1. ## Prove

Hi,

I'm stuck on the following problem:

Let $h \ \epsilon \ [0,1]$
Show that for every $n \ \epsilon \ N$

$(1 + h)^n \leq 1 + (2^n -1)h$

is true..

I tried with math induction:

$(1) \ \ \ n = 1$
$1 + h \leq 1 + (2 - 1)h$
$1 + h \leq 1 + h \ \ \ \surd$

$(2) \ \ \ n \Rightarrow n + 1$

$(1 + h)^{h+1} \leq 1 + (2^{n+1} - 1)h$

$(1+h) (1+h)^n \leq 1 + (2 \cdot 2^n - 1)h$

$(1+h) (1+h)^n \leq 1 + 2h \cdot 2^n - h$

$(1+h) (1+h)^n \leq 1 - h + 2(h \cdot 2^{n-1})$

now i'm lost. don't know how to continue.

By the way, how do I write the "is element of" symbol? did i use the correct one above?
and is there any way i could skip the [tex][/ math] in every line?

2. Originally Posted by metlx
Hi,

I'm stuck on the following problem:

Let $h \ \epsilon \ [0,1]$
Show that for every $n \ \epsilon \ N$

$(1 + h)^n \leq 1 + (2^n -1)h$

is true..

I tried with math induction:

$(1) \ \ \ n = 1$
$1 + h \leq 1 + (2 - 1)h$
$1 + h \leq 1 + h \ \ \ \surd$

$(2) \ \ \ n \Rightarrow n + 1$

$(1 + h)^{h+1} \leq 1 + (2^{n+1} - 1)h$

$(1+h) (1+h)^n \leq 1 + (2 \cdot 2^n - 1)h$

$(1+h) (1+h)^n \leq 1 + 2h \cdot 2^n - h$

$(1+h) (1+h)^n \leq 1 - h + 2(h \cdot 2^{n-1})$

now i'm lost. don't know how to continue.

By the way, how do I write the "is element of" symbol? did i use the correct one above?
and is there any way i could skip the [tex][/ math] in every line?
So in the inductive step, we assume $(1 + h)^k \leq 1 + (2^k -1)h$ for some $k$.

Then $(1 + h)^{k+1} = (1 + h)^k (1+h) \leq (1 + (2^k -1)h)(1+h)$ and $(1 + (2^k -1)h)(1+h) = 1 + 2^kh + (2^k - 1)h^2$

Now $1 + 2^kh + (2^k - 1)h^2 \leq 1 + 2^kh + (2^k - 1)h = 1 + 2^kh + 2^kh - h = 1 + (2^{k+1 } - 1)h$. Where this was true $(2^k - 1)h^2 \leq (2^k - 1)h$ since $h \in [0,1]$

Therefore $(1+h)^{k+1} \leq 1 + (2^{k+1 } - 1)h$, as required.

Oh and also, the element of symbol in latex is simply " \in ". So I used that above, you can click and see. And I don't know a shortcut to get around the constant math tags, sorry.