User:Jeandavid54

How transclusion works

test of transclusion

Atonishing identities

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.