Question

What is the antiderivative of

(sinx)^{1/2}?

Answer 1

@Zarkon
@amistre64
@TuringTest
@dpaInc
@satellite73

Answer 2

go fish for this one
just because you write down a function does not mean it has a nice closed form for an anti - derivative

Answer 3

The antiderivative of \[\huge \sin(x)^{1/2}\]

is asking for \[\huge \int \sqrt{\sin(x)} \ dx\]

As far as I know, this has no answer in terms of compositions of elementary functions...

Answer 4

what mathteacher said

Answer 5

in case you didn't know this, \(most\) functions are non-integrable

Answer 6

actually none are

Answer 7

none are non-integrable you say sat? could you elaborate plz?

Answer 8

measure is zero

Answer 9

in other words if you do a thought experiment, where you put all the functions (say on [0,1] into a hat, and pick out one at random, the probability it is integrable is 0

Answer 10

same as the probability you pick a rational number between 0 and 1 is also zero

Answer 11

of course that doesn't mean there are no rational numbers, so i was kind of being silly when i said none are, i didn't mean "none" i meant "almost all" are not

Answer 12

oh and i see from your response that maybe i was not clear. i didn't mean none are not integrable, i means none ARE integrable

Answer 13

ookay I gotchya now
That's a nice point about the probability thing though, which I know you are quite good at.

interesting, thanks :)

Answer 14

yw
you can do an easy thought experiment so show that the probability you pick a rational number at random between zero and one is zero.
imagine you pick your number at random by flipping a coin forever and writing down the binary decimal expansion, say head are 1 and tail are 0
in order for the number to be rational, you would have to get an infinite repeating pattern, which is not possible if the coin is fair.

Answer 15

>in order for the number to be rational, you would have to get an infinite repeating pattern, which is not possible if the coin is fair.

This is slippery argument. If each flip is independent of the previous, there is nothing to say that any sequence is any less likely than any other sequence. To assume that there is a greater chance of H if you have T T T T T T T T T T T T..... a miillion Ts .... T T T is false. the chance is precisely 1/2.

Such a sequence is very unlikely to occur, but not impossible.

Answer 16

it is impossible if it goes on forever, at least it would be if the coin is fair

Answer 17

btw it would not have to be repeating T to be rational, it could be any repeating pattern for example repeating HHHTHHHTHHHT...
but the point is the pattern would have to repeat infinitely often, and that is at least intuitively impossible for a fair coin.
you are right, this argument is not exactly precise, but it can be made precise. it is however (to my mind at least) a decent explanation that requires no measure theory and only rudimentary probability

Answer 18

I would like to see this proven rigorously.

Answer 19

Oh wait a minute. Wait a minute. It's coming back to me. Lebesgue integration gives us that the integral from x = 0 to x = 1 of the piecewise function
x = 1 if x is rational and
x = 0 if x is irrational

equals exactly 1 due to there being countably many rationals in [0,1] yet uncountably many irrationals in that same interval. I've not heard it argued from a probability standpoint, but it seems reasonable one is possible.

Answer 20

your integral would be 0

Answer 21

Grr, right because x = 0 if x is irrational.