- #1

- 17

- 15

- Homework Statement
- Let ##n_1## be the smallest positive integer ##n## for which the inequality ##(1+x)^n > 1 + nx + nx^2 ## is true for all ##x >0##. Compute ##n_1##, and prove that the inequality is true for all integers ## n \ge n_1##.

- Relevant Equations
- N/A

For calculating ##n_1## I had no problems as ##n_1 = 3##.

PF: Show ##(1+x)^n > 1 + nx + nx^2 ## is true for all ##x \ge 3##.

Let ##n = 3, (1+x)^3 > 3x^2 + 3x + 1 ##

##x^3 +3x^2 +3x + 3 > 3x^2 + 3x + 1 ## is true.

Assume ## n = k ## such that ##(1+x)^k > 1+kx + kx^2, k \ge 3## is true.

We want to show ## n = k + 1, n \ge 3 ## is true.

##(1+x)^{k + 1} > 1 + (k+1)x + (k+1)x^2##

##(1+x)^k(1+x) > kx^2 +x^2 + kx + x + 1##

##(1+x)^k(1+x) > (kx^2 +kx + 1) + x^2 + x ##

After a couple hours of algebra I came back to this spot above and think I am in the right place, because I noticed I have my assumption from the inductive step as part of the equation. However, I am unsure of how to show the product on the LHS is greater than the sum on the RHS.

Additionally, I worked to the following:

##(1+x)^k > \frac{kx^2 + kx + 1}{(1+x)} + x ##

Which makes me feel even better about the claim, but again under of how to show that LHS > RHS. I really fill like this is the spot, more than above as I have made the RHS of my inductive step smaller, while I am only add x to its total.

I went back through the previous problems to see if I proved something that would help solve this problem but I am not seeing anything.

Additionally, there is an issue that I believe just for new users. For MathJax to work you have to log out once you confirm your account and then log back in otherwise MathJax will not work. I spent way too long trying to get it to work and searching for help, there was a post which stated the fix. This should be added to the LaTex Guide!

Jonathan

PF: Show ##(1+x)^n > 1 + nx + nx^2 ## is true for all ##x \ge 3##.

Let ##n = 3, (1+x)^3 > 3x^2 + 3x + 1 ##

##x^3 +3x^2 +3x + 3 > 3x^2 + 3x + 1 ## is true.

Assume ## n = k ## such that ##(1+x)^k > 1+kx + kx^2, k \ge 3## is true.

We want to show ## n = k + 1, n \ge 3 ## is true.

##(1+x)^{k + 1} > 1 + (k+1)x + (k+1)x^2##

##(1+x)^k(1+x) > kx^2 +x^2 + kx + x + 1##

##(1+x)^k(1+x) > (kx^2 +kx + 1) + x^2 + x ##

After a couple hours of algebra I came back to this spot above and think I am in the right place, because I noticed I have my assumption from the inductive step as part of the equation. However, I am unsure of how to show the product on the LHS is greater than the sum on the RHS.

Additionally, I worked to the following:

##(1+x)^k > \frac{kx^2 + kx + 1}{(1+x)} + x ##

Which makes me feel even better about the claim, but again under of how to show that LHS > RHS. I really fill like this is the spot, more than above as I have made the RHS of my inductive step smaller, while I am only add x to its total.

I went back through the previous problems to see if I proved something that would help solve this problem but I am not seeing anything.

Additionally, there is an issue that I believe just for new users. For MathJax to work you have to log out once you confirm your account and then log back in otherwise MathJax will not work. I spent way too long trying to get it to work and searching for help, there was a post which stated the fix. This should be added to the LaTex Guide!

Jonathan