Question

HELP!

Answer 1

you need help? this is a first!

Answer 2

How do I solve this integral.

\[\int\limits_{\int\limits_{\int\limits_{...}^{...}f(x)dx}^{\int\limits_{...}^{...}f(x)dx}f(x)dx}^{\int\limits_{\int\limits_{...}^{...}f(x)dx}^{\int\limits_{...}^{...}f(x)dx}f(x)dx}f(x)dx\]

It's going on for infinity that's what the dot dot dots mean. The limits of each integral is the limits of each integral.

Answer 3

damn

Answer 4

please fix your integral LOL

Answer 5

The difference between this and an improper integral is?

Answer 6

Will it not be zero? The limits are the same.

Answer 7

PK pls, one infinity is bigger than the other.

Answer 8

Hmm, I'm trying to see how that occurs.

Answer 9

kanui, how is that kind of problem going to solve world poverty and help lower gas prices?

Answer 10

So is it 0 haha?

Answer 11

What if f(x) is the dirac delta function? http://en.wikipedia.org/wiki/Dirac_delta_function#Definitions

Answer 12

y u do dis

Answer 13

idk

Answer 14

What if we defined the derivative as: \[f'(x)=\lim_{h \rightarrow 0} \int\limits^{f(x+h)}_{f(x)}\frac{dx}{h}\]

Answer 15

See when the limit h approaches 0, then the limits of the integral become the same. But we know the derivative isn't 0. So maybe we can apply that here since the limits are also the same. Bwahahohehe!

Answer 16

guess i would start by analyzing what happens to f(x) = 1

Answer 17

somehow this integral reminds me of complements :
....999999999999
+1
-------------------
...000000000000

Answer 18

So what's the second derivative look like?

\[f''(x)=\lim_{h \rightarrow 0}\int\limits^{f'(x+h)}_{f'(x)}\frac{dx}{h}\]

Now let's go deeper.\[f''(x)=\lim_{h \rightarrow 0}\int\limits^{\int\limits^{f(x+2h)}_{f(x+h)}\frac{dx}{h}}_{\int\limits^{f(x+h)}_{f(x)}\frac{dx}{h}}\frac{dx}{h}\]

It looks like the infinitieth derivative of some random function in x will satisfy this nonsense. Wth?

Answer 19

I don't know, I'm not gonna lie, I was just making stuff up. But now I'm curious lol. XD

Answer 20

wow ! looks we can expand the branches forever and set it equal to that nth derivative xD

Answer 21

Yeah and all the limits and 1/h can actually come out front. The nth derivative will be the limit as h approaches zero of the insane integral divided by 1/h^n.

Answer 22

they're at different levels right

Answer 23

how can 1/h^n be pulled out

Answer 24

\[f''(x)=\lim_{h \rightarrow 0}\frac{1}{h}\int\limits^{f'(x+h)}_{f'(x)}dx=\lim_{h \rightarrow 0}\frac{1}{h}\int\limits^{\lim_{h \rightarrow 0}\frac{1}{h}\int\limits^{f(x+2h)}_{f(x+h)}dx}_{\lim_{h \rightarrow 0}\frac{1}{h}\int\limits^{f(x+h)}_{f(x)}dx}dx\]

However we can move them because suppose:

\[\lim_{h \rightarrow 0}\frac{1}{h}\int\limits^{\lim_{h \rightarrow 0}\frac{1}{h}\int\limits^{f(x+2h)}_{f(x+h)}dx}_{\lim_{h \rightarrow 0}\frac{1}{h}\int\limits^{f(x+h)}_{f(x)}dx}dx=\lim_{h \rightarrow 0}\frac{1}{h}[\lim_{h \rightarrow 0}\frac{1}{h}\int\limits^{f(x+2h)}_{f(x+h)}dx-\lim_{h \rightarrow 0}\frac{1}{h}\int\limits^{f(x+h)}_{f(x)}dx]\]

I think we can just combine these two separate h's into one h, since they're both approaching zero. It might be a slightly unjustified thing to do in general, but I don't think it's that big of a deal. \[\lim_{h \rightarrow 0}\frac{1}{h^2}[\int\limits^{f(x+2h)}_{f(x+h)}dx-\int\limits^{f(x+h)}_{f(x)}dx]\] This can then be put back up into the original form I had written with the stuff out of the limits of integration. rofl

Answer 25

whoa ! 1/h^n part makes sense :) are you suggesting the earlier infinite branching tree collapses to a telescoping(kind of) series whose terms are definite integrals ?

Answer 26

ahh nvm i see they're equivalent !!

Answer 27

Well all I'm really sort of saying is: \[f^{(n)}=\lim_{ h \rightarrow 0} \frac{1}{h^n}\int\limits^{\int\limits^{\int\limits^{...}_{...} dx}_{\int\limits^{...}_{...} dx} dx}_{\int\limits^{\int\limits^{...}_{...} dx}_{\int\limits^{...}_{...} dx} dx} dx\] take the limit as n approaches infinity.

I'm saying maybe now its possible that f(x)=1 in the original question I asked. lol

Answer 28

Perhaps as an added bonus, let's say f(x)=e^x since we know that will remain constant with respect to its derivative... \[\frac{d}{dn}\left( \frac{d^n}{dx^n} e^x\right)=0\]

Ok maybe that's not very interesting or important, I just thought it would be fun to say.

Answer 29

That way we know our limits of integration aren't going to do anything different, if that makes sense, because we're essentially doing something with an arbitrary function f(x) and it sort of disappears when we go to infinity and it sort of needs to be infinitely differentiable.

Perhaps f(x)=0 would have been just as good?

Answer 30

I would be wary about taking any of the answers we come up with here very seriously. This very well could be something along the lines of how:

S=1+2+4+8+...
(S-1)/2 = 1+2+4+8+...
S=(S-1)/2
S=-1

lol

Answer 31

But don't let that stop us from having fun, the ways of thinking we can sort of play with here may be applicable to other things which will give us an interesting and unique insight on them hopefully. Don't be afraid to be wrong is what I say haha.

Answer 32

even the nonsense makes sense if we persist long enough lol

Answer 33

I think you're right; I wonder some times about how we can interpret the "other side" of the geometric series for values greater than 1. I feel like it has some meaning there.

Answer 34

true^ f(x) = 0 vanishes immediately right ? It looks e^x is the only smooth function for which derivative with respect to n becomes 0..

Answer 35

We have an infinite number of possible answers. It could be 0 (which is the normal answer we would have suspected normally), which will be true for all polynomials except for ones that can be expressed as a taylor series.

Essentially the answer of this polynomial is the infinieth derivative of a function, so an interesting choice might be:

\[\Large f(x)=e^{\frac{x}{2}}\] which is cool because our answer to the integral will be \[\Large f^{(\infty)}(x)=2e^{\frac{x}{2}}\]

or maybe in general we could write \[\Large f^{(\infty)}(x)=\frac{n}{n-1}e^{\frac{x}{n}}\]

Answer 36

For fun, let's try sine.\[\Large f(x)=\sin(ax+b) \\ \Large f^{(n)}(x)=\ a^nsin(ax+b+\frac{n \pi }{2}) \]

If a<1 then we can say that the integral will converge to 0 since lim as n approaches infinity of a^n =0. But that's not very interesting in the sense that 0 isn't an answer we didn't have before.

Answer 37

one sec, im still trying to figure out how u said
http://prntscr.com/3ty9w7

:(

Answer 38

That's because it's wrong, whoops my bad. I was thinking that multiplying every time you took the derivative would give us a geometric series! Nevermind! >_>

Answer 39

ohhk.. 1/2*1/2*1/2*1/2... = 1/2^n = 0 (as n->inf)

so e^(x/n) also gives 0 is it

Answer 40

Yeah, unless n=1 of course. Otherwise, it blows up to infinity or to zero.

Answer 41

omg! yes the monster is just a nth derivative problem now xD and the original series or whatever it is - it converges to some non zero value only for the function f(x) = e^x. Oh there might be other smooth functions as well ?

Answer 42

I'm not sure, I'm investigating! What functions have a nonzero, finite, infinitieth derivative lol.

Answer 43

infinitieth derivative of sin(ax+b) is undefined

Answer 44

hyperbolics may work ?

Answer 45

nvm they are also dancing lol

Answer 46

Yeah haha.

Answer 47

Looks like the most general function might be:
\[f(x)=Ae^x\]
Hmm...

Answer 48

\(\large f^{(n)}(x)= \lim_{h\rightarrow 0} \int_{f^{(n-1)(x)}}^{f^{(n-1)(x-h)}} \frac{dx}{h^n} \)
so what next ?
do u mean that
\(\large f^{ }(x)= \lim_{h\rightarrow 0} \int_{f^{(n )(x)}}^{f^{(n )(x-h)}} f^{n}(x) dx\)

Answer 49

if its like this , then you can convert it into Nonlinear Integral equation

Answer 50

Hmm, I don't think it turns out like what you describe, but can you show me what nonlinear integrals are?

Answer 51

ok lol im not sure of that too , but nonlinear Integral equation is
\(u(x)=f(x)+\lambda \int_{a}^{b}K(x,t)u^{n}(t) dt \)

so if u have this equation then u can find u(x)


but i dint study it mmm

Answer 52

but i dint learn it yet ^^

Answer 53

hmm interesting. Never heard of this before. What's k(x,t)?

Answer 54

it could be any function :D

Answer 55

like this
K(x,t)=g(x)h(t)

Answer 56

its called the kernal

Answer 57

Hmm where did you hear of this? I want to try to find stuff about this so I can learn it.

Answer 58

from a book ( A First Course In Integral Equation )