Let k∈N, Show that there is i∈N s.t (1−(1/k))^i − (1−(2/k))^i ≥ 1/4

[idobido]

Homework Statement

not done

Relevant Equations

Bernoulli's inequality
Newton's Binomial
Taylor expansion

let $k \in\mathbb{N},$ Show that there is $i\in\mathbb{N}$s.t $\ \left(1-\frac{1}{k}\right)^{i}-\left(1-\frac{2}{k}\right)^{i}\geq \frac{1}{4}$

I tried to use Bernoulli's inequality and related inequality for the left and right expression but i the expression smaller than 1/4 for any i or k.
I also tried Newton's Binomial Theorem, Taylor expansion with no success.

 

[Mark44]

Homework Statement: not done
Relevant Equations: Bernoulli's inequality
Newton's Binomial
Taylor expansion

let $k \in\mathbb{N},$ Show that there is $i\in\mathbb{N}$s.t $\ \left(1-\frac{1}{k}\right)^{i}-\left(1-\frac{2}{k}\right)^{i}\geq \frac{1}{4}$

I tried to use Bernoulli's inequality and related inequality for the left and right expression but i the expression smaller than 1/4 for any i or k.
I also tried Newton's Binomial Theorem, Taylor expansion with no success.

Since you haven't had any success with the methods you've tried so far, it might be helpful to do some paper and pencil exploration. Start by setting k to 2, and seeing what happens for i = 1, i = 2, and so on.
Then set k = 3, and see what you get for i = 1, i = 2, and so on. You might be able to discover some pattern that will give you insight into a proof.

 

[idobido]

for k = 1 or 2 the solution is trivial, i tried online calculator and it seems the result is about i = k*ln(2), but i can not show exactly how, the expression after setting i = k*ln(2), and his first derivative is very complicated to deal with.
https://math.stackexchange.com/ques...t1-frac1?noredirect=1#comment10074764_4748040

 

[WWGD]

Maybe plot the function :
$f(x,y):=(1-\frac {1}{x})^y-(1- \frac {2}{x})^y-1/4$?

 

[idobido]

already did, it seems the f(k, k*ln2) >=0, i.e. k*ln2 is the solution, but this needed to be shown analytically,
and k*ln2 is not a natural number.

 

[pasmith]

Let f_i(x) = (1 - x)^i - (1 -2x)^i with x \in [0,1]. This is continuous with f_i(0) = 0 for all i \geq 1, so the problem comes for large k.

 

[WWGD]

And you have continuity + compactness, which guarantee extrema.

 

[idobido]

ok, only need to show now that

$\alpha\left(k\right)\ :=\ \left(1-\tfrac{1}{k}\right)^{\ln\left(2\right)k}-\left(1-\tfrac{2}{k}\right)^{\ln\left(2\right)k}$

is decreasing for k>=3, i tried to show the first derivative non-positive but the expression i got is really complicated.

 

[haruspex]

what’s an approximation for $(1-1/x)^i$ for large x?

 

[idobido]

what’s an approximation for $(1-1/x)^i$ for large x?

1

 

[haruspex]

ok, only need to show now that

$\alpha\left(k\right)\ :=\ \left(1-\tfrac{1}{k}\right)^{\ln\left(2\right)k}-\left(1-\tfrac{2}{k}\right)^{\ln\left(2\right)k}$

is decreasing for k>=3, i tried to show the first derivative non-positive but the expression i got is really complicated.

I note that the original problem statement requires $i\in\mathbb{N}$, so you cannot set $i=k\ln(2)$.
The approximation I was thinking of was $(1-1/x)^i\approx e^{-i/x}$. Using that, it is easy to show that $i\approx k\ln(2)$ near equality of the target inequality. It is not clear from the thread whether you already deduced that or merely observed it numerically.

 

[idobido]

I note that the original problem statement requires $i\in\mathbb{N}$, so you cannot set $i=k\ln(2)$.
The approximation I was thinking of was $(1-1/x)^i\approx e^{-i/x}$. Using that, it is easy to show that $i\approx k\ln(2)$ near equality of the target inequality. It is not clear from the thread whether you already deduced that or merely observed it numerically.

i observed i = k*ln(2) numerically, i cant use approximation, i need to show that exactly. trying to follow te comments here:
https://math.stackexchange.com/ques...t1-frac1?noredirect=1#comment10074764_4748040

i got stuck at showing that $\alpha\left(k\right)\ :=\ \left(1-\tfrac{1}{k}\right)^{\ln\left(2\right)k}-\left(1-\tfrac{2}{k}\right)^{\ln\left(2\right)k}$ is decreasing for k>=3

 

[haruspex]

i cant use approximation

No, I know you can't use that simple approximation to answer the question. My point is that you can use it to show that the appropriate value of $i$ must be close to $k\ln(2)$.
In principle, you might be able to use a more precise version of the approximation to solve the problem, though.
Note that $k\ln(2)=k/2+k/12+k/30+…$.

I note this comment at the cited link "analyzing the relevant first derivatives/a well-known limit." Maybe that limit is $\lim _{x\rightarrow\infty}(1-1/x)^i$.