Katex

**Olly** · 10-30-2014, 03:18 AM

IDK what any of this means, but you show math as a image now!!

f'(x) = \lim_{h \to 0}(\frac{2x^2 + 4xh + 2h^2 - 2x^2}{h})

Say thanks to core!

**The Mayor** · 10-30-2014, 03:22 AM

Originally Posted by Olly

IDK what any of this means, but you show math as a image now!!

f'(x) = \lim_{h \to 0}(\frac{2x^2 + 4xh + 2h^2 - 2x^2}{h})

Say thanks to core!

Nice!

E: y no work for me

**Ian** · 10-30-2014, 03:34 AM

f(x) = (thanks+core)

**NKN** · 10-30-2014, 10:39 AM

x = x+3

did I do it?

**slacky** · 10-30-2014, 11:35 AM

Originally Posted by Olly

IDK what any of this means, but you show math as a image now!!

Correction: It's not a image.

@core: Issue with division (frac and dfrac), the dividing line is two high up:

\frac{x}{y}

Anyways, tyvm for adding this.

**KeepBotting** · 10-30-2014, 11:55 AM

yolo = (69 + 420)

wat

rj · 10-30-2014, 02:05 PM

f(x) = 1/1+(1/x_o - 1)a^-rt

how do i make it work

**masterBB** · 10-30-2014, 02:16 PM

f(x) = 1/1+(1/xo - 1)a^{-rt}

**Kevin** · 10-30-2014, 02:23 PM

Originally Posted by Robert

f(x) = 1/1+(1/x_o - 1)a^-rt

how do i make it work

Hit quote on anyone that did it, and then you'll see it fine.

Kevin = NKN * Sirenia - Sjoe ^TK + Hoodz * i

Edited: Many times over.

**masterBB** · 10-30-2014, 03:40 PM

Originally Posted by Kevin

Hit quote on anyone that did it, and then you'll see it fine.

Damn. it doesn't like fake functions. (Quote me to see what I wanted to impose

)

Kevin = NKN * Sirenia - Sjoe - TK + Hoodz * i

I just doesn't accept the &

**Kevin** · 10-30-2014, 05:52 PM

Originally Posted by masterBB

Kevin = NKN * Sirenia - Sjoe - TK + Hoodz * i

I just doesn't accept the &

So it dislikes bit-wise operators, and I can't figure out how to get multiple characters qualified with a power '^'.

**Ian** · 10-30-2014, 05:54 PM

Originally Posted by Kevin

So it dislikes bit-wise operators, and I can't figure out how to get multiple characters qualified with a power '^'.

Kevin = NKN * Sirenia - Sjoe ^(TK) + Hoodz * i

Edit: my attempt to use ^(TK) has failed

**masterBB** · 10-30-2014, 05:55 PM

Kevin = NKN * Sirenia - Sjoe ^{TK} + Hoodz * i

**masterBB** · 10-30-2014, 06:00 PM

WTF, why if I refresh all katex like things stop working. Can't even edit my post above without me having to repost it.

**The Killer** · 10-30-2014, 08:59 PM

Originally Posted by masterBB

Kevin = NKN * Sirenia - Sjoe ^{TK} + Hoodz * i

Sjoe to the power of TK... damn thats a big number

@The Mayor I posted... are you happy?

**Kyle** · 10-30-2014, 09:02 PM

Originally Posted by The Killer

@The Mayor I posted... are you happy?

I know I am

**The Mayor** · 10-30-2014, 09:14 PM

Originally Posted by The Killer

Sjoe to the power of TK... damn thats a big number

@The Mayor I posted... are you happy?

Because you've been missing in action?

**Daniel** · 10-31-2014, 03:16 AM

Originally Posted by Kevin

Hit quote on anyone that did it, and then you'll see it fine.

Kevin = NKN * Sirenia - Sjoe ^TK + Hoodz * i

Damn. it doesn't like fake functions. (Quote me to see what I wanted to impose

)

\text{Kevin} = \text{NKN} \times \text{Sirenia} - \text{Sjoe}^{\text{TK}} + \text{Hoodz} \times i

**Kevin** · 10-31-2014, 03:00 PM

Originally Posted by The Killer

Sjoe to the power of TK... damn thats a big number

@The Mayor I posted... are you happy?

Indeed. I'm honestly not sure how any of it is valued, and the only part of it in relation to me that I can make sense of is Hoodz * i.... Because I have never touched OSR so that's imaginary for me.

**core** · 10-31-2014, 05:49 PM

Originally Posted by slacky

Issue with division (frac and dfrac), the dividing line is two high up

No clue what is causing that. I think it's some css conflict but no chance to solve at the moment. If someone wants to figure it out and tell me, that'd be great.

Originally Posted by masterBB

WTF, why if I refresh all katex like things stop working. Can't even edit my post above without me having to repost it.

Because of the way I implemented it (lazily), you need to refresh the entire page for the script to execute properly, which doesn't happen if you just edit or make a post. Oops.

**Sin** · 10-31-2014, 07:45 PM

Originally Posted by Olly

IDK what any of this means, but you show math as a image now!!

f'(x) = \lim_{h \to 0}(\frac{2x^2 + 4xh + 2h^2 - 2x^2}{h})

Say thanks to core!

The derivative of

f(x) = 2x^2 + c

if i'm doing my integrals correctly.

**masterBB** · 10-31-2014, 09:15 PM

Originally Posted by Sin

The derivative of

f(x) = 2x^2 + c

if i'm doing my integrals correctly.

uhm

limit h -> 0 means there is no solution if h = 0 , since you divide it by zero. However when I rewrote it, I removed the h:

f'(x) = \frac{2x^2 + 4xh + 2h^2 - 2x^2}{h}

f'(x) = \frac{4xh + 2h^2}{h}

f'(x) = 4x + 2h

f(x) = 2x^2 + 2hx + C

But I think the h was added with a reason, as a paremeter for function fitting or something similar

**Sin** · 10-31-2014, 11:26 PM

Originally Posted by masterBB

uhm

limit h -> 0 means there is no solution if h = 0 , since you divide it by zero. However when I rewrote it, I removed the h:

f'(x) = \frac{2x^2 + 4xh + 2h^2 - 2x^2}{h}

f'(x) = \frac{4xh + 2h^2}{h}

f'(x) = 4x + 2h

f(x) = 2x^2 + 2hx + C

But I think the h was added with a reason, as a paremeter for function fitting or something similar

Well it's a limit isn't it? The only reason h cant be 0 is because there's a h in the denominator, but once you cancel out you're allowed to replace all the h's with a 0. You end up with

f(x) = 4x

And i'm pretty sure h = 0 doesn't mean that it doesn't have a solution. It's simply the limit as h approaches to 0. If I try and use the difference quotient on the 2x^2:

\lim_{h /to 0}(\frac{f(x+h) - f(x)}{h})

\lim_{h /to 0}(\frac{2(x+h)^2 - 2x^2}{h})

\lim_{h /to 0}(\frac{2(x^2 + 2xh + h^2) - 2x^2}{h})

\lim_{h /to 0}(\frac{2x^2 + 4xh + h^2 - 2x^2}{h})

\lim_{h /to 0}(\frac{4xh + h^2}{h})

\lim_{h /to 0}(4x + h)

=4x

So yeah, the function in the OP is the derivative of 2x^2

**masterBB** · 10-31-2014, 11:56 PM

Originally Posted by Sin

Well it's a limit isn't it? The only reason h cant be 0 is because there's a h in the denominator, but once you cancel out you're allowed to replace all the h's with a 0. You end up with

f(x) = 4x

And i'm pretty sure h = 0 doesn't mean that it doesn't have a solution. It's simply the limit as h approaches to 0. If I try and use the difference quotient on the 2x^2:

\lim_{h /to 0}(\frac{f(x+h) - f(x)}{h})

\lim_{h /to 0}(\frac{2(x+h)^2 - 2x^2}{h})

\lim_{h /to 0}(\frac{2(x^2 + 2xh + h^2) - 2x^2}{h})

\lim_{h /to 0}(\frac{2x^2 + 4xh + h^2 - 2x^2}{h})

\lim_{h /to 0}(\frac{4xh + h^2}{h})

\lim_{h /to 0}(4x + h)

=4x

So yeah, the function in the OP is the derivative of 2x^2

Here you make a mistake

\lim_{h /to 0}(\frac{2(x^2 + 2xh + h^2) - 2x^2}{h})

\lim_{h /to 0}(\frac{2x^2 + 4xh + h^2 - 2x^2}{h})

Here you moved the 2 in front of

x^2 + 2xh + h^2

but forgot to apply it to the

h^2

correct would be:

\lim_{h /to 0}(\frac{2x^2 + 4xh + 2h^2 - 2x^2}{h})

continueing:

\lim_{h /to 0}(\frac{4xh + 2h^2}{h})

\lim_{h /to 0}(4x + 2h)

But you can't just drop the 2h, that is only that case when it is the other way around. Now we are integrating:

f(x) = 2x^2 + 2hx

e:
missed the part where you said you used the difference quotient.

**Sin** · 11-01-2014, 12:15 AM

Originally Posted by masterBB

Here you make a mistake

\lim_{h /to 0}(\frac{2(x^2 + 2xh + h^2) - 2x^2}{h})

\lim_{h /to 0}(\frac{2x^2 + 4xh + h^2 - 2x^2}{h})

Here you moved the 2 in front of

x^2 + 2xh + h^2

but forgot to apply it to the

h^2

correct would be:

\lim_{h /to 0}(\frac{2x^2 + 4xh + 2h^2 - 2x^2}{h})

continueing:

\lim_{h /to 0}(\frac{4xh + 2h^2}{h})

\lim_{h /to 0}(4x + 2h)

But you can't just drop the 2h, that is only that case when it is the other way around. Now we are integrating:

f(x) = 2x^2 + 2hx

e:
missed the part where you said you used the difference quotient.

Ah yeah I admit I did miss that, but wouldn't my point remain the same? 2h is still 0 when h is 0.

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