What's 1+1? (1 Viewer)

sadpwner · Oct 27, 2015

I wouldn't know because I don't do math anymore nor school? Time to slowly forget everything I've learned.

InteGrand · Oct 27, 2015

sadpwner said:
I wouldn't know because I don't do math anymore nor school? Time to slowly forget everything I've learned.

0 (mod 2).

FlyingKanga · Oct 27, 2015

minisu · Oct 27, 2015

0
Imagine that you have zero cookies and you split them evenly among zero friends. How many cookies does each person get? See? It doesn't make sense. And Cookie Monster is sad that there are no cookies, and you are sad that you have no friends.

life is a struggle

Drsoccerball · Oct 27, 2015

1+1! =2 but idk what 1+1? is equal to...

InteGrand · Oct 27, 2015

Drsoccerball said:
1+1! =2 but idk what 1+1? is equal to...

$1+1 = \sum_{k=0}^{\infty} (-1)^k \frac{\left(\frac{\pi}{2} \right)^{2k+1}}{(2k+1)!}+ \sum_{k=0}^{\infty} (-1)^k \frac{\left(2\pi \right)^{2k}}{(2k)!}$

turnerloos · Oct 27, 2015

1 + 1 = fuck the hsc. aint doing no math

Drsoccerball · Oct 27, 2015

turnerloos said:
1 + 1 = fuck the hsc. aint doing no math

how was physics?

Drsoccerball · Oct 27, 2015

InteGrand said:
$1+1 = \sum_{k=0}^{\infty} (-1)^k \frac{\left(\frac{\pi}{2} \right)^{2k+1}}{(2k+1)!}+ \sum_{k=0}^{\infty} (-1)^k \frac{\left(2\pi \right)^{2k}}{(2k)!}$

lol why do you know that?

sadpwner · Oct 27, 2015

Drsoccerball said:
how was physics?

Hardest paper I've ever witness. I can only pray.

Drsoccerball · Oct 27, 2015

sadpwner said:
Hardest paper I've ever witness. I can only pray.

last time I said a physics exam was easy i got crap sooo not saying anything

InteGrand · Oct 27, 2015

Drsoccerball said:
lol why do you know that?

A magical thing called Taylor Series.

Drsoccerball · Oct 27, 2015

InteGrand said:
A magical thing called Taylor Series.

What were your functions ? cos and sin ?

InteGrand · Oct 27, 2015

Drsoccerball said:
What were your functions ? cos and sin ?

sin and cos.

Drsoccerball · Oct 27, 2015

InteGrand said:
sin and cos.

Why do derivatives give you the function

InteGrand · Oct 27, 2015

Drsoccerball said:
Why do derivatives give you the function

?

Drsoccerball · Oct 27, 2015

InteGrand said:
?

When you're finding a taylor series we keep differentiating to find the function... why?

InteGrand · Oct 27, 2015

Drsoccerball said:
When you're finding a taylor series we keep differentiating to find the function... why?

$We want to say a function $f(x)$ can be written as $f(x) = c_0 + c_1 x + c_2 x^2 + \cdots $ (Taylor series centred about 0, i.e. Maclaurin series, where $|x|<R$, where $R$ is the radius of convergence). If it can be written this way, then differentiating $k$ times we have$

$$f^{(k)}(x) = 0 + 0 + \cdots + 0 + k! c_k+ c_{k+1} (k+1)k(k-1)\ldots 2x^{k+1}+\text{etc}$, where the etc. means further higher order terms in $x$, which will vanish if we put $x=0$. So putting $x=0$ gives $f^{(k)}(0) = k! c_k + 0 + 0 +\cdots \Rightarrow c_k = \frac{f^{(k)}}{k!}$, for $k=0,1,2,\ldots$, where $f^{(k)}(x)$ is the $k$th derivative of $f$ evaluated at $x$.$

$This doesn't necessarily mean that $f$ \textit{can} be written this way, it just means that \textit{if} it can, then the coefficients are as given above ($k$th derivative at $x$ divided by $k$ factorial). There are other theorems regarding whether the function can be written that way.$

jackleung34 · Oct 28, 2015

InteGrand said:
$We want to say a function $f(x)$ can be written as $f(x) = c_0 + c_1 x + c_2 x^2 + \cdots $ (Taylor series centred about 0, i.e. Maclaurin series, where $|x|<R$, where $R$ is the radius of convergence). If it can be written this way, then differentiating $k$ times we have$

$$f^{(k)}(x) = 0 + 0 + \cdots + 0 + k! c_k+ c_{k+1}x^{k+1}+\text{etc}$, where the etc. means further higher order terms in $x$, which will vanish if we put $x=0$. So putting $x=0$ gives $f^{(k)}(0) = k! c_k + 0 + 0 +\cdots \Rightarrow c_k = \frac{f^{(k)}}{k!}$, for $k=0,1,2,\ldots$, where $f^{(k)}(x)$ is the $k$th derivative of $f$ evaluated at $x$.$

$This doesn't necessarily mean that $f$ \textit{can} be written this way, it just means that \textit{if} it can, then the coefficients are as given above ($k$th derivative at $x$ divided by $k$ factorial). There are other theorems regarding whether the function can be written that way.$

yes i agree

turnerloos · Oct 28, 2015

Drsoccerball said:
how was physics?

bruhh.. the final time bos fucks me.. doesnt use a condom.

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