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

$$\int_C P dx + \int_C Q dy$$

which is just abbreviated as

$$ \int_C P dx + Q dy$$.

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

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

$$x\in\mathbb{R}$$