(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
a)
 
a)
  
g(x)+h(x)=0
+
<math>g(x)+h(x)=0</math>
  
g(x) even h(x) odd
+
<math>g(x)</math> even <math>h(x)</math> odd
  
 
g is both even and odd
 
g is both even and odd
  
g(x)=g(-x)=-g(x)
+
<math>g(x)=g(-x)=-g(x)</math>
  
 
b)
 
b)
  
<math>f(x)=f$_{e}$(x)+f$_{0}$(x)</math>
+
<math>f(x)=f_{e}(x)+f_{0}(x)</math>
  
<math>f(-x)=f$_{e}$(-x)+f$_{0}$(-x)=f$_{e}$(x)-f$_{0}$(x)</math>
+
<math>f(-x)=f_{e}(-x)+f_{0}(-x)=f_{e}(x)-f_{0}(x)</math>
  
<math>solve for f$_{e}$(x) and f$_{0}$(x)</math>
+
solve for <math>f_{e}(x)</math> and <math>f_{0}(x)</math>
  
<math>f$_{e}$(x)= (f(x)+f(-x))/2</math>
+
<math>f_{e}(x)= (f(x)+f(-x))/2</math>
  
<math>f$_{0}$(x)= (f(x)-f(-x))/2</math>
+
<math>f_{0}(x)= (f(x)-f(-x))/2</math>

Latest revision as of 07:43, 6 October 2008

a)

$ g(x)+h(x)=0 $

$ g(x) $ even $ h(x) $ odd

g is both even and odd

$ g(x)=g(-x)=-g(x) $

b)

$ f(x)=f_{e}(x)+f_{0}(x) $

$ f(-x)=f_{e}(-x)+f_{0}(-x)=f_{e}(x)-f_{0}(x) $

solve for $ f_{e}(x) $ and $ f_{0}(x) $

$ f_{e}(x)= (f(x)+f(-x))/2 $

$ f_{0}(x)= (f(x)-f(-x))/2 $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett