Question

joe that thingy that we were talking about earlier
i proved it by induction
is there another way?

Answer 1

i'll post what i turned in for my test. one sec, let me find it.

Answer 2

im fixing to switch to my laptop
since my desktop isn't portable

Answer 3

Apparently I missed a nice problem?

Answer 4

Product and sum of the roots of a polynomial.

Answer 5

its part b) somewhere in the middle of page 2. The original problem was:
Given a polynomial:
\[a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0 = 0\]
Prove that the sum of the roots is:
\[-\frac{a_{n-1}}{a_n}\]
for the curious, part a was to prove the Rational Roots Theorem,

Answer 6

an inductive proof sounds really interesting...im gonna look into that later :)

Answer 7

we want to show this (where r1,r2,..,rn are roots of Pn)
\[P_n=(x-r_1)(x-r_2)(x-r_3) \cdots (x-r_n)\]
can be wriiten as
\[P_n=x^n-(r_1+r_2+r_3+ \cdots +r_n)x^{n-1}+b_{n-2}x^{n-2}-b_{n-3}x^{n-3}+ \cdots +(-1)^n(r_1 r_2 r_3 \cdots r_n)\]
where the b's are some coefficients that we dont really care about
\[P_1=x-r_1\] has root r1
\[P_2=(x-r_1)(x-r_2)=x^2-(r_1+r_2)x+r_1 r_2\]
anyways lets assume it is true for k
for even k first
then we have
\[P_k=(x-r_1)(x-r_2) \cdots (x-r_k)=x^n-(r_1+r_2+r_3+ \cdots r_k)x^{k-1}+b_{k-2}x^{k-2} \cdots +(r_1 r_2 \cdots r_k)\]
now we want to show it is true for k+1 where k is even
\[P_{k+1}=(x^k-(r_1+r_2+\cdots +r_k)x^{k-1}+gibberish+ \cdots + (r_1 r_2 \cdots r_k))(x-r_{k+1})\]
\[=x^{k+1}-(r_1+r_2+ \cdots + r_k)x^k+(gibberish)x+ \cdots (r_1 r_2 \cdots r_k)x\]\[-r_{k+1}x^{k}+r_{k+1}(r_1+r_2+ \cdots + r_k)x^{k-1}+(gibberish)r_{k+1}+ \cdots -(r_1 r_2 \cdots r_k r_{k+1})\]
now all we have to do is put it tgether

Answer 8

so we can do this for odd k too

Answer 9

right right. nice :)

Answer 10

At first i had some points taken off my proof because I didnt show that a polynomial with roots r_1......r_n has a unique representation.

Answer 11

thats madness lol

Answer 12

i ended up proving it to him in his office when we were talking about the test so he gave them back <.< lolol I wouldnt have run into that problem if i had done it this way.

Answer 13

this way isn't long if you are writing it down on paper
it took me forever to use that equation thing

Answer 14

i was waiting in suspense!! lol

Answer 15

and after all that i couldn't brink my else to put my like terms togeher but you guys know how to do that

Answer 16

bring*

Answer 17

very nice indeed, i'll have to work it out myself so i dont forget it.

Answer 18

induction when possible to use i think is the easiest method

Answer 19

I know of a problem thats not better with induction >.< i found that out the hard way lolol

Answer 20

show me

Answer 21

prove that:
\[\left(\begin{matrix}n \\ 0\end{matrix}\right)+\left(\begin{matrix}n \\ 1\end{matrix}\right)+\left(\begin{matrix}n \\ 2\end{matrix}\right)+\cdots \left(\begin{matrix}n \\ n\end{matrix}\right) = 2^n\]

Answer 22

oh
it may not be easy to use induction here lol

Answer 23

but i might try to try it

Answer 24

binomial theorem

Answer 25

brb

Answer 26

its possible with induction using the fact that:
deleted since you are going to try it.
but there is a much much much much much nicer way to do it.

Answer 27

which zarkon spoiled!!!

Answer 28

lol...sorry

Answer 29

lolol, np, the Induction proof is important to know as well. its helped me with other similar problems.

Answer 30

I just wrote out the induction proof and it is really not that bad

Answer 31

just need to use

\[{n+1\choose k}={n\choose k-1}+{n\choose k}\]

Answer 32

i was just trying to use
\[\left(\begin{matrix}n \\ k\end{matrix}\right)=\frac{n!}{k!(n-k)!}\]
i should probably just use what you wrote zarkon lol

Answer 33

I think it would make life easier :)

Answer 34

yes indeed

Answer 35

the inductive proof isnt bad, but when you see the binomial theorem proof your just like ".........>.>"