User:Jeandavid54/Identities at infinity bounds

Demonstration 1 : Extraordinary identity
We all know the reamarkable identity :


 * $$\displaystyle a^2-b^2 = (a - b)( a + b)

$$

We can generalize to the power of $$\displaystyle p $$ to give the following identity:


 * $$a^p - b^p = (a^{\frac p 2} - b^{\frac p 2})( a^{\frac p 2} + b^{\frac p 2})$$

Then we can see that the first term of the right member $$(a^{\frac p 2} - b^{\frac p 2})$$ can be factorized as followed.


 * $$(a^{\frac p 2} - b^{\frac p 2}) = (a^{\frac p 4} - b^{\frac p 4})( a^{\frac p 4} + b^{\frac p 4})$$

That gives :



a^p - b^p = (a^{\frac p 4} - b^{\frac p 4})( a^{\frac p 4} + b^{\frac p 4})( a^{\frac p 2} + b^{\frac p 2})$$

We can operate $$\displaystyle n$$ times until we get the next general formula :


 * $$a^p - b^p = (a^{\frac p {2^n}} - b^{\frac p {2^n}})(a^{\frac p {2^n}} + b^{\frac p {2^n}}) \cdots (a^{\frac p 8} + b^{\frac p 8})(

a^{\frac p 4} + b^{\frac p 4})( a^{\frac p 2} + b^{\frac p 2})$$

or again :


 * $$a^p - b^p = (a^{\frac p {2^n}} - b^{\frac p {2^n}}) \prod_{i=1}^n( a^{\frac p {2^i}} + b^{\frac p {2^i}})$$

It's interesting to see that $$(a^{\frac p {2^n}} - b^{\frac p {2^n}})$$ becomes zero when $$\displaystyle n$$ approaches infinity.

Indeed, we have :


 * $$ \lim_{n \to \infty}(a^{\frac p {2^n}} - b^{\frac p {2^n}})=0$$

So the left member of the equation is also zeroed.


 * $$ \displaystyle a^p - b^p=0$$

for all values of a, b et p.

Astonishing, isn't it ?

Demonstration 2 : Any number is equal to 1
Here is another example.

Any number $$\displaystyle a$$ can be written as a power $$\displaystyle n$$ of its nth-root, $$\displaystyle n$$ can be as great as you want..


 * $$\displaystyle a=(\sqrt[n] a)^n=\underbrace{(\sqrt[n] a).(\sqrt[n] a)\cdots(\sqrt[n] a) }_{n\;times}$$

In maths, we write nth-root of a number in 2 ways :


 * $$\displaystyle \sqrt[n] a$$

or as a power of an unit fraction,
 * $$\displaystyle a^{\frac 1 n}$$

So, we can write :


 * $$\displaystyle a=(a^{\frac 1 n})^n=\underbrace{(a^{\frac 1 n}).(a^{\frac 1 n})\cdots(a^{\frac 1 n}) }_{n\;times}$$

The limit of each factor $$ a^{\frac 1 n}$$, when n goes towards infinity, is equal to 1 :


 * $$ \lim_{n \to \infty}a^{\frac 1 n}=1$$

So:


 * $$\displaystyle a = 1$$

Any number is equal to 1.