# User:Chela5808/My sandbox Sam2 HandStyled

 Before Styling Sample 1 Sample 2 Sample 3 Sample 4 Sample 5 Sample 6 After Styling Sample 1 Sample 2 Sample 3 Sample 4 Sample 5 Sample 6

Styling by hand Sample 2

• Text: Wiki Markup;
• Formulae: $\mathrm{L\!\!^{{}_{A}} \!\!\!\!\!\;\; T\!_E} \! X} \$ some took directly from corresponding .tex file.

• Hand styling time : 45 min

The function deﬁned in (??) has amazing properties. Let’s multiply two such series together, using two diﬀerent values of $x\,$ (call them $x=p\,$ in one series and $x=q\,$ in the other):

 \begin{align} f(p)f(q) & = \left (1+p+\frac{p^2}{2!}+... \right)\left (1+q+\frac{q^2}{2!}+... \right) \\ & = 1+(p+q)+ \left (\frac{p^2}{2!}+ pq+ \frac{q^2}{2!}\right)+... \\ & = 1+(p+q)+ \frac{(p+q)^2}{2!}+...\;, \\ \end{align} (1)

– including terms only up to the ‘second degree’ (i.e. those with not more than two variables multiplied together). The result seems to be just the same function (??), but with the new variable $x=p+q\,$. And if you go on, always putting together products of the same degree, you’ll ﬁnd the next terms are

 $(p+q)^3/{3!}=(p^3+3p^2q+3pq^2+q^3)/{3!}\,$ (third degree)

and

 $(p+q)^4/{4!}=(p^4+4p^3q+6p^2q^2+4pq^3+q^4)/{4!}\,$ (degree 4.)

As you can guess, if we take more terms we’re going to get the result

 $f(p)f(q)=1+(p+q)+\frac{(p+q)^2}{2!}+\frac{(p+q)^3}{3!}+...=f(p+q)\,$ (2)

To get a proof of this result is much harder: you have to