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Why do I get an error when I try to begin an equation?

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0

This is what I have typed so far in latex:

documentclass[10pt,letter]article


usepackageamsmath
usepackageamsthm
usepackagemathtools


usepackagegraphicx


usepackagesetspace
usepackage[left=1.5in, right=1.5in, top=0.5in]geometry


onehalfspacing



begindocument



titleHomework Chapter 5

author




maketitle 



section*Section 5.2: 4,5,7

paragraph4.
Let $alpha$ be a complex number. Show that if $(1+z)^alpha$ is taken as $e^alpha operatornameLog(1+z)$, then for $|z|< 1$\

$(1+z)^alpha = 1 + displaystylefracalpha1z + fracalpha(alpha -1)1cdot2z^2 + fracalpha(alpha-1)(alpha-2)1cdot2cdot3z^3 + cdots$\

In general, 

beginequation

displaystylefracd^jdz^j(1+z)^alpha=fracalpha !(1+z)^alpha - j(alpha - j)!

endequation

I get an error with the following code:

beginequation

displaystylefracd^jdz^j(1+z)^alpha=fracalpha !(1+z)^alpha - j(alpha - j)!

endequation

I was wondering if someone could clarify why this happens?

equations

share

asked 3 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You can't have a blank line inside an equation environment.
  
  – Phelype Oleinik
  1 min ago
  

  
  
  
  

add a comment | 

0

This is what I have typed so far in latex:

documentclass[10pt,letter]article


usepackageamsmath
usepackageamsthm
usepackagemathtools


usepackagegraphicx


usepackagesetspace
usepackage[left=1.5in, right=1.5in, top=0.5in]geometry


onehalfspacing



begindocument



titleHomework Chapter 5

author




maketitle 



section*Section 5.2: 4,5,7

paragraph4.
Let $alpha$ be a complex number. Show that if $(1+z)^alpha$ is taken as $e^alpha operatornameLog(1+z)$, then for $|z|< 1$\

$(1+z)^alpha = 1 + displaystylefracalpha1z + fracalpha(alpha -1)1cdot2z^2 + fracalpha(alpha-1)(alpha-2)1cdot2cdot3z^3 + cdots$\

In general, 

beginequation

displaystylefracd^jdz^j(1+z)^alpha=fracalpha !(1+z)^alpha - j(alpha - j)!

endequation

I get an error with the following code:

beginequation

displaystylefracd^jdz^j(1+z)^alpha=fracalpha !(1+z)^alpha - j(alpha - j)!

endequation

I was wondering if someone could clarify why this happens?

equations

share

asked 3 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You can't have a blank line inside an equation environment.
  
  – Phelype Oleinik
  1 min ago
  

  
  
  
  

add a comment | 

This is what I have typed so far in latex:

documentclass[10pt,letter]article


usepackageamsmath
usepackageamsthm
usepackagemathtools


usepackagegraphicx


usepackagesetspace
usepackage[left=1.5in, right=1.5in, top=0.5in]geometry


onehalfspacing



begindocument



titleHomework Chapter 5

author




maketitle 



section*Section 5.2: 4,5,7

paragraph4.
Let $alpha$ be a complex number. Show that if $(1+z)^alpha$ is taken as $e^alpha operatornameLog(1+z)$, then for $|z|< 1$\

$(1+z)^alpha = 1 + displaystylefracalpha1z + fracalpha(alpha -1)1cdot2z^2 + fracalpha(alpha-1)(alpha-2)1cdot2cdot3z^3 + cdots$\

In general, 

beginequation

displaystylefracd^jdz^j(1+z)^alpha=fracalpha !(1+z)^alpha - j(alpha - j)!

endequation

I get an error with the following code:

beginequation

displaystylefracd^jdz^j(1+z)^alpha=fracalpha !(1+z)^alpha - j(alpha - j)!

endequation

I was wondering if someone could clarify why this happens?

equations

share

asked 3 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

documentclass[10pt,letter]article


usepackageamsmath
usepackageamsthm
usepackagemathtools


usepackagegraphicx


usepackagesetspace
usepackage[left=1.5in, right=1.5in, top=0.5in]geometry


onehalfspacing



begindocument



titleHomework Chapter 5

author




maketitle 



section*Section 5.2: 4,5,7

paragraph4.
Let $alpha$ be a complex number. Show that if $(1+z)^alpha$ is taken as $e^alpha operatornameLog(1+z)$, then for $|z|< 1$\

$(1+z)^alpha = 1 + displaystylefracalpha1z + fracalpha(alpha -1)1cdot2z^2 + fracalpha(alpha-1)(alpha-2)1cdot2cdot3z^3 + cdots$\

In general, 

beginequation

displaystylefracd^jdz^j(1+z)^alpha=fracalpha !(1+z)^alpha - j(alpha - j)!

endequation

beginequation

displaystylefracd^jdz^j(1+z)^alpha=fracalpha !(1+z)^alpha - j(alpha - j)!

endequation

share

asked 3 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share

asked 3 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 mins ago

K.MK.M

1305

New contributor

K.M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 mins ago

K.MK.M

1305