Question

Theory time: started improper integrals today. I get that the area under the curve from 0 to infinity for 1/x is infinity. As x approaches infinity 1/x gets close to zero but never quite gets there so that is divergent.
But 1/x^2 -- which also never quite gets to zero is convergent...seems to me that 1/1000000000^2 while pretty close to zero...still isn't zero and should still be divergent. In fact, that 1/really big number squared is on the same line at 1/really big number not squared, ain't it?
Or should I just get over it and say okay I'll calculate it out even if I don't believe?

Answer 1

Do you know how to prove that integral 1/x diverges and 1/x^2 does not?
that would help you believe

Answer 2

...by the way the integral is usually done from 1 to infinity because 1/0 is undefined of course

Answer 3

Ya we did that in class...it just didnt make any sense. Right but really it doesn't matter where you start it is the tail end that doesn't make any sense.

Answer 4

Neither function actually reaches zero --- so you would keep adding some to the area and even a little bit of infinity is a lot

Answer 5

This is the same problem that everybody has with the geometric proof that\[\lim_{x \rightarrow 0}\frac{\sin x}x=1\]people often say 'but it's not defined at 0/0' but this doesn't matter.

Limits ask about the behavior of a function near a point. When we evaluate an improper integral we ask how the function behaves in the limit. In this case we are asking how fast the function diminishes as we let x approach infinity.
Limits, as you probably know, do not need to be defined at the actual value in question, they just need to get arbitrarily close (remember the delta-epsilon thing).

It turns out that the behavior of 1/x^p in the limit decreases fast enough to converge whenever p>1. How fast a function approaches zero makes all the difference in a case like this.

I'm sorry if I can't make it obvious for you, you need to develop an intuition on this kind of thing.

Answer 6

How fast a function approaches zero changes answers enormously.

this is why 0^0 is not defined

0^0 has no meaning, but we can define it in terms of limits\[\large \lim_{x \rightarrow \infty}(\frac1x)^{e^{-x}}\]has a different value than\[\large \lim_{x \rightarrow 0}x^x\]even though both go toward 0^0, they approach it in a different way, and so we get different values.

Hence the speed of convergence on a value is important in limits

Answer 7

here's a specific example\[\lim_{x \rightarrow 0^+}x^0=1\]\[\lim_{x \rightarrow 0^+}0^x=0\]this illustrates the difference made by how we approach our limit, and how the values behave in that region very close to the x-value in question

again, developing intuition is required. this is not obvious, nor easy to explain without hard proof

Answer 8

for a more intuitive example think of zeno's paradox
it leads to the idea that\[\sum_{n=1}^{\infty}\frac1{2^n}=1\]how can an infinite series of non-zero, positive terms add to a finite number?
how does the fox catch the rabbit? (or whatever version you know)

Answer 9

I guess I am just used to the math making sense when translated into english. In this case, the graph of 1/x^2 does not get to zero so even way out at 1/reallybignumber^2 you would still have to add something to the area under that curve. I'm ok with just going with the "what I'm told". And as for developing an intuition with math. I'm resigned to the fact that that boat has sailed. I'm just trying to survive through calc 3.

But I'll go back to reread the limits sections -- maybe I'll find something there. But limit 1/x as x goes to infinity is zero; and 1/x^2 as x approaches infinity is also zero....

\[\int\limits\limits\limits_{1}^{\infty}1/x dx = \ln x] \ln \infty = \infty\]
and infinity -1 is still pretty big

\[\int\limits_{1}^{\infty}1/x^2 = -1/x]\]

-1/ reallybignumber is almost zero; almost zero minus negative one is almost one.

but don't understand why 1/(reallybig n)^2 + 1/(reallybig n+1)^2 doesn't keep adding up while 1/(reallybig n) + 1/(reallybig n+1) does. Maybe because I don't understand the difference between infinity and infinity^2.

Answer 10

Maybe with is newton letting me down. I never did trust that guy since I found out about the alchemy and the mercury madness.


thanks for the time and effort

Answer 11

with is? this is

Answer 12

remember that infinity is not really a number, so there is no difference between 1/infty and 1/(infty)^2

also remember that we are not asking about the lim x to infty of 1/x and 1/x^2, but the limit of their partial sums.

Forget about what happens to the function AT infinity, that point does not exist.

maybe a comparison with an interpretation of zeno's paradox can help:

let a pie chart represent the distance traveled by the moving thing in the paradox
1/2+1/4+1/8+1/16+...
looks like thisas you add each successive distance you can see we are only getting closer and closer to filling the pie (which means we are summing to 1) and we can never pass it.

Try that same pie chart trick with the sequence 1/2+1/3+1/4+1/5...
and you will notice that you never approach a particular value, you just go around and around the circle forever....