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November 21[edit]

Antiderivative of the Weierstrass function[edit]

Since all continuous functions are integrable, the Weierstrass must be integrable. What does its indefinite integral look like? (unfortunately I don't have any program that can graph it and don't fancy doing by hand!). Thanks 24.92.74.238 (talk) 00:53, 21 November 2012 (UTC)[reply]

Nothing special: it's just a sin function (you can integrate term by term in the summation). image. 70.162.4.242 (talk) 06:49, 21 November 2012 (UTC)[reply]

why is arithmetic meaningful[edit]

If I give you a formula for how long it takes to repay the principal given a certain interest rate and monthly installment amount, and, for example you pay $200 per month. But when you plug that and the principal and interest rate into the formula, you get a division by zero. You realize something is wrong.

Well, maybe the interest payment come to $200 per month exactly. So you never pay it off - fair enough that it's a division by 0. But my question is about all the OTHER cases where you don't stop to think 'wait a minute'. If the formula gives you 24 months, then you "believe" it. My question is that how do I know that mathematics is credible in its answers, given that it can require rethinking the input when giving a bad output. In other words, who is to say that any formula is ever actually correct? Maybe they are all incorrect, all the time, and just so "happen" to be right whenever they prove to be, i.e. whenever we check. Much like a fallacious answer of "18 months" might 'prove' to be correct if 18 months from now your debt is forgiven. --178.48.114.143 (talk) 21:26, 21 November 2012 (UTC)[reply]

Mathematicians tend to prove their arithmetic formulas using algebra. If you do not understand the proof, then you may check simple cases by inserting small numbers into the formula, and then you may feel increasingly convinced (provided these checks turnes out to be correct) even if the formula is not really proved to your satisfaction. If you really understand the proof, then you will not doubt the correctness of the formula. Bo Jacoby (talk) 22:35, 21 November 2012 (UTC).[reply]

Where things really matter one can use computer aid in proving systems conform to specifications. It is far less likely then that a plane will flip upside down when it goes over the equator as the specification of what's to be ensured is usually much shorter than the lines of code but it is still possible as anyone can make a mistake. Dmcq (talk) 23:19, 21 November 2012 (UTC)[reply]

Proof (mathematics)? Don't worry, I'll be more specific.

In the example, you would draw some connections, and try to "see what the formula actually does". Let's say you have a credit of $800 and an annual interest rate of 25%. That's $200 for interest and nil for debt reduction. Obviously, the formula will not work out.

But let's look what happens if you're less than $800 in debt. Let's call the amount between $800 and your debt x. Now if your debt is 800 minus x, the interest you'll have to pay 0.25 times (800 - x) , which simplifies to 200 - x/4, and so the remaining x/4 goes into debt reduction, and after one year, you are at a debt of 800 - x - x/4 which is 800 minus (5/4)x.

The point is that it doesn't matter what x is, as long as it's positive and <$800. x increases by a factor of 5/4 per year. So the question is, "how many times do I have to multiply by 5/4 until x equals/exceeds $800?" , or to put it in a formula, the smallest integer n satisfying (5/4)ⁿ x ≥ 800. If you divide by x, you get

(5/4)ⁿ ≥ 800/x, and that's where the division by zero comes from if you don't actually reduce your debt.

What the common interest formulas do is just that. They compute something a physicist would call an event horizon, namely the debt which could never be reduced because the payments equal the interest ($800 in the example), then they verify if you are on the right side of it, and calculate the exponential divergence from that point using a logarithm.

Logarithms turn the formula we have into another one with only n on one side; in our example that would be

n = _(5/4)log (800 / x). This is not always an integer but you can round up.

If you generalize the formulae even more, you can get the following,

n = _(1+r)log (D / x).

D is the critical debt, and r is the interest rate, 25% = 0.25. It is quite easy to see that D = p/r if p is the annual payment. You can either lump all this together for a huge formula or compute D and x and then apply the formula above. In practice, I find the D and x approach more useful, as I can see how "bad" the debt really is. In a homework problem or similar (without real debt issues), both work equally well but I find the D and x method still more intuitive than learning a huge formula. YMMV. - ¡Ouch! (^{hurt me} / _{more pain}) 13:46, 23 November 2012 (UTC)[reply]

There is an important insight hidden in the original question: You cannot "trust" a naked formula to work for a given application problem. There are always assumptions, and you need to understand them, or at least be aware of the limitations. The simple high-school formula for gravitational potential energy strictly only apply if the gravitational field is constant. It works well enough for heights of a few kilometers on earth, but not for computing the escape velocity. --Stephan Schulz (talk) 16:14, 23 November 2012 (UTC)[reply]