Cambridge 4U Ex4.2 Q8 (1 Viewer)

[Pyrobooby]

 

[Bored Of Fail 2]

just equate the polynomials

bn z^n + b (n-1) z^(n-1) + ..... = c n z^n + c (n-1) z^(n-1) + ....

take everything over to one side and group the z^n, z^(n-1) etc

ie ( b n -c n) z^n + ( b (n-1) - c (n-1) ) z^(n-1) + ...... =0

now this can only be true if all the coefficents of z ^n , z^(n-1) , z^(n-2) etc... are all zero

therefore the result comes out

 

[Bored Of Fail 2]

the " are equal for more than n values of z" , comes from the fact that a nth degree equation , will have n solutions
( be they real or complex or a mixture )

when we equate the two polynomials we will get a polynomial equation of degree n , and by fundamental theorm of algebra this will have n solutions ( n values of z for which equality holds ) , the only way for their to be more than n values of z for which it is true is if they are infact the same polynomial

 

[Pyrobooby]

Hmm, wow okay thanks for the help.

 

[Bored Of Fail 2]

Hmm, wow okay thanks for the help.

its actually quite trivial, that wont be on the hsc paper , gauranteed

 

[cutemouse]

its actually quite trivial, that wont be on the hsc paper , gauranteed

Could be a one marker in the earlier questions. You never know.

 

[cookieeater1234]

the only way for their to be more than n values of z for which it is true is if they are infact the same polynomial

Why is this?

 

[khorne]

This is not a trivial problem, berks. This is an important theorem you need to know how to prove:

Consider them P(z) and Q(z)

Now consider P(z) - Q(z) as a polynomial which has more than n zeros. But the degree of P - Q is obviously less than or equal to n. So the only posibility is if P(z) - Q(z) = 0

I.e P(z) = Q(z), and then equate the coeffs.

 

[Aindan]

wait, isn't this just equating coefficients?

 

[Bored Of Fail]

wait, isn't this just equating coefficients?

u are a genius man

 

[Aindan]

It just seemed wierd to me, this seems too easy to be posted up. Maybe it was all the wording that confused pyro.