[ 2 posts ] 
 Exercise [04.06] 
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Joined: 11 May 2008, 20:07
Posts: 11
Location: france
Post Exercise [04.06]
Let's say we have the complex polynomial :

P_1=a_0+a_1z+a_2z^2+...+a_nz^n

If we name the solutions of P_1=0 , b_1,b_2,...b_n, we can rewrite it as :

P_1=a_n(z-b_1)(z-b_2)...(z-b_n)

With

a_0=a_nb_1b_2...b_n

a_1=a_n(\sum_{k-1=1}^n \frac{b_1b_2...b_n}{b_k})

a_2=a_n(\sum_{k_1,k_2=1;k_1<k_2}^n\frac{b_1b_2...b_n}{b_{k_1}b_{k_2}})

...

a_{n-1}=a_n(\sum_{k_1,k_2,...k_{n-2}=1;k_1<k_2<...<k_{n-2}}^n\frac{b_1b_2...b_n}{b_{k_1}b_{k_2}...b_{k_{n-2}}})

a_n=a_n

As we can see (if you really pay attention because it is rather small!) every a_k is a "function" of every b_k (except a_n). So as Mr penrose says in this exercise, if we divide P_n with z-b_k every a_k will change (except once more a_n) and the polynomial will have a different fully developed formula.


13 May 2008, 16:25

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Exercise [4.6]
Alternative prooof:
P(z)=a_0+a_{1}z+a_{2}z^2+...+a_{n}z^n
From the discussion in section 4.2, we can infer that P(z) will have at least one complex solution.
Call it b_1
Now we have to prove that z-b_1 is a factor of P(z).
Assume (z-b_1)is not a factor of P(z).
From the division algorithm we have,
P(z)=P_{1}(z)\ (z-b_1)+R(z)where R(z) is the remainder which must be non zero for all z because of our assumption.
Setting z=b_1 in this identity
P(b_1)=P_{1}(b_1)(b_1-b_1)+R(b_1)
orR(b_1)=0(b_1is a zero of P(z))which is a contradiction.
Hence,(z-b_1)is a factor of P(z) which means
P(z)=P_{1}(z)\ (z-b_1)
Using the same argument we have,
P_{1}(z)=P_{2}(z)\ (z-b_2)
P_{2}(z)=P_{3}(z)\ (z-b_3)
.
.
.
P_{k-1}(z)=P_{k}(z)\ (z-b_k)where b_kis the solution ofP_{k}(z)
.
P_{n-1}(z)=P_{n}(z)\ (z-b_n)
Now
P(z)=P_{1}(z)\ (z-b_1)=P_{2}(z)\ (z-b_1)(z-b_2)=...=P_n(z)\ (z-b_1)(z-b_2)...(z-b_n)
Now little thought revealsp_{k}(z)is apolynomial of degree n-k
Therefore P_{n}(z)is a constant as it has degree n-n=0
On multiplying out the factors the coefficient of z^n is P_{n}(z).
Equating this coefficient with the one in original expression for P(z) we have
P_{n}(z)=a_n
Hence,
P(z)=a_{n}(z-b_1)(z-b_2)(z-b_3)...(z-b_n)


16 May 2008, 12:00
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