/sci/ - Science & Math » Thread #11197365

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Help Wed 04 Dec 02:26:07 2019 No.11197365 View Reply Original Report

Quoted By: >>11197379

I need a formula.

I need a formula where
x > 1000= 1000
but if x < 1000 = x

I can't use IF functions, I need a math formula, is this even possible with like absolute values and whatnot? Sorry I suck at explaining.

Anonymous

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Anonymous Wed 04 Dec 2019 02:30:20 No.11197376 Report

Quoted By: >>11197419 >>11197809

$f(x)=x+(1000-x)\cdot u(x-1000)=x+(1000-x)\cdot u_{1000}(x)$
where $u(x)$ is the Heaviside step function

Anonymous

Anonymous Wed 04 Dec 2019 02:30:34 No.11197379 Report

Quoted By: >>11197419 >>11197425 >>11197430 >>11197489

>>11197365
Ultimately, IF is as much a math formula as anything else. But max(x, 1000) might be what you're after here.

>is this even possible with like absolute values and whatnot
Yes. If max() is also unavailable, try
>(1000 + x) / 2 - abs((1000 - x) / 2)

Help

Help Wed 04 Dec 2019 02:47:57 No.11197419 Report

Quoted By: >>11197483

>>11197376
This is too big brained for me

What is u

>>11197379
I can't use excel stuff like max :(

Help

Help Wed 04 Dec 2019 02:50:35 No.11197425 Report

Quoted By:

>>11197379
>>(1000 + x) / 2 - abs((1000 - x) / 2)
Wow this is exactly what I was looking for !

Help

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Help Wed 04 Dec 2019 02:52:33 No.11197430 Report

Quoted By: >>11197815

>>11197379
Thanks this would have taken me forever to figure out

Anonymous

Anonymous Wed 04 Dec 2019 03:10:44 No.11197483 Report

Quoted By: >>11197489 >>11197538

>>11197419
u(x-c)=1 when x>c
u(x-c)=0 when x<c

Help

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Help Wed 04 Dec 2019 03:12:23 No.11197489 Report

Quoted By: >>11197815

>>11197483
Thanks but this guy >>11197379
did a better formula

Anonymous

Anonymous Wed 04 Dec 2019 03:32:30 No.11197538 Report

Quoted By: >>11197605

>>11197483
Hahaha fucking loser

Anonymous

Anonymous Wed 04 Dec 2019 03:59:27 No.11197605 Report

Quoted By:

>>11197538
look, im not going to give the integral of the dirac delta definition

Anonymous

Anonymous Wed 04 Dec 2019 05:17:26 No.11197809 Report

Quoted By: >>11197879

>>11197376
Not OP, but neat.

Anonymous

Anonymous Wed 04 Dec 2019 05:20:44 No.11197815 Report

Quoted By:

>>11197489
>>11197430
Imagine the smell

Anonymous

Anonymous Wed 04 Dec 2019 05:49:00 No.11197873 Report

Quoted By:

Suppose
$f_k(x)=k\ \ \text{when}\ -\frac{1}{k}\leq x\leq\frac{1}{k}\ ;\ \ f_k(x)=0\ \ \text{otherwise}$
and suppose
$\delta(x)=\lim_{k\to\infty}f_k(x)$
therefore
$\int_{-\infty}^{\infty}\exp(-sx)\cdot\delta(x)\text{ d}x=\mathcal{L}\big(\delta(x)\big)=1$
and
$u(x)=\int_{-\infty}^x\delta(t)\text{ d}t$

Anonymous

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Anonymous Wed 04 Dec 2019 05:50:05 No.11197879 Report

Quoted By: >>11197954

>>11197809
Suppose
$f_k(x)=k\ \ \text{when}\ -\frac{1}{k}\leq x\leq\frac{1}{k}\ ;\ \ f_k(x)=0\ \ \text{otherwise}$
and suppose
$\delta(x)=\lim_{k\to\infty}f_k(x)$
therefore
$\int_{-\infty}^{\infty}\exp(-sx)\cdot\delta(x)\text{ d}x=\mathcal{L}\big(\delta(x)\big)=1$
and
$u(x)=\int_{-\infty}^x\delta(t)\text{ d}t$

Anonymous

Anonymous Wed 04 Dec 2019 06:45:36 No.11197954 Report

Quoted By:

>>11197879
I spent too much time TeXing this not to get validation from strangers on the internet.

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