Very easy! but very interesting

[fu^2]

This is my fist post in this room!
It's beautiful!!

so

proof that

$(1-a)^n<=1/(1+na)$, with $0

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Risposte

fu^2

19 set 2007, 22:19

good!

thanks!

good night

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Camillo

19 set 2007, 21:48

Summation is the translation for sommatoria.

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fu^2

19 set 2007, 21:33

a..
hem
i've not understood the link!

whatever the answer is:
yes!!!

ghghg this was the first example of induction's proof on inequality (i think that inequality is disequazioni in english ) that the teacher had proposed to us.

it's very interesting thing! it's a new world ! before i knew the induction only to proof the "sommatorie"...

(one question, how can i translate "sommatorie" in english?)

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amel3

19 set 2007, 19:47

"Camillo":I guess that for " beautiful climate " or excellent atmosphere Amel refers to your first lessons at Bicocca and to how you enjoyed !

Of course!

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Camillo

19 set 2007, 17:03

I guess that for " beautiful climate " or excellent atmosphere Amel refers to your first lessons at Bicocca and to how you enjoyed !

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fu^2

19 set 2007, 16:37

"amel":As usual, before beginning, I'm sorry for my bad English...


Is the post a result of the "beautiful climate" you found?

I don't know "beautiful climate"... what's it?

"zorn":
Note: the inequality is strict by n>=1

of course!
I'm stupid, i have forgotten this important strict!
sorry


whatever your proof are good!

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zorn1

19 set 2007, 11:49

It's easy by induction!

The inequality is trivial by $n=0$, and we suppose that's true for $1,2,...,n-1$.

Then:
$(1-a)^n=(1-a)(1-a)^(n-1)<=(1-a)*1/(1+(n-1)a)=(1-a+na-na)/(1+(n-1)a)=1-(na)/(1+(n-1)a)<1-(na)/(1+na)=1/(1+na)$

and this prove the inequality for all $n in NN$.


Note: the inequality is strict by n>=1

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amel3

19 set 2007, 11:18

As usual, before beginning, I'm sorry for my bad English...


Is the post a result of the "beautiful climate" you found?


It could be probably seen as a consequence of Bernoulli inequality, but I'm not sure...
However, I don't want to waste time, so I'll prove directly the assertion using mathematical induction.

$(1-a)^(n+1)<=1/(1+na)$,

$0


Proof.

The inequality is obviously true when $n=1$:
$(1-a)<=1/(1+a)$ $=>$ $1-a^2<=1$ ... true

Now we assume the result will be true for a fixed natural number $n$ and now we'll prove the result is valid also for $n+1$.
$(1-a)^(n+1)=(1-a)^n (1-a)<=1/(1+na) (1-a)=(1-a^2)/((1+na)(1+a))=(1-a^2)/(1+a+na+na^2)<=1/(1+a+na+na^2)<=1/(1+a+na)=1/(1+(n+1)a)$

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[fu^2]

good!

thanks!

good night

[Camillo]

Summation is the translation for sommatoria.

[fu^2]

a..
hem
i've not understood the link!

whatever the answer is:
yes!!!

ghghg this was the first example of induction's proof on inequality (i think that inequality is disequazioni in english ) that the teacher had proposed to us.

it's very interesting thing! it's a new world ! before i knew the induction only to proof the "sommatorie"...

(one question, how can i translate "sommatorie" in english?)

[amel3]

"Camillo":I guess that for " beautiful climate " or excellent atmosphere Amel refers to your first lessons at Bicocca and to how you enjoyed !

Of course!

[Camillo]

I guess that for " beautiful climate " or excellent atmosphere Amel refers to your first lessons at Bicocca and to how you enjoyed !

[fu^2]

"amel":As usual, before beginning, I'm sorry for my bad English...


Is the post a result of the "beautiful climate" you found?

I don't know "beautiful climate"... what's it?

"zorn":
Note: the inequality is strict by n>=1

of course!
I'm stupid, i have forgotten this important strict!
sorry


whatever your proof are good!

[zorn1]

It's easy by induction!

The inequality is trivial by $n=0$, and we suppose that's true for $1,2,...,n-1$.

Then:
$(1-a)^n=(1-a)(1-a)^(n-1)<=(1-a)*1/(1+(n-1)a)=(1-a+na-na)/(1+(n-1)a)=1-(na)/(1+(n-1)a)<1-(na)/(1+na)=1/(1+na)$

and this prove the inequality for all $n in NN$.


Note: the inequality is strict by n>=1

[amel3]

As usual, before beginning, I'm sorry for my bad English...


Is the post a result of the "beautiful climate" you found?


It could be probably seen as a consequence of Bernoulli inequality, but I'm not sure...
However, I don't want to waste time, so I'll prove directly the assertion using mathematical induction.

$(1-a)^(n+1)<=1/(1+na)$,

$0


Proof.

The inequality is obviously true when $n=1$:
$(1-a)<=1/(1+a)$ $=>$ $1-a^2<=1$ ... true

Now we assume the result will be true for a fixed natural number $n$ and now we'll prove the result is valid also for $n+1$.
$(1-a)^(n+1)=(1-a)^n (1-a)<=1/(1+na) (1-a)=(1-a^2)/((1+na)(1+a))=(1-a^2)/(1+a+na+na^2)<=1/(1+a+na+na^2)<=1/(1+a+na)=1/(1+(n+1)a)$