracD2\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endalign
answered Jul 22 at 6:25
A DiyanatA Diyanat
4192 silver badges8 bronze badges
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[compat=1.1.0]tikz-feynman
usepackagemathtools
usepackageamsthm, amssymb,amsfonts
usepackagegraphicx
usepackagetcolorbox
setcounterMaxMatrixCols20
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
begindocument
beginframeEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original
Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of
poles to get.
beginmultline*
G_1 = (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3\
timesfracGamma(a_1+a_2+a_3-fracD2)
Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)
Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)\
times _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endmultline*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
enddocument
answered Jul 22 at 6:28
Red-CloudRed-Cloud
4,9852 silver badges17 bronze badges
add a comment |
3
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[compat=1.1.0]tikz-feynman
usepackagemathtools
usepackageamsthm, amssymb,amsfonts
usepackagegraphicx
usepackagetcolorbox
setcounterMaxMatrixCols20
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
begindocument
beginframeEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original
Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of
poles to get.
beginmultline*
G_1 = (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3\
timesfracGamma(a_1+a_2+a_3-fracD2)
Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)
Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)\
times _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endmultline*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
enddocument
answered Jul 22 at 6:28
Red-CloudRed-Cloud
4,9852 silver badges17 bronze badges
add a comment |
3
3
3
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[compat=1.1.0]tikz-feynman
usepackagemathtools
usepackageamsthm, amssymb,amsfonts
usepackagegraphicx
usepackagetcolorbox
setcounterMaxMatrixCols20
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
begindocument
beginframeEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original
Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of
poles to get.
beginmultline*
G_1 = (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3\
timesfracGamma(a_1+a_2+a_3-fracD2)
Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)
Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)\
times _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endmultline*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
enddocument
answered Jul 22 at 6:28
Red-CloudRed-Cloud
4,9852 silver badges17 bronze badges
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[compat=1.1.0]tikz-feynman
usepackagemathtools
usepackageamsthm, amssymb,amsfonts
usepackagegraphicx
usepackagetcolorbox
setcounterMaxMatrixCols20
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
begindocument
beginframeEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original
Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of
poles to get.
beginmultline*
G_1 = (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3\
timesfracGamma(a_1+a_2+a_3-fracD2)
Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)
Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)\
times _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endmultline*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
enddocument
answered Jul 22 at 6:28
Red-CloudRed-Cloud
4,9852 silver badges17 bronze badges
answered Jul 22 at 6:28
Red-CloudRed-Cloud
4,9852 silver badges17 bronze badges
answered Jul 22 at 6:28
Red-CloudRed-Cloud
4,9852 silver badges17 bronze badges
answered Jul 22 at 6:28
Red-CloudRed-Cloud
4,9852 silver badges17 bronze badges
4,9852 silver badges17 bronze badges
add a comment |
add a comment |
2
An alignment on 3 lines, and various improvements as to the size of fractions, with the medium-size fractions frm nccmath. Also, I simplified the code, deleting the multiply loaded packages, and replacing amsmath with its extension mathtools, for its matrix* environments, which accept an optional argument for the columns alignment.
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[utf8]inputenc
usepackagemathtools, nccmath
usepackageamsthm, amssymb
usepackagegraphicx
usepackagetcolorbox
usepackagetikz
usepackage[compat=1.1.0]tikz-feynman
setcounterMaxMatrixCols20
%usepackagebreqn
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
usecolortheme
date
begin document
%
begin frame
setlengthleftmargini12pt
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
beginalign*
G_1 =&(-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3times \
& fracGammaBigl(a_1+a_2+a_3-mfracD2Bigr) GammaBigl(mfracD2-a_1-a_2Bigr) GammaBigl(mfracD2-a_1-a_3Bigr)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3) times \
& _2F_1
Biggl(
beginmatrix*[r]
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-mfracD2\[0.5ex]
1+a_1+a_3-mfracD2
endmatrix*
Biggm | mfrac-M^2Q^2
Biggr)
endalign*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
end document
answered Jul 22 at 9:53
BernardBernard
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add a comment |
2
An alignment on 3 lines, and various improvements as to the size of fractions, with the medium-size fractions frm nccmath. Also, I simplified the code, deleting the multiply loaded packages, and replacing amsmath with its extension mathtools, for its matrix* environments, which accept an optional argument for the columns alignment.
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[utf8]inputenc
usepackagemathtools, nccmath
usepackageamsthm, amssymb
usepackagegraphicx
usepackagetcolorbox
usepackagetikz
usepackage[compat=1.1.0]tikz-feynman
setcounterMaxMatrixCols20
%usepackagebreqn
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
usecolortheme
date
begin document
%
begin frame
setlengthleftmargini12pt
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
beginalign*
G_1 =&(-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3times \
& fracGammaBigl(a_1+a_2+a_3-mfracD2Bigr) GammaBigl(mfracD2-a_1-a_2Bigr) GammaBigl(mfracD2-a_1-a_3Bigr)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3) times \
& _2F_1
Biggl(
beginmatrix*[r]
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-mfracD2\[0.5ex]
1+a_1+a_3-mfracD2
endmatrix*
Biggm | mfrac-M^2Q^2
Biggr)
endalign*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
end document
answered Jul 22 at 9:53
BernardBernard
186k7 gold badges84 silver badges220 bronze badges
add a comment |
2
2
2
An alignment on 3 lines, and various improvements as to the size of fractions, with the medium-size fractions frm nccmath. Also, I simplified the code, deleting the multiply loaded packages, and replacing amsmath with its extension mathtools, for its matrix* environments, which accept an optional argument for the columns alignment.
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[utf8]inputenc
usepackagemathtools, nccmath
usepackageamsthm, amssymb
usepackagegraphicx
usepackagetcolorbox
usepackagetikz
usepackage[compat=1.1.0]tikz-feynman
setcounterMaxMatrixCols20
%usepackagebreqn
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
usecolortheme
date
begin document
%
begin frame
setlengthleftmargini12pt
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
beginalign*
G_1 =&(-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3times \
& fracGammaBigl(a_1+a_2+a_3-mfracD2Bigr) GammaBigl(mfracD2-a_1-a_2Bigr) GammaBigl(mfracD2-a_1-a_3Bigr)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3) times \
& _2F_1
Biggl(
beginmatrix*[r]
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-mfracD2\[0.5ex]
1+a_1+a_3-mfracD2
endmatrix*
Biggm | mfrac-M^2Q^2
Biggr)
endalign*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
end document
answered Jul 22 at 9:53
BernardBernard
186k7 gold badges84 silver badges220 bronze badges
An alignment on 3 lines, and various improvements as to the size of fractions, with the medium-size fractions frm nccmath. Also, I simplified the code, deleting the multiply loaded packages, and replacing amsmath with its extension mathtools, for its matrix* environments, which accept an optional argument for the columns alignment.
documentclassbeamer
newcommandFontvifontsize24selectfont
usepackage[utf8]inputenc
usepackagemathtools, nccmath
usepackageamsthm, amssymb
usepackagegraphicx
usepackagetcolorbox
usepackagetikz
usepackage[compat=1.1.0]tikz-feynman
setcounterMaxMatrixCols20
%usepackagebreqn
newtheoremrulesRule
usethemeAntibes
newcommandbracket[1]leftlangle #1 rightrangle
usecolortheme
date
begin document
%
begin frame
setlengthleftmargini12pt
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get three series for three set of poles to get.
beginalign*
G_1 =&(-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3times \
& fracGammaBigl(a_1+a_2+a_3-mfracD2Bigr) GammaBigl(mfracD2-a_1-a_2Bigr) GammaBigl(mfracD2-a_1-a_3Bigr)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3) times \
& _2F_1
Biggl(
beginmatrix*[r]
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-mfracD2\[0.5ex]
1+a_1+a_3-mfracD2
endmatrix*
Biggm | mfrac-M^2Q^2
Biggr)
endalign*
For poles: $ z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
end document
answered Jul 22 at 9:53
BernardBernard
186k7 gold badges84 silver badges220 bronze badges
answered Jul 22 at 9:53
BernardBernard
186k7 gold badges84 silver badges220 bronze badges
answered Jul 22 at 9:53
BernardBernard
186k7 gold badges84 silver badges220 bronze badges
answered Jul 22 at 9:53
BernardBernard
186k7 gold badges84 silver badges220 bronze badges
186k7 gold badges84 silver badges220 bronze badges
add a comment |
add a comment |
2
I'd split the thing into three lines.
documentclassbeamer
usethemeAntibes
begindocument
beginframe
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full
solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get
three series for three set of poles to get.
beginequation*
beginsplit
G_1 =& (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3
\
×
frac
Gamma(a_1+a_2+a_3-fracD2)
Gamma(fracD2-a_1-a_2)
Gamma(fracD2-a_1-a_3)
Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)
\
× _2F_1
left(
beginsmallmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\
1+a_1+a_3-fracD2
endsmallmatrix
;middle|; frac-M^2Q^2
right)
endsplit
endequation*
For poles: $z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
enddocument
Note the before the symbol for the hypergeometric function, that's meant to avoid the subscript being attached to times. Note also middle.
answered Jul 22 at 10:12
egregegreg
759k90 gold badges1986 silver badges3332 bronze badges
add a comment |
2
I'd split the thing into three lines.
documentclassbeamer
usethemeAntibes
begindocument
beginframe
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full
solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get
three series for three set of poles to get.
beginequation*
beginsplit
G_1 =& (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3
\
×
frac
Gamma(a_1+a_2+a_3-fracD2)
Gamma(fracD2-a_1-a_2)
Gamma(fracD2-a_1-a_3)
Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)
\
× _2F_1
left(
beginsmallmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\
1+a_1+a_3-fracD2
endsmallmatrix
;middle|; frac-M^2Q^2
right)
endsplit
endequation*
For poles: $z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
enddocument
Note the before the symbol for the hypergeometric function, that's meant to avoid the subscript being attached to times. Note also middle.
answered Jul 22 at 10:12
egregegreg
759k90 gold badges1986 silver badges3332 bronze badges
add a comment |
2
2
2
I'd split the thing into three lines.
documentclassbeamer
usethemeAntibes
begindocument
beginframe
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full
solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get
three series for three set of poles to get.
beginequation*
beginsplit
G_1 =& (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3
\
×
frac
Gamma(a_1+a_2+a_3-fracD2)
Gamma(fracD2-a_1-a_2)
Gamma(fracD2-a_1-a_3)
Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)
\
× _2F_1
left(
beginsmallmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\
1+a_1+a_3-fracD2
endsmallmatrix
;middle|; frac-M^2Q^2
right)
endsplit
endequation*
For poles: $z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
enddocument
Note the before the symbol for the hypergeometric function, that's meant to avoid the subscript being attached to times. Note also middle.
answered Jul 22 at 10:12
egregegreg
759k90 gold badges1986 silver badges3332 bronze badges
I'd split the thing into three lines.
documentclassbeamer
usethemeAntibes
begindocument
beginframe
frametitleEquivalence with Original Method of Brackets
beginitemize
item Single solution of Modified Method of Brackets contains the full
solution obtained from Original Method of Bracket
item If we apply Cauchy's Residue Theorem to the last 1-MB then we get
three series for three set of poles to get.
beginequation*
beginsplit
G_1 =& (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3
\
×
frac
Gamma(a_1+a_2+a_3-fracD2)
Gamma(fracD2-a_1-a_2)
Gamma(fracD2-a_1-a_3)
Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)
\
× _2F_1
left(
beginsmallmatrix
1+a_1+a_2+a_3-D, & a_1+a_2+a_3-fracD2\
1+a_1+a_3-fracD2
endsmallmatrix
;middle|; frac-M^2Q^2
right)
endsplit
endequation*
For poles: $z_2=-n-a_1-a_2-a_3+d/2$
enditemize
endframe
enddocument
Note the before the symbol for the hypergeometric function, that's meant to avoid the subscript being attached to times. Note also middle.
answered Jul 22 at 10:12
egregegreg
759k90 gold badges1986 silver badges3332 bronze badges
answered Jul 22 at 10:12
egregegreg
759k90 gold badges1986 silver badges3332 bronze badges
answered Jul 22 at 10:12
egregegreg
759k90 gold badges1986 silver badges3332 bronze badges
answered Jul 22 at 10:12
egregegreg
759k90 gold badges1986 silver badges3332 bronze badges
759k90 gold badges1986 silver badges3332 bronze badges
add a comment |
add a comment |
0
You can divide your equation into three part. Consider the following code:
beginalign
G_1 =& (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3nonumber\
×
fracGamma(a_1+a_2+a_3-fracD2)Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)
nonumber\
× _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, a_1+a_2+a_3-fracD2\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endalign
answered Jul 22 at 6:25
A DiyanatA Diyanat
4192 silver badges8 bronze badges
add a comment |
0
You can divide your equation into three part. Consider the following code:
beginalign
G_1 =& (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3nonumber\
×
fracGamma(a_1+a_2+a_3-fracD2)Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)
nonumber\
× _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, a_1+a_2+a_3-fracD2\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endalign
answered Jul 22 at 6:25
A DiyanatA Diyanat
4192 silver badges8 bronze badges
add a comment |
0
0
0
You can divide your equation into three part. Consider the following code:
beginalign
G_1 =& (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3nonumber\
×
fracGamma(a_1+a_2+a_3-fracD2)Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)
nonumber\
× _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, a_1+a_2+a_3-fracD2\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endalign
answered Jul 22 at 6:25
A DiyanatA Diyanat
4192 silver badges8 bronze badges
You can divide your equation into three part. Consider the following code:
beginalign
G_1 =& (-1)^-D/2(Q^2)^D/2-a_1-a_2-a_3nonumber\
×
fracGamma(a_1+a_2+a_3-fracD2)Gamma(fracD2-a_1-a_2)Gamma(fracD2-a_1-a_3)Gamma(a_2)Gamma(a_3)Gamma(D-a_1-a_2-a_3)
nonumber\
× _2F_1
left(
beginmatrix
1+a_1+a_2+a_3-D, a_1+a_2+a_3-fracD2\
hspace0.5cm1+a_1+a_3-fracD2
endmatrix
Bigg frac-M^2Q^2
right)
endalign
answered Jul 22 at 6:25
A DiyanatA Diyanat
4192 silver badges8 bronze badges
answered Jul 22 at 6:25
A DiyanatA Diyanat
4192 silver badges8 bronze badges
answered Jul 22 at 6:25
A DiyanatA Diyanat
4192 silver badges8 bronze badges
answered Jul 22 at 6:25
A DiyanatA Diyanat
4192 silver badges8 bronze badges
4192 silver badges8 bronze badges
add a comment |
add a comment |
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