Hard Inequalities Help (1 Viewer)

[king.rafa]

Prove the following
a)1+x+x^2>0
b)1+x+x^2+x^3+x^4>0
c) For all positive integers n, 1+x+x^2+x^3+x^4+...x^(2n-1)+x^(2n)>0
d)x=-1 is the only zero of 1+x+x^2+x^3+x^4+...x^(2n-1)

I do not have a clue as to how to start any of these questions. Can anyone please help me with the solutions? I'm just going through the extension questions of the Cambridge 3 unit book and these ones are killers

 

[lolokay]

a)1+x+x^2>0
x^2+x+1/4>-3/4
(x+1/2)^2>-3/4
therefore true

b)1+x+x^2+x^3+x^4>0
x(x^3+1) + (x^3+1)>-x^2
(x+1)^2(x^2-x+1)>-x^2
true if
x^2 - x + 1 > 0
(x - 1/2)^2 > -3/4
therefore true

too tired to do the other 2

 

[minijumbuk]

a)1+x+x^2>0
x^2+x+1/4>-3/4
(x+1/2)^2>-3/4
therefore true

b)1+x+x^2+x^3+x^4>0
x(x^3+1) + (x^3+1)>-x^2
(x+1)^2(x^2-x+1)>-x^2
true if
x^2 - x + 1 > 0
(x - 1/2)^2 > -3/4
therefore true

too tired to do the other 2

Neither of a) nor b) make any sense. This question was either poorly worded, or some hints were not shown. How can anyone know what the value of x is? I mean, sure you can solve it and then come up with a square root of negative, but that still only shows the value of x, not actually proving the equation given.

 

[tommykins]

回复: Re: Hard Inequalities Help

Neither of a) nor b) make any sense. This question was either poorly worded, or some hints were not shown. How can anyone know what the value of x is? I mean, sure you can solve it and then come up with a square root of negative, but that still only shows the value of x, not actually proving the equation given.

It makes perfect sense, inequalities assume that x is a real number, and any number squared is positive. The first one proves it by saying the square of x+1/2 is bigger than -3/4, which is a negative number, which makes it true.

b)1+x+x^2+x^3+x^4>0
x(x^3+1) + (x^3+1)>-x^2 -> makes sense
(x+1)^2(x^2-x+1)>-x^2 -> negative number as -x^2 is always negative
true if
x^2 - x + 1 > 0since we know (x+1)^2 > 0 , we only need to cover the other bracket
(x - 1/2)^2 > -3/4 same technique as a
therefore true

They make sense and are correct.

 

[king.rafa]

you dont make any sense



if x > 0, clearly true
if x = 0, LHS = 1, therefore true
1+x+x^2+x^3+x^4+...x^(2n-1)+x^(2n) = [1 - x^(2n + 1)] / (1 - x)
if x < 0, numerator is positive as (2n + 1) is odd for all positive integers n, and denominator is positive, therefore true



y = 1+x+x^2+x^3+x^4+...x^(2n-1)
= [1 - x^(2n)] / (1 - x)

for all positive integers n:
when -1 < x <1
1 - x^(2n) > 0
(1 - x) > 0
.'. y > 0
when x = 1, y = 1 + 1 + 1 +...+ 1 > 0
when x > 1
1 - x^(2n) < 0
(1 - x) < 0
.'. y > 0
when x < -1
1 - x^(2n) < 0
(1 - x) > 0
.'. y < 0

.'. y>0 for x>-1, and y<0 for x<-1, since y is continuous, x=-1 is the only zero

for your proof, when you use the difference of the powers, and come up with [1 - x^(2n)] / (1 - x), do you have to worry about discontinuities, or because there weren't any in the original, ie. y = 1+x+x^2+x^3+x^4+...x^(2n-1), do you just disregard it? like cant you just make x-1=(x-1)(1-x) / (1-x) and have a discontinuity at x=1?

 

[lolokay]

for your proof, when you use the difference of the powers, and come up with [1 - x^(2n)] / (1 - x), do you have to worry about discontinuities, or because there weren't any in the original, ie. y = 1+x+x^2+x^3+x^4+...x^(2n-1), do you just disregard it? like cant you just make x-1=(x-1)(1-x) / (1-x) and have a discontinuity at x=1?

x-1=(x-1)(1-x) / (1-x) if x=/=1, so they're not the exact same thing. Similarly, [1 - x^(2n)] / (1 - x) is the not the exact same expression as the one that the inequality gives, as it is not true for x=1. So for x=1 you have to use the original expression.