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Verifying Binomials 
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#1
Mar3014, 08:34 AM

P: 686

I was verifying that [tex]\\x^2y^2=(xy)(x+y) \\x^3y^3=(xy)(x^2+xy+y^2)[/tex] and I realized that can there is a formulation more general like the theorem binomial... my question is: exist a general theorem for sum or difference of terms^n ?



#2
Mar3014, 08:57 AM

PF Gold
P: 354

You can always proceed with the division of x^ny^n by xy.
Then you can guess and then demonstrate what the general solution is. For example: (x^11  y^11)/ (x  y) = x^10 + x^9 y + x^8 y^2 + x^7 y^3 + x^6 y^4 + x^5 y^5 + x^4 y^6 + x^3 y^7 + x^2 y^8 + x y^9 + y^10 Try to be specific and by considering multiple examples, you can often find the path to a generalization. Never try to be general too early. 


#3
Mar3014, 07:18 PM

HW Helper
P: 1,965

(x^{n}  y^{n}) = (x  y)(x^{n1} + x^{n2}y + x^{n3}y^{2} + ... + y^{n1})
is considered fairly elementary, but often useful. It is fairly easy to see it is true if you just multiply the x of the first bracket by the second bracket on one line and y from the first bracket by the second bracket on the second line you will see. A connection you should not fail to observe is that this gives you the answer to getting the sum of a geometric series which is 1 + x + x^{2} + x^{n1} (I have made the final term x^{n} for easy comparison, but you you'll be able to see what the sum is if the final term is x^{n}). The most useful of all applications of this is when x < 1 and n is infinite. 


#4
Mar3014, 08:25 PM

P: 686

Verifying Binomials
And which the formula for x^{n} + y^{n} and for x^{n} + y^{n} + z^{n}? 


#5
Mar3114, 12:55 AM

Mentor
P: 21,216

If there's a formula for x^{n} + y^{n} + z^{n} I'm not aware of it. 


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