Invisible wrote:
>> If it's a limit, just compute it for numbers as close as possible to 
>> the limiting value (or really really big numbers if the limiting value 
>> is infinite).  The closer (or the bigger) the number you compute it 
>> for the closer your answer will be.
> 
> Wouldn't that be increadibly unstable, numerically?

How much a a problem this is depends on the limit.  For most limits that 
I've ever numerically computed it's worked fine.  If you're taking the 
limit of a series (as opposed to the limit of a function) then it'll 
almost certainly be fine.


>> If it's an integral do something like this: 
>> http://en.wikipedia.org/wiki/Riemann_sum
> 
> I see. (Although I'm still not sure how you compute an infinite integral 
> this way...)

Combine this definition with how you compute the limit of an infinite 
sum.  So you'll have two approximations, one in approximating the 
integral with an infinite sum, and another with approximating the 
infinite sum with a finite one.  If you're doing this, it'd be wise to 
prove that these approximations both converge to the correct value.

>>> But you can't compute an infinite product.
>>
>> If the product converges (which it does, otherwise you couldn't define 
>> a number/function with it) then you can by definition get an 
>> arbitrarily good approximation by computing the product of the first n 
>> terms for a large enough n.
> 
> I don't see how that is the case.
> 
> If you have an infinite *sum*, then as long as the terms get 
> progressively more tiny and never get larger again, you can disregard 
> all the terms after a certain point. But if you're taking a *product* 
> then any term, anywhere in the series could radically alter the final 
> result.

The same is true for a sum actually.  The point is proving that an 
infinite sum or product *converges* is to show that this doesn't happen. 
  Basically you show that the partial sums/products form a convergent 
sequence (see the "Limit of a sequence" Wikipedia article I linked). 
Then you know by definition that you can get a good approximation by 
computing a large enough partial sum/product.

In the case of a sum this requires that the terms progressively get 
closer to zero, for a product it'll require they get closer to one.


>> Out of curiosity have you ever had a calculus class?
> 
> I've never had *any* maths class!
> 
> (Unless you count what we did at school. This simply involved filling 
> out hundreds of thousands of pages of long-division problems over a 
> 7-year period...)
> 
> Hypothetically I shouldn't be able to do algebra at all...

That's too bad, it seems like you would have greatly enjoyed being able 
to do some math that was more advanced than long division.

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