Question

I came across an interesting limit, tell me what you think.

Answer 1

\[\lim_{x \rightarrow 0} (\frac{ 1 }{ x^2 } + \frac{ - \sin x }{ x^3 })\]

What's the answer?

Answer 2

sum property

Answer 3

is the answer -1?

Answer 4

you can just use l'hospital a bunch a few times

Answer 5

write as one fraction of course

Answer 6

then do the whole l'hospital thing

Answer 7

The answer's not -1 @Flop

Answer 8

can you not multiply by x^2 so you would get 1-sin(x)/x = 1-1=0

Answer 9

its actually 0?

Answer 10

It is zero, yeah.

Answer 11

Nope @Jemurray3

Answer 12

Oh my fault.. 1

Answer 13

hint: it is something between 0 and 1

Answer 14

i agree with @myininaya 's idea
we have a 0/0 form by substitution so l'hopital's rule looks good (in two cases)

Answer 15

the divergence cancels out, was the point.

Answer 16

1/3

Answer 17

not 1/3

Answer 18

but getting warmer

Answer 19

l'hospital would work but there has to be a shorter way

Answer 20

l'hospital isn't long here
it is actually very short and cute

Answer 21

oh balls I can't divide. 1/6

Answer 22

that's great

Answer 23

Despite my apparent inability to do fractions, the main principle is what I meant -- expanding in a power series cancels the divergence and you just pick up the leftovers.

Answer 24

\[-\sin x=-x+\frac{ x^3 }{ 3! }-...\] \[-\sin x+x=\frac{ x^3 }{ 3! }-...\]
\[\frac{-\sin x+x}{x^3}=\frac{ 1 }{ 3! }-...\]

The rest of the terms are dependent on x so they go away. =P

Answer 25

^ That's what I get for trying to do eight things at once... but I'm a big supporter of power series over l'Hospital when possible.

Answer 26

now that is creative

Answer 27

Just playing around with trig stuff and came across that and thought I'd share it. If anyone else wants to show interesting/fun trig/powerseries/limit stuff go for it!

Answer 28

I've got a nice integration trick, something cool about power series, and an unusual limit/iterated function/approximation trick. Take your pick.

Too much of that rhymed.

Answer 29

ALL!

Answer 30

Or your favorite, I have time lol. Integration if you still can't pick.

Answer 31

lets go with the integral first.....i should be doing physics

Answer 32

i do like integral tricks, although all of the above is still as good a choice as any. >.>

Answer 33

Alrighty -- I'll just keep going in this thread until I get yelled at by the establishment.

Answer 34

So we are imagining that we have an integral of the form

\[ \int dx \space f(g(x)) \cdot g(x) \]
Notice that the second function is g(x), not g'(x) -- otherwise this would be a straightforward antiderivative.

Answer 35

what I'm going to do is invent a parameter lambda, so I write my integral as a function of lambda:

\[ I(\lambda) = \int dx \space f(\lambda \cdot g(x)) \cdot g(x) \]

Answer 36

I'm listening... =)

Answer 37

I notice something lovely - that I can rewrite the integrand as follows:
\[ I(\lambda) = \int dx \frac{\partial}{\partial \lambda} f(\lambda\cdot g(x)) \]
and since the parameter lambda and the integration variable x are completely independent, I'm free to exchange the integral sign and the partial derivative:
\[I(\lambda) = \frac{d}{d\lambda} \int dx \space f(\lambda\cdot g(x)) \]

Answer 38

Wait a sec... Are you going to take the derivative wrt lambda and ultimately set lamda = 1? I'm

Answer 39

That's the idea, yes! Though it should be noted that you can do other things with this as well.

Answer 40

Yeah, I'm curious how that goes, since you'll have t integrate and slve fr a constant of integration and wow this keyboard is broken.

Answer 41

The point is, if you are able to perform the remaining integral, then differentiation wrt lambda and setting lambda = 1 solves the problem. It may perhaps seem artificial, but allow me to give you an example:

Answer 42

there's one way to do it without series or l'hopital, change x = 3y

Answer 43

Hmm that's really clever, I was thinking of something slightly different.

Answer 44

It's well known that
\[ \int_{-\infty}^\infty dx \space e^{-x^2} = \sqrt{\pi} \]
It can similarly be shown straightforwardly that
\[ \int_{-\infty}^\infty dx \space e^{-ax^2} = \sqrt{\pi/a} \]

So what happens when we have (suppressing the limits, which go from - infinity to infinity)
\[ \int dx \space x^2 e^{-x^2} \]
?

Answer 45

You can use integration by parts to solve this, but what I'm gonna show in a second will demonstrate IBP's limitations in this problem.

Answer 46

\[ I(\lambda) = \int dx \space x^2 \cdot e^{-\lambda x^2} = -\frac{d}{d\lambda} \sqrt{\pi/\lambda} = \frac{\sqrt{\pi}}{2\lambda^{3/2}}\]
so
\[I(\lambda) = \frac{\sqrt{\pi}}{2} \]

Just like you said. Now let's stretch our wings a little bit--- what about
\[ \int dx \space x^n \cdot e^{-\lambda x^2} \]

Answer 47

Because of the parity of the problem, it's clear that if n is odd, the integral is equal to zero. But if it's even, and we define m = n/2, then we can write down the answer:

\[ I(\lambda) = \frac{d^m}{d\lambda^m} \sqrt{\frac{\pi}{\lambda}} \]

Answer 48

Which you can simplify however you like. Now let's go for the big one: What about
\[\int dx \space f(x) e^{-x^2} \]
?

Let's assume we know the power series of f(x) around the point x = 0. I.e.
\[ f(x) = \sum_{n = 0}^\infty \frac{f^{(n)}(0)}{n!}x^n \]

We can immediately write down our integral:

\[\int dx \space f(x) e^{-x^2} = \sum_{m = 0,2,4...} \frac{f^{(n)}(0)}{n!}\frac{d^m}{d\lambda^m} \sqrt{\frac{\pi}{\lambda}} |_{\lambda = 1}\]

Answer 49

Oops! Also, there should be a -1 term in there:

\[ \int dx \space f(x)e^{-x^2} = \sum_{\text{even m}} (-1)^m\frac{f^{(m)}}{m!} \frac{d^m}{d\lambda^m} \sqrt{\frac{\pi}{\lambda}}|_{\lambda = 1} \]

Whew! Sorry for the numerous typos, but that should be right.

Answer 50

Well damn, that's clever!

Answer 51

But does the power series simplify further into something nicer?

Answer 52

Sigh.. one more typo on that last line, it should be
\[ f^{(m)}(0) \]

And it depends entirely on the function f(x), which we haven't specified.

Answer 53

That's the nice part -- that represents the integral for *any* function f(x) that has a continuous power series. So really, just analytic functions.... but still, that's quite a variety you could have there.

Answer 54

I get the reasoning though, makes perfect sense since the power series turns it into a polynomial and only the even terms won't be zero, so we're good there.

What about if we know their fourier series, can we do anything interesting there?

Answer 55

Hmm... if we knew the Fourier series, we'd still end up with a bunch of sines and cosines, which would be difficult to integrate.

Answer 56

Here's an interesting question which I just thought of, and will be spending some time on ... if f has a power series with a particular radius of convergence, i.e. it DIVERGES past a certain point, could we still use it here? The exponential part would kill off all of the polynomial terms as they go to infinity, so almost everything would converge in principle.

Answer 57

Have you seen this before? If a is a constant greater than or equal to zero.

\[1=\int\limits_{0}^{\infty} ae^{-ax}dx\] This integral is equal to 1 by simply solving it. Pull out the constant a and you get:\[a^{-1}=\int\limits_{0}^{\infty} e^{-ax}dx\]differentiate both sides with respect to a: \[a^{-2}=\int\limits_{0}^{\infty} xe^{-ax}dx\] the negative signs cancel! \[2a^{-3}=\int\limits_{0}^{\infty} x^2e^{-ax}dx\]\[3!*a^{-4}=\int\limits_{0}^{\infty} x^3e^{-ax}dx\]keep going... \[\frac{ n! }{ a^{n+1} }=\int\limits_{0}^{\infty} x^n e^{-ax}dx\] and of course for a=1\[n!=\int\limits_{0}^{\infty} x^n e^{-x}dx\]

Answer 58

I've never seen that before -- that's an awesome way to arrive at the gamma function.

Answer 59

@Jemurray3 Actually isn't that kind of the idea behind the Laplace transform? I think you're right as long as you can show that e^{-x^2} dies out faster.

Answer 60

Have you ever heard of a pade approximant?

Answer 61

Nope, what's that?

Answer 62

Let me start another thread and introduce it to you quickly. It's an active area of research and pretty damn mysterious, but it is related to the subtlety of the question I asked.

Answer 63

Hmm sounds awesome, thanks!