poly question (1 Viewer)

[crono]

Can anyone do exercise 4.4 ,10(a) from cambridge?

Solve the equation 16x^5 - 20x^3 + 5x - 1 = 0 and deduce the exact values of cos 2pi/5 and cos 4pi/5.

 

[Slidey]

Let z=cos@+i.sin@
z^5=(cos@+isin@)^5=cos5@+isin5@
Let cos@=c, sin@=s
z^5=(cos@+isin@)^5=c^5 + 5c^4.i.s + 10c^3.i^2.s^2 + 10c^2.i^3.s^3 + 5c.i^4.s^4 + s^5
Re(z^5)=cos5@=c^5 - 10c^3.s^2 + 5c.s^4
That is:
cos5@=cos^5(@) - 10cos^3(@)sin^2(@) + 5cos@sin^4(@)
cos5@=cos^5(@) - 10cos^3(@)(1-cos^2(@)) + 5cos@(1-cos^2(@))^2
cos5@=cos^5(@) - 10cos^3(@) + 10cos^5(@) + 5cos@-10cos^3(@)+5cos^5(@))
cos5@=16cos^5(@)-20cos^3(@)+5cos@

For 16x^5 - 20x^3 + 5x - 1 = 0,
16x^5 - 20x^3 + 5x = 1
Let x=cos@:
16cos^5(@)-20cos^3(@)+5cos@=1
.'. cos5@=1
So 5@=pi, 2pi, 3pi, 4pi, 5pi ...
@=pi/5, 2pi/5, 3pi/5, 4pi/5, 5pi/5
.'. roots of
16x^5 - 20x^3 + 5x - 1 = 0 are:
cos(pi/5), cos(2pi/5), cos(3pi/5), cos(4pi/5), cos(pi)

Go from there.

 

[dawma88]

good work slide rule ....for a LOSER

and crono u have to practise more if u wanna get get top 20 in the state (like me)

 

[martin]

.'. cos5@=1
So 5@=pi, 2pi, 3pi, 4pi, 5pi ...

Actually, if cos5@ = 1
then 5@=0, 2Pi, 4Pi, 6Pi, 8Pi, 10Pi.... (Note that cosPi = -1).
So then @=0, 2Pi/5, 4Pi/5, ....

But the hard part of this question is solving the original polynomial (16x^5 - 20x^3 + 5x - 1 = 0) in terms of surds. (x - 1) is a factor but on dividing out I get 16x^4 + 16x^3 - 4x^2 - 4x + 1 which I don't really know what to do with.

 

[crono]

still stuck on the exact value bit

 

[KFunk]

working from what slide rule had: cos5@=1
so you know that @ = 0, ±2π/5, ±4π/5 are solutions

hence cos(0), cos(±2π/5), cos(±4π/5) are solutions to 16x⁵ - 20x³ + 5x - 1 = 0

using the product of roots 5 at a time you know that:

cos(0)cos(2π/5)cos(-2π/5)cos(4π/5)cos(-4π/5) = 1/16

cos²(2π/5)cos²(4π/5) = 1/16

Get some simultaneous stuff happening with summing roots and you should get some exact values.

 

[martin]

cos²(2π/5)cos²(4π/5) = 1/16

Get some simultaneous stuff happening with summing roots and you should get some exact values.

let the roots be 1,a,a,b,b then

a²b² = 1/16 (product of roots) and
2a + 2b + 1 = 0 (sum of roots) so
b=-1/2-a

we know that one of a and b is +ve and one -ve because they are cos(2Pi/5) and cos(4Pi/5) so ab is -ve.

therefore ab = -1/4

so a(-1/2-a) = -1/4

solving this as a quadratic
a=(-1 ± √ 5) / 4
so b = the same with a minus plus instead of a plus minus

so we see that
cos(2Pi/5) = (-1 + √ 5) / 4 and
cos(4Pi/5) = (-1 - √ 5) / 4

Similarly if you solve 16x^5 - 20x^3 + 5x + 1 = 0 then you can get exact values for cos(Pi/5) and cos(3Pi/5).

 

[Slidey]

"Similarly if you solve 16x^5 - 20x^3 + 5x + 1 = 0 then you can get exact values for cos(Pi/5) and cos(3Pi/5)."

cos(3Pi/5) = (1 - √ 5) / 4
cos(Pi/5) = (1 + √ 5) / 4