/sci/ - Science & Math

[Anonymous No.4925546]

>Implying most people on /sci/ know what they preach.

>Bonus points: Name a militaristic way to use derivatives on the battlefront.

[Anonymous No.4925548]

>>4925546
Binomial Theorem

[Anonymous No.4925561]

use the limit definition of the derivative.

[Anonymous No.4925565]

dy/dx=2x
dy=2dx^2 (times both sides by dx)
y=2x^2 (divide by d)

Hmm, that's as close as I could get.

[Anonymous No.4925567]

dy/dx=2x
dy=2dx^2 (times both sides by dx)
y=2x^2 (divide by d)
x^2=2x^2 (since y=x^2)
1=2 (divide by x^2)

Since this is a contradiction it shows that dy/dx != 2x

[Anonymous No.4925570]

To prove dy/dx = 2x, we must first prove the universe.

[Anonymous No.4925573]

It's nonsensical to ask us what it means since we don't know what x stands for.

[Anonymous No.4925576]

>>4925573
Not sure if serious.

[Anonymous No.4925581]

times both sides by dy = 2x dx
integrate both sides
y = 2^x
divide both sides by zero

[Anonymous No.4925582]

>>4925546
calculate speed and therefore force of impact of a projectile as it strikes a target. Calculus and terminal ballistics are a match made in heaven .

[Anonymous No.4925583]

http://en.wikipedia.org/wiki/Power_rule

Therefore,

dy/dx = 2x^(2-1) = 2x^(1) = 2x

[Anonymous No.4925586]

>>4925548
>>4925561
>>4925583

The reason we have these equations and "rules" is becaise, yes, these have already been proven before your time.

The question was to prove it yourself using your own math, rather than merely stating or copying a proof online, to see if you remember (or should I say, can determine) how to prove it.

As such, none of you completed the point.

Also, >>4925567
you're not even close.

On your second line:
>dy=2dx^2 (times both sides by dx)
You multiply the "dx" term on both sides to give dy on the left. However, you can't add the x off of "dx" to give you x^2. "dx" isn't d * x, it's the change in x.

It isn't a contradiction, it's wrong mathematics. Note in the picture, I italicized the x in "dx" to help differentiate the two.

[Anonymous No.4925587]

>>4925567

>mfw I read this comment

>Captcha: pKa ruchiin
>Oh /sci/...

[Anonymous No.4925588]

>>4925567

>mfw I read this comment

>Captcha: pKa ruchiin
>Oh /sci/...

[Anonymous No.4925589]

<div class="math">y+dy = (x+dx)^2 = x^2 + 2xdx + dx^2 \to dy = 2xdx + dx^2\to\frac{\mathrm{d} y}{\mathrm{d} x} = 2x + \textup{very small stuff}</div>

[Anonymous No.4925590]

Let <span class="math">y=f(x)[\math]. The derivative at x is defined as <span class="math"> lim_{h\to\0}\frac{f(x+h)-f(x)}{h} = lim_{h\to\0}\frac{(x+h)^{2}-x^{2}}{h} = lim_{h\to\0}\frac{x^{2}+h^{2}+2xh-x^{2}}{h} = lim_{h\to\0}\frac{h(h+2x)}{h} = lim_{h\to\0}\{h+2x} = 2x [\math]

So THERE.[/spoiler][/spoiler]

[Anonymous No.4925591]

Let <span class="math">y=f(x)[/spoiler]. The derivative at x is defined as <span class="math"> lim_{h\to\0}\frac{f(x+h)-f(x)}{h} = lim_{h\to\0}\frac{(x+h)^{2}-x^{2}}{h} = lim_{h\to\0}\frac{x^{2}+h^{2}+2xh-x^{2}}{h} = lim_{h\to\0}\frac{h(h+2x)}{h} = lim_{h\to\0}\{h+2x} = 2x [/spoiler]

So THERE.

[Anonymous No.4925595]

>>4925590

>Let Unknown control sequence '\math'

>>4925591

>Let Unknown control sequence '\math'

This is quite hilarious to watch.

[Anonymous No.4925593]

introduce epsilon delta defn of limit
define dy/dx as limit as h->0 of [f(x+h) - f(x)]/h
apply to f(x) = x^2

[Anonymous No.4925596]

Let <span class="math">y=f(x)[\math]. The derivative at x is defined as <span class="math"> lim_{h\to\{0}}\frac{f(x+h)-f(x)}{h} = lim_{h\to\{0}}\frac{(x+h)^{2}-x^{2}}{h} = lim_{h\to\{0}}\frac{x^{2}+h^{2}+2xh-x^{2}}{h} = lim_{h\to\{0}}\frac{h(h+2x)}{h} = lim_{h\to\{0}}\{h+2x} = 2x [/spoiler]

So THERE.[/spoiler]

[Anonymous No.4925599]

>>4925596

[Anonymous No.4925631]

>>4925590
>>4925591
>>4925596

I am quite disappointed in you /sci/.

[Anonymous No.4925636]

>>4925586
The problem is that it's so simple I and many others thought this was a troll. I mean, you may as well have asked us to prove that (a+b)(a-b) = a^2 - b^2. Anyone can easily figure out the proof in their head but few people would care enough to type it out.

[Anonymous No.4925654]

>>4925636
Nice try, but no.

[Anonymous No.4925657]

The proof is:
((x+h)^2 - x^2)/h = (2xh + h^2) / h = 2x + h which limits to 2x as h approaches 0.

[Anonymous No.4925665]

>mfw someone says dy/dx = y/x

[Anonymous No.4925672]

Let <span class="math">f:]0,\infty[ \rightarrow \mathbb{R}, f(x)=\sqrt{x}.[/spoiler] Then <span class="math">f'(x)=\frac{1}{2\sqrt{x}}.[/spoiler] As <span class="math">f[/spoiler] is a bijection, it is invertible with inverse <span class="math">g(x) = x²[/spoiler] By the inverse function theorem, <span class="math">g[/spoiler] is differentiable with derivative <span class="math">g'(x) = \frac{1}{f(g(x))} = \frac{1}{\frac{1}{2\sqrt{x^2}}} = 2x. [/spoiler].

[Anonymous No.4925677]

Let <span class="math">f:]0,\infty[ \rightarrow \mathbb{R}, f(x)=\sqrt{x}.[/spoiler] Then <span class="math">f'(x)=\frac{1}{2\sqrt{x}}.[/spoiler] As <span class="math">f[/spoiler] is a bijection, it is invertible with inverse <span class="math">g(x) = x²[/spoiler] By the inverse function theorem, <span class="math">g[/spoiler] is differentiable with derivative <span class="math">g'(x) = \frac{1}{f'(g(x))} = \frac{1}{\frac{1}{2\sqrt{x^2}}} = 2x. [/spoiler].

[Anonymous No.4925678]

Let <span class="math">f:]0,\infty[ \rightarrow \mathbb{R}, f(x)=\sqrt{x}.[/spoiler] Then <span class="math">f'(x)=\frac{1}{2\sqrt{x}}.[/spoiler] As <span class="math">f[/spoiler] is a bijection, it is invertible with inverse <span class="math">g(x) = x².[/spoiler] By the inverse function theorem, <span class="math">g[/spoiler] is differentiable with derivative <span class="math">g'(x) = \frac{1}{f(g(x))} = \frac{1}{\frac{1}{2\sqrt{x^2}}} = 2x. [/spoiler].

[Anonymous No.4925679]

Let <span class="math">f:]0,\infty[ \rightarrow \mathbb{R}, f(x)=\sqrt{x}.[/spoiler] Then <span class="math">f'(x)=\frac{1}{2\sqrt{x}}.[/spoiler] As <span class="math">f[/spoiler] is a bijection, it is invertible with inverse <span class="math">g(x) = x².[/spoiler] By the inverse function theorem, <span class="math">g[/spoiler] is differentiable with derivative <span class="math">g'(x) = \frac{1}{f'(g(x))} = \frac{1}{\frac{1}{2\sqrt{x^2}}} = 2x. [/spoiler].

[Anonymous No.4925683]

Let <span class="math">f:]0,\infty[ \rightarrow \mathbb{R}, f(x)=\sqrt{x}.[/spoiler] Then <span class="math">f'(x)=\frac{1}{2\sqrt{x}}.[/spoiler] As <span class="math">f[/spoiler] is a bijection, it is invertible with inverse <span class="math">g(x) = x².[/spoiler] By the inverse function theorem, <span class="math">g[/spoiler] is differentiable with derivative <span class="math">g'(x) = \frac{1}{f'(g(x))} = \frac{1}{\frac{1}{2\sqrt{x^2}}} = 2x. [/spoiler]

[Anonymous No.4925684]

>>4925657

I was gonna post this, but since this guy beat me to it then I will post a geometric insight/representation that I made (several pictures).

f(x) = x^2

[Anonymous No.4925686]

>>4925684

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

[Anonymous No.4925687]

can you explain the bonus question a bit more OP? sounds kinda interesting actually.

[Anonymous No.4925689]

>>4925686
f(x+h)-f(x)

[Anonymous No.4925692]

Its simple: let <span class="math"> f = (x \mapsto x^2) [/spoiler], and let <span class="math"> \epsilon [/spoiler] be an infinitesimal. By definition, we have
<div class="math">
f'(x) = st \left( \frac{f(x+\epsilon) - f(x)}{\epsilon} \right) = st(2 x + \epsilon) = 2x
</div>
Where <span class="math"> st(y) [/spoiler] denotes the standard part of <span class="math"> y [/spoiler], of course.

[Anonymous No.4925695]

>>4925689

taking the limit as h approaches 0

[Anonymous No.4925696]

>>4925695

[Anonymous No.4925698]

>>4925696

We divide by h, note that the h^2 section has effectively disappeared. We're left with 2x.

The rate of change of x^2 is 2x. This can be seen by growing a square by an infinitesimally small amount through the process we just did (differentiation).

[Anonymous No.4925700]

>>4925698

There are similar insights for circles, cones, pretty much any geometric object. Calculus is supposed to change your perception of reality once you truly understand it. Probably an overstatement, but it is still true that the differentials and integrals of any object can be perceived without having to necessarily be computed.

[Anonymous No.4925969]

this
>>4925692
while the other stuff is like, school/first semester of university, this is pretty advanced

[Anonymous No.4925975]

>>4925969
>non-standard analysis
>advanced
pick exactly one

[Anonymous No.4925994]

I posted >>4925692

>>4925969
Dude, I have never made a serious effort to learn non-standard analysis, I had just heard of it. It took me exactly 24 seconds to google and copy this proof from wikipedia. Non-standard analysis might be advanced, what I wrote down is not. And I think obscure is a better way to describe non-standard analysis, because tbh, it doesn't look too difficult.

[Anonymous No.4926004]

>>4925692
ew infinitesimals

Here is the proof for the general rule
<div class="math"> \lim_{j\to0} \Big[ \frac{1}{j} \Big( \sum_{k= 1}^{n}\big(~ _nC_k \cdot x^{n-k}j^k \big) \Big) \Big] = \lim_{j\to0} \Big( \sum_{k= 1}^{n}\big(~ _nC_k \cdot x^{n-k}j^{k-1} \big) \Big) = \lim_{j\to0} \Big(~ _nC_1 \cdot x^{n-1} j^0+ \sum_{k= 2}^{n}\big(~ _nC_k \cdot x^{n-k}j^{k-1} \big) \Big) = n \cdot x^{n-1} + \sum_{k=2}^{n} 0 = n \cdot x^{n-1} </div>
The first part is kind of confusing, but it's just <span class="math"> \lim_{j\to 0} (x+j)^n/j [/spoiler]. The proof only works for integer n
I could get more specific, but that would be work. It would be pretty easy to prove the special case, even if you went all the way to delta-epsilon limits.

[Anonymous No.4926021]

>>4925975
Students actually did better when they were taught non-standard analysis in place of calculus.

[Anonymous No.4926024]

>>4926021
>better
sounds like hard science there

[Anonymous No.4926025]

>>4926004
>hating on infinitesimals

[Anonymous No.4926028]

>>4926021
>thinks non-standard analysis replaces calculus

calculus is pretty much taught in a non-rigorous way, but it's closer to non-standard than standard analysis

[Anonymous No.4926042]

My calculus teacher was in an artillery division in the army. He picked it because it had the most math. You use calculus a lot in artillery, as well as geometry.

[Anonymous No.4926054]

>>4925657
That is not a proof. You need to use the definition of a limit to prove it.

[Anonymous No.4926065]

>>4926054
and the construction of real number, and of course the naturals

if you want to make a pie from scratch you first have to do set theory

[Anonymous No.4926070]

>>4926065
>>4926054
there is a reason Principia Mathematic is 3 massive volumes :)

[Anonymous No.4926078]

im
>>4925692
maybe the proof is pretty simple in nonstandard-analysis but i saw ops problem asking for the derivation of x² and the idea not to use the (standard) definition but using the fact that
first: nonstandardmodels of R exist and
second: the techniques are equivalent (at least for special functions like x² in real numbers) is pretty andvanced.
but you guys are right, assuming that the definition of the derivation uses infinitesimal numbers its pretty simple..

[Anonymous No.4926091]

>>4926078
That is why Leibniz and Newton used infinitesimals, even though they were not formalized until much later. In between that, Weierstrass etc. redid everything with limits.