polynomials question (sorta) (1 Viewer)

[bleakarcher]

What expression would give the sum of the following taken four at a time: tan(+/-x), tan(+/-y) and tan(+/-z)

Thanks guys

 

[barbernator]

What expression would give the sum of the following taken four at a time: tan(+/-x), tan(+/-y) and tan(+/-z)

Thanks guys

are you implying that there are 6 roots to the polynomial, tan(x), tan(-x), tan(y), tan(-y), tan(z), tan(-z)?

 

[bleakarcher]

yep

 

[barbernator]

ok, well we know from trig that tan(-x)=-tan(x). So factorise the 6 roots into 3 quadratic roots, all of which will be difference of 2 squares. Then just selectively expand to get the coefficient of x^2 term and that will be your sum.

the solution will be.

sum of the roots 4 at a time = tan^2(x)tan^2(y)+tan^2(y)tan^2(x)+tan^2(x)tan^2(z)

 

[bleakarcher]

^ that is right but do you mind going through the process mathematically?

 

[barbernator]

^ that is right but do you mind going through the process mathematically?

<a href="http://www.codecogs.com/eqnedit.php?latex=a~monic~polynomial~with~roots~as~such\\ \\ (x-tan(x))(x-tan(-x))(x-tan(y))(x-tan(-y))(x-tan(z))(x-tan(-z))\\ \\ =(x-tan(x))(x@plus;tan(x))(x-tan(y))(x@plus;tan(y))(x-tan(z))(x@plus;tan(z))\\ \\ =(x^{2}-tan^{2}(x))(x^{2}-tan^{2}(y))(x^{2}-tan^{2}(z))\\ \\ shouldn't~have~used~x~as~both~variable~and~root~but~oh~well\\ \\ after~expanding~the~co-efficient~of~x^{2}~is~equal~to\\ \\ tan^{2}(x)tan^{2}(y)@plus;tan^{2}(x)tan^{2}(z)@plus;tan^{2}(y)tan^{2}(z)\\ \\ therefore\\ \\ \sum abcd=\frac{e}{a}\\ =tan^{2}(x)tan^{2}(y)@plus;tan^{2}(x)tan^{2}(z)@plus;tan^{2}(y)tan^{2}(z)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?a~monic~polynomial~with~roots~as~such\\ \\ (x-tan(x))(x-tan(-x))(x-tan(y))(x-tan(-y))(x-tan(z))(x-tan(-z))\\ \\ =(x-tan(x))(x+tan(x))(x-tan(y))(x+tan(y))(x-tan(z))(x+tan(z))\\ \\ =(x^{2}-tan^{2}(x))(x^{2}-tan^{2}(y))(x^{2}-tan^{2}(z))\\ \\ shouldn't~have~used~x~as~both~variable~and~root~but~oh~well\\ \\ after~expanding~the~co-efficient~of~x^{2}~is~equal~to\\ \\ tan^{2}(x)tan^{2}(y)+tan^{2}(x)tan^{2}(z)+tan^{2}(y)tan^{2}(z)\\ \\ therefore\\ \\ \sum abcd=\frac{e}{a}\\ =tan^{2}(x)tan^{2}(y)+tan^{2}(x)tan^{2}(z)+tan^{2}(y)tan^{2}(z)" title="a~monic~polynomial~with~roots~as~such\\ \\ (x-tan(x))(x-tan(-x))(x-tan(y))(x-tan(-y))(x-tan(z))(x-tan(-z))\\ \\ =(x-tan(x))(x+tan(x))(x-tan(y))(x+tan(y))(x-tan(z))(x+tan(z))\\ \\ =(x^{2}-tan^{2}(x))(x^{2}-tan^{2}(y))(x^{2}-tan^{2}(z))\\ \\ shouldn't~have~used~x~as~both~variable~and~root~but~oh~well\\ \\ after~expanding~the~co-efficient~of~x^{2}~is~equal~to\\ \\ tan^{2}(x)tan^{2}(y)+tan^{2}(x)tan^{2}(z)+tan^{2}(y)tan^{2}(z)\\ \\ therefore\\ \\ \sum abcd=\frac{e}{a}\\ =tan^{2}(x)tan^{2}(y)+tan^{2}(x)tan^{2}(z)+tan^{2}(y)tan^{2}(z)" /></a>

 

[Carrotsticks]

We have 6 terms and we want to put them into groups of 4 ie: there will be C(6,4)=15 terms.

HOWEVER, we note that the majority of these terms will CANCEL each other out due to the simple fact that tan(-x) = -tan(x).

Consider the 4-group tan(x)tan(y)tan(z)XXXXX

The XXXX must be either tan(-x), tan(-y) or tan(-z) (basic form of the Pigeonhole Principle).

Suppose we pick tan(-x) to be our XXXXX, our 4-group will be:

tan(x)tan(-x)tan(y)tan(z) = -tan^2(x)tan(y)tan(z)

HOWEVER, won't this cancel with the term:

tan(x)tan(-x)tan(-y)tan(z) = tan^2(x)tan(y)tan(z) ???

So we can quite clearly see that if we have 3 distinct terms, it can be cancelled by another term.

THEREFORE, the ONLY way we can have the 4-group sums is when we only have 2 distinct terms (we can't have 1 distinct term nor can we have 4).

And so the only way we have 2 distinct terms is if we pick tan(x)tan(-x)tan(y)tan(-y) etc etc, eventually leading to the final answer:

tan^2(x)tan^2(y)+tan^2(y)tan^2(x)+tan^2(x)tan^2(z)

And a tip barbernator, use $ text $ tags instead of the ~~~ that you use.

 

[bleakarcher]

thanks alooooooooooooooooooooooooooooooott guys. you're awesome