/sci/ - Science & Math

<div class="math"> x^2 = x +x+x+x+...~~ (x ~times) </div>
<div class="math"> \frac{d}{dx}x^2 = 2x </div>
<div class="math"> \frac{d}{dx}x = 1 </div>

<div class="math"> 2x = \left[1 +1+1+1+...~~ (x ~times) \right]+\left[x+x+x+x+...~~(1~times) \right] </div>
<div class="math"> 2x = \left[x \right]+\left[x \right] = 2x </div>
Check... wait...

x=(1-1)

0=x+x+x+x+x...to infinity
0=(1-1)+(1-1)+(1-1)+(1-1)...to infinity
and because of the associative property of addition
0=1-(1+1)-(1+1)-(1+1)-(1+1)...to infinity
if we turn all -(1+1)'s into (1-1), then
0=1+(1-1)+(1-1)+(1-1)+(1-1)...to infinity
turn all (1-1)'s into 0
0=1+0+0+0+0+0+.....to infinity
0=1

>>6296435

So? All you proved is that this shit is impossible.

0/10 trolling. Stick with .999 repeating equals 1 threads or PEMDAS threads, those are always good fun.

>>6296457
>turn -(1+1) into (1-1)
>turning -2 into 0
pls
stahp
pls

>>6296464
>turning cant read mat notation into our problem

>>6296465
-(1+1) = (-1-1) you fool.
Learn to distribute

>>6296468
LMAO

-1+1 = -(1-1)

>>6296475
0/10
-1+1 =/= -(1+1)
Again, learn to distribute.
If you think you "broke math", you're just wrong

>>6296476
so -1+1 is not -(1-1)?

>>6296482
It is.
But -(1+1) is not -1+1.
His problem states that
0 = 1-(1+1)-(1+1)-(1+1)-...
You're implying he said
0 = 1-1+1-1+1-1+1-1+1...
They are not the same, which is why he is wrong

>>6296488
those brackets show visual ordering, not operational ordering

>>6296492
Once you factor out the negative, which he did in the prior step, it does become operational ordering.
You're refusing to acknowledge one of the most rudimentary principles of math, factoring and distribution.
You can use them for visual ordering but when you factor out the negative you have to distribute it appropriately, which he did not do.

2x = -1/12

>>6296435
first line is wrong
Pi^2= Pi + Pi + Pi... "Pi times" ? -> only work for integers.
And then you're doomed to derivate.

>>6296492
1-1+1-1+1-1+1... = 1-(1+1)-(1+1)-(1+1)...
Subtract 1,
-1+1-1+1-1+1... = -(1+1) - (1+1) - (1+1)...
0 = -2 - 2 - 2 - 2...
0 = -infinity
This is why he's wrong

>>6296435
The amount of 1's varies with x as you noted, yet you have not accounted for that variance when differentiating. Assuming you're really interested in why it doesn't work.

Nice one OP!

f(n,x)=nx
f_n(n,x) = x
f_x(n,x) = n
g(x)=f(x,x)=x^2
g'(x)=f_n(x,x)+f_x(x,x)=x+x=2x

OP forgets to add f_n(x,x).

>(d/dx)x^2 = 2x

>>6296504
>>6296508
>>6296516
retards.

>>6296499

>>6296528

x is an integer.

>>6296530
X is a variable
It can be an integer, but that doesn't mean it is

>>6296449
Where does the
>+x+x+x+x+x+x+x+x+x
part come from and also, +x+x+x+x+x+x+x is like a gazillion x and not just one x.

>>6296547

[x+x+x+x...x] (1 time) = x

>>6296551
x=x
x+x=2x
x+x+x=3x
x+x+x+x+x+x+x .... =x
ok

>>6296547
the x +x +x +x is the original x + x+ x +x

>>6296567

You're misunderstanding what the notation means, it simply means the pattern [x+x+x...x], 1 time, aka x.

<div class="math">\frac{\mathrm{d} }{\mathrm{d} x}\sum_{1}^{x} \neq \sum_{1}^{x}\frac{\mathrm{d} }{\mathrm{d} x}</div>

You can't pull the derivative inside if the summing index is the variable itself...Also >>6296499

You're forgetting that "x times" is an additional x dependence which you're ignoring by simply differentiating each term in the sum.

>>6296599
To clarify, imagine you have f = x + x + ... + x (y times) = xy, then ∂f/∂x = y = x in the special case when y = x numerically.

However, ∂f/∂x is -not- the total derivative with respect to x. By the chain rule df/dx = ∂f/∂x ∂x/∂x + ∂f/∂y ∂y/∂x . In the case of y = x, that equals y*1 + x* ∂y/∂x. When y = x that evaluates to x + x = 2x.

>>6296435
let x=100
x^2-9900 = x
d/dx x^2-9900 = 2x
d/dx x = 1

200 = 1

Do I fit in yet?

Am I retarded enough for /sci/?

Your identity only holds if x is an integer, as stated.

If you restrict the domain of x^2 to Z, then there are no accumulation points and hence you cannot talk about limits (and therefore derivatives).

>>6296435

First line doesn't make sense. What if x is a negative number, then the left side will be positive and the right side will be negative.

Always be wary when you see the "..." notation.

To do it properly you would write out the sum like so:
<div class="math">\sum_{n=1}^{x}x</div>

So when you take the derivative it's not just 1 summed x times, because you forgot that your upper limit is x, and you're finding out what happens as x changes, so your upper limit is changing.

what if x < 0?

>>6296435
except x^2 = |x| + |x| + ... + |x| (|x| times)
and differentiating this piecewise at x=0 will blow up in our faces since d/dx |x| does not exist at x=0.

>>6296530
>>6296534
>>6296888
Nope, it holds only if x is a natural number.

>>6296930
Makes sense. How do you correctly take the derivative of such a sum (if it's possible)?

>>6297114
I really don't know

The first thing that comes to mind is
http://en.wikipedia.org/wiki/Leibniz_integral_rule

But I think the idea of a derivative involving the limit in a sum is just nonsense.
Because a derivative involves infinitesimal change by definition, so what would be the sum from n=1 to N+ε as ε->0?
this doesn't make sense for whole numbers

>>6297114
Use the limit form: Lim_{c->0} [sum(x+c,{0,x+c}) - sum(x,{0,x})]/c
=sum(x,{0,x+c})/c+sum(c,{0,x+c})/c-sum(x,{0,x})/c
=sum(x,{0,x})/c+sum(x,{x,x+c})/c+sum(1,{0,x+c})-sum(x,{0,x})/c
=sum(x,{x,x+c})/c+x+c
=x*c/c+x+c [because sum x from x to x+c is c*x for small c]
=2*x (+c->0)

>>6296435
Stop triggering me

>>6296435

>2x=x
>2x-x = x-x
>x=0

You divided by zero you nigger

Herp? A derivative is a rate of change. The function x^2 changes at a different rate than x+ x+ x + x +x ... + x. You can easily tell this graphically as x+ x+ x+x..x is a line and x^2 is parabolic curve...

>>6298997
>x+ x+ x+x..x is a line
Lrn2graph

>>6296435
High-school teacher here. This is easy.

I thought some of /sci/ would be in college at least..

>>6296435
wtf, did you drop out or something?

>>6301605
>math in Word

>>6301605
thanks

>>6301605
Very good, A+.

I get a ban for asking about tripcodes and this shit doesn't?!! fuck you faggot mods

>>6301699
What did you post?