Inequalities question (1 Viewer)

[middlemarch]

Can anybody help me with this question:

if x>0, n a positive integer>1 and nx<1 show that
1+nx < (1+x)^n < [1 / (1-nx) ]

Thanks

 

[damo676767]

(1 + X)ⁿ = 1 + ⁿC₁X + ⁿC₂X² + ... + ⁿC_nXⁿ
= 1 + nX + ⁿC₂X² + ⁿC₂X³ + ... + ⁿC_nXⁿ
1 + nX = (1 + X)ⁿ - (ⁿC₂X² + ⁿC₂X³ + ... + ⁿC_nXⁿ)

1 + nX < (1 + X)ⁿ



(1 + X)^-n > 1 *as 1 + X > 1

0 < nX *as n and X are both positive
-nX < 0
1 - nX < 1

(1 + X)^-n > 1 - nX
(1 + X)ⁿ < 1/(1 - nX)

1 + nX < (1 + X)ⁿ < 1/(1 - nX)


do you want me to explain any of that, or does that make scence

 

[middlemarch]

Thanks damo...spoke too soon
nah that'll be fine

Middle

 

[robbo_145]

(1 + X)^-n > 1 *as 1 + X > 1

as 1 + X > 1
1/(1+X)ⁿ < 1
hence
(1 + X)^-n < 1

instead
1/(1-nX) - (1+X)ⁿ
= [ 1 - (1+X)ⁿ + nX(1+X)ⁿ] / ( 1-nX) * common denominator

using the first inequality (1+X)ⁿ > 1 + nX it becomes

> [1 - (1+nX) + nX(1 + nX)] / (1-nX)
> nX/1-nX
> 0 as 1-nX > 0 and nX >0

hence
1/(1-nX) - (1+X)ⁿ > 0
1/(1-nX) > (1+X)ⁿ

 

[justchillin]

You could also tackle it graphically...just drawing what the curves are like for values up to n... might be quicker than other methods?

 

[robbo_145]

maybe, but would you have to find 1st, 2nd derivitives, intercepts etc? as well as controlling n and x :s

 

[DRAGONZ]

> [1 - (1+nX) + nX(1 + nX)] / (1-nX)
> nX/1-nX
> 0 as 1-nX > 0 and nX >0

Is it just me, or should that line ACTUALLY read:
> (n^2)(x^2) / (1 - nx)

In the end, it doesn't matter to the solution ... just being pedantic?

 

[robbo_145]

yeah it probably should
ta for picking that up