Rational roots of a polynomial (1 Viewer)

[undalay]

So i would have to prove it on a HSC paper. Ok thanks.

they can have heaps of lead on parts specifying any method; memorising one is not recommended

 

[tommykins]

回复: Re: Rational roots of a polynomial

the method they gave was integrating t.e^-t or something a rather.

 

[vds700]

Re: 回复: Re: Rational roots of a polynomial

the method they gave was integrating t.e^-t or something a rather.

whats this for?

 

[midifile]

no im saying for that equation is cant be fractions (show me a monic polynomial with integer coefficients that has fraction roots) .
Obviously fractions are rational.

Any my post wasn't meant to be a proof, rather its suppose to stimulate thoughts/intuition.
proofs are lameee.

Yea i get what you mean. I thought you were saying that rational rots couldnt be fractions [i didnt actually read the question].

 

[tommykins]

Re: 回复: Re: Rational roots of a polynomial

whats this for?

Proof of e's irrationality.

 

[vds700]

Re: 回复: Re: Rational roots of a polynomial

Proof of e's irrationality.

ah i see

 

[shaon0]

Re: 回复: Re: Rational roots of a polynomial

Proof of e's irrationality.

I saw a proof that used infintie series by Joseph Fourier. But it was quite short.
Are we allowed to produce a proof like that for 4unit HSC?
Question: In integration if you get a value pi to substitute into cosx or sinx do we change pi into 180 degrees?

 

[vds700]

Re: 回复: Re: Rational roots of a polynomial

I saw a proof that used infintie series by Joseph Fourier. But it was quite short.
Are we allowed to produce a proof like that for 4unit HSC?
Question: In integration if you get a value pi to substitute into cosx or sinx do we change pi into 180 degrees?

pi is 180 degrees, but its best to leave it in radians.

Its pretty easy, all u need to know is

sin(pi) = 0
cos(pi) = -1

 

[Iruka]

I doubt you would be asked to use Fourier series to prove the irrationality of pi.

I think the irrationality proof for e that you are talking about used the Taylor series expansion for the exponential function.

When doing calculus, you should always assume that angles are measured in radians.

 

[AMorris]

no im saying for that equation is cant be fractions (show me a monic polynomial with integer coefficients that has fraction roots) .
Obviously fractions are rational.

Any my post wasn't meant to be a proof, rather its suppose to stimulate thoughts/intuition.
proofs are lameee.

but why is it obvious that a monic polynomial can't have a non-integer rational root? The only way I can think of proving that is by my method where you show that q has to be +-1?

 

[shaon0]

Re: 回复: Re: Rational roots of a polynomial

pi is 180 degrees, but its best to leave it in radians.

Its pretty easy, all u need to know is

sin(pi) = 0
cos(pi) = -1

then, why does sin(pi)=0.0548.... and cos(pi)=0.9985.... instead of sin(pi)=sin(180)=0 and cos(pi)=cos(180)=1?

 

[shaon0]

I doubt you would be asked to use Fourier series to prove the irrationality of pi.

I think the irrationality proof for e that you are talking about used the Taylor series expansion for the exponential function.

When doing calculus, you should always assume that angles are measured in radians.

Sorry for my mistake.

 

[Templar]

then, why does sin(pi)=0.0548.... and cos(pi)=0.9985.... instead of sin(pi)=sin(180)=0 and cos(pi)=cos(180)=1?

Changing your calculator to radian mode would give the required answer.

but why is it obvious that a monic polynomial can't have a non-integer rational root? The only way I can think of proving that is by my method where you show that q has to be +-1?

If (x-p/q) is a solution you could deduce that (qx-p) is a factor of the polynomial.

 

[shaon0]

Changing your calculator to radian mode would give the required answer.



If (x-p/q) is a solution you could deduce that (qx-p) is a factor of the polynomial.

thanks for the advice.