Ln(x) = -ln(1/x)

I think you mean x/sqrt(x) = sqrt(x).

Nice troll. You know what he meant

[math]\frac{\mathrm{d} }{\mathrm{d} x}\int_{a}^{x}f(t)dt = f(x)[/math]

Yeah exponent rules blew my mind

Yeah your faggotry blows my mind too

how about sqrt(X)^3 = Xsqrt(X)

>1/0 = ∞

[math]1\cdot 1^{-1}\cdot x^{-1}=x^{-1}=\frac{1}{x}\neq x[/math]

Are you fucking retarded?

[math]\frac{1}{1+\sqrt{2}}=\sqrt{2}-1[/math]

fucking reet

>this thread

2^4=4^2

how does −ln(∞)=−∞ if
e^(−ln(∞))=0

are you memeing mate

cos(x)+cos(x)=2*cos(x)+(tan(x)-tan(x))

e^iπ - 1 = 0

HFW

>x=x

e^(-ln (infinity))=e^(-infinity)=1/e^infinity=0

1+1=2

i^2 + 1^2 = 0^2

2+2=4
4-1=3
You can't get quicker than this

Wrong, correct is: [math]e^{i\pi} + 1 = 0[/math]

>i = (-1^-1)^.5

Anyone know a good textbook on logarithms? I know its hs level math but I didnt really pay attention to class back then so now whenever I see expressions with logarithms its a pain in the ass.

very cute so far

watch out brainlets:

[eqn]\frak{for} \hspace{0.2cm}x_{1} = 1 \hspace{0.2cm} \frak{consider} \hspace{0.2cm} x_{n+1} = \sqrt{ 1 + x_{n} }. \hspace{0.2cm} \frak{ \lim x_{n+1} = \varphi } [/eqn]

Logs are piss tho.

tutorial.math.lamar.edu/Classes/Alg/LogFunctions.aspx
all you ever have to know really, just do the examples and you'll be fine

...

You thought you could best me kid?
*teleports behind you*
I've found the chink in your armor
*thrusts masamune into your back*
*whispers:*
You didn't prove it converges first.

well 1/0 is not mathematically defined in R or
R bar. You obviously saw one of the reasons why

im not him but unless i don't understand the task you said [math] \lim\limits_{n \rightarrow \infty}{Xn+1} = /varphi [\math]. Therefore it must converge otherwise a lim wouldn't exist

That's like saying that I need to prove an infinite continued fraction converges.

had a good point too. Why would you ask the question if phi didn't exist?

>Therefore it must converge otherwise a lim wouldn't exist
Yes, however in his workings there was no PROOF of this convergence. Without it, the calculations are meaningless. I merely gave the answer prior to show the neat result. On an exam or something you'd just be given the recursion and asked "does it converge, if so what is the limit".

...

>prod(f(x)) = e^(sum(ln(f(x))))

Thanks, that actually was pretty useful.
Yeah theyre very simple now that I actually sit down and learn about them.

U aint got shit on me son.
d/dx(e^x) = e^x = integral e^xdx

U aint got shit on me son.
d/dx(e^x) = e^x = integral e^xdx

What does this mean? This is the most beautiful thing I've ever seen. Why don't they teach this in a math undergrad?

[math]f(x) = \sqrt{1+x}[/math] is a contraction on [math][0, \infty)[/math], as f is closed on the same interval and differentiable with derivative [math]\frac{1}{2\sqrt{1+x}}[/math] bounded by 1/2. Then apply the Banach fixed point theorem.

>I have discovered a truly remarkable proof of this theorem which this post is too small to contain

Wouldn't it be more precise to call it f(x,t)?

what the fuck

>using an nuclear bomb instead of C4
dude, just use monotone convergence theorem. two induction proofs and you're good to go.

>[math]1+2=7^2-\frac{96}{2}+2\cdot{}5-\sqrt{64}[/math]

[math] \infty > \infty [/math]

2^4=4^2

[math]ln: (\mathbb{R}^+, \cdot) \Rightarrow (\mathbb{R}, +)[/math] is a group isomorphism.

A dozen, a gross, and a score
Plus 3 times the square root of 4
Divided by 7
Plus 5 times 11
Is 9 squared and not a bit more

>if the additive identity of a ring has a multiplicative inverse, the ring has exactly one element

[math]i^i = e^{-{π \over 2}} \approx 0.20787957635 [/math]

Although you are right, I must point out that (lim x -> -inf) e^x goes toward 0, so it's not an intuitive counterexample in this case.

142857
758241

1/7 = 0,14285714285714285714285714285714

yeah it blows mind a little

You are like a little baby

aᵖ = a (mod p)

look at the definition, look at the common operations and prove to yourself all that shit. write as logax=y as a^y=x if in doubt

[eqn]10^2 + 11^2 + 12^2 = 13^2 + 14^2[/eqn]

[eqn]21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2[/eqn]

[eqn]\forall n \in \mathbb{N}, \sum _{i=2 n^2+n}^{2 n^2+2 n} i^2=\sum _{i=2 n^2+2 n+1}^{2 n^2+3 n} i^2[/eqn]

[math] \inf\limits_{\bigcup\limits_{j=1}^\infty A_j \supseteq [0,1]} \sum\limits_{j=1}^\infty m(A_j) = 1 [/math]

The guys proof is simpler.

No. Why would it be?

Why is this funny to me

because you're a brainlet

Convincing proof you got there.

...

are you serious mate?

this upsets the mathematician

no way

>says the brainlet

Why is this thread filled with disgusting high school faggots who say "ln" instead of log. Immediately delete yourself from R^3, via any nearby firearm.

i^i is not a function, it has infinite values

noice

φ-1 = φ/1

lots of sarcasm here but proving simple shit like this is actually really interesting tbqh

...

Y'all ever realize the division symbol is just a fraction with two openings on the top and bottom

>I am the observed relation between myself and observing myself
Defines "I" as the eigenform of the transformation
T(x)=the observed relation between x and observing x
If a solution to the equation arises
I=TI

An "I" becomes
Now think about this in terms of life itself, life is produced by the actions of the system it defines, but is not part of such systems.
Take that materialists

...

Doesn't apply to negative numbers.

had to laugh

>implying we live in R^3
Can you tell me where (0,0,0) is located? Or what the sum of two points in space is?

>What is less, 2/5 or 4/10?
>Believe it or not, they have the same absolute. It's just an optical illusion bro