User talk:Bitshift

Ok so the important thing to remember is that the expression

$\int _{C}{\vec {F}}\cdot d{\vec {r}}$

is totally equivalent to the expression

$\int _{a}^{b}{\vec {F}}({\vec {r}}(t))\cdot {\vec {r}}'(t)dt$

where the vector equation is parameterized with respect to some $t$. But if we let ${\vec {F}}=\langle P,Q\rangle$, then it turns out that the previous two integrals are also equivalent to

$\int _{C}Pdx+\int _{C}Qdy$

which is just abbreviated as

$\int _{C}Pdx+Qdy$.

testing:

$\cos(x)=0$

$\sum _{0}^{\infty }$

$\pi =4\cdot \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}=4\cdot \left({\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots \right)$

and of course

$\cos(x)=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\cdots$

$x\in \mathbb {R}$

Start a discussion with Bitshift

Talk pages are where people discuss how to make content on Wikipedia the best that it can be. Start a new discussion to connect and collaborate with Bitshift. What you say here will be public for others to see.

Start a discussion