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+5
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avatar+1491 

l'm left to figure out if equations are homogeneous or not.

 

l do know that a method would be to convert the function from f(x,y) to f(tx,ty). Afterwards l would symplify the equation and try to get my answer to be f(tx,ty) = t^n(f(x,y))

 

Of course it takes trig... to trigger me! (bum dum tss!)

 

Anyways, l have the equation:

 

f(x,y) = tan (x + y)

 

Anyone think they can help me UNDERSTAND?

 

(HSC got so triggered that he has been on this problem for 30 mins. Don't be HSC, and actually take notes instead of falling asleep.)

 Nov 30, 2016

Best Answer 

 #2
avatar+33653 
+15

The function is homogeneous if f(tx,ty) = f(x,y) where t is any number.

 

Do you think  tan(tx + ty) = tan(x + y) or not?

 Nov 30, 2016
 #1
avatar+118654 
+15

Sorry HSC I have no idea.  This is probably one for Heuraka or Alan ....

 

I just watch a Khan Academy  youtube clip on homogenous equations but it didn't help with your question :(

 Nov 30, 2016
 #2
avatar+33653 
+15
Best Answer

The function is homogeneous if f(tx,ty) = f(x,y) where t is any number.

 

Do you think  tan(tx + ty) = tan(x + y) or not?

Alan Nov 30, 2016
 #3
avatar+1084 
0

tan x+y = (tan x + tan y) / (1 - tan x tan y)

 Maby? Idk

 Nov 30, 2016
 #5
avatar+1084 
0

Sucks to be incorrect

Deathrises101  Dec 1, 2016
 #4
avatar+1491 
+10

Thanks guys... turns out the problem was not homogeneous at all.

 

You could not distribute "t" out of tan(tx + ty).

 

That being said it could not fufill f(tx,ty)= t^n(f(x,y))

 Nov 30, 2016
 #6
avatar
0

lmao wow 

 Dec 1, 2016

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