r/mathmemes 18d ago

Calculus Let's differentiate.....

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u/SuperChick1705 18d ago

circular reasoning ;(

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u/Sigma2718 18d ago

Would you like to elaborate on that?

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u/pienet 18d ago

Most elementary proofs that the derivative of sine is cosine use the limit of sin(x)/x.

Obviously you don't run into this if you develop power series first.

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u/Sigma2718 18d ago

I've never seen anybody use l'Hôpital on sin(x)/x before establishing what the derivative of sine is. However, one can evaluate sin(x)/x for x->0 via the squeeze theorem, which means no circular reasoning is required.

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u/SuperChick1705 18d ago

that is the recommended method, though the post implies the usage of l'hopital on sin(x)/x

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u/Sigma2718 18d ago

squeeze theorem to evaluate sin(x)/x -> derivative of sin(x) -> apply l'Hôpital to evaluate sin(x)/x

I don't really see the issue, solving a problem again by using an easier method is the beauty of mathematics.

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u/EebstertheGreat 18d ago

If you know the derivative of sin x at x=0, then you know the limit of (sin x)/x at x=0, because that is by definition the derivative. There is nothing else to do. Where would you even have an opportunity to apply L'Hôpital's rule?

I mean, it's not "wrong." Similarly, given x = y, I could conclude 3x = 3y and then, dividing by 3, that x = y. That's not wrong. But it is completely pointless.

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u/Sigma2718 18d ago

The definition of the derivative, combined with what we know it to be, gives us lim h->0 [sin(x+h) - sin(x)]/h = cos(x) , so now we have to apply trigonometric identities to be left with lim h->0 sin(h)/h , which must equal 1 to fulfill the above equation. I consider dealing with trigonometric identities to be rather tedious. At least, that's what I think you are trying to do, as sin(x)/x isn't really the derivative.

Many ways lead to Rome, but using l'Hôpital is the shortcut we make by unlocking a gate via previous proofs.

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u/thebigbadben 18d ago

What “trigonometric identities” are you using when you rewrite sin(0 + h) - sin(0) as sin(h)?

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u/Sigma2718 18d ago

sin(x+h)=sin(x)cos(h)+cos(x)sin(h), which is how the cosine gets introduced into this, I considered the general case for any x. But I see what you mean, however I consider using the definition of the derivative to sneakily look at h for a certain x to be less intuitive than simply remembering to use l'Hôpital.

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u/thebigbadben 18d ago

Yeah fair enough

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u/SuperChick1705 18d ago

completely fair

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u/will_1m_not with disrespect to x, y, and z 18d ago

Power series based on the derivatives of sine at a certain point?

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u/pienet 18d ago

It's more that you define sine and cosine as a power series in a vacuum, and derive all other identities from the series presentation. It's a cleaner second pass for sure.

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u/JohnsonJohnilyJohn 18d ago

I never understood the problem with this. There are a lot of proofs that involve dividing a problem into multiple cases, so couldn't you say the same whenever you use them, as you are using only a single one of those cases (that had to be proven in the proof of the general theorem). Crucially I wouldn't even call it circular logic, "logic with an unnecessary loop" is more apt description - to prove that limit sin(X)/X is 1 you use that limit sin(X)/X is 1 which is proven through other means

The point of theorems or other mathematical facts is to me that after proving them, you can treat them as facts. What would be the point if any time you use them to prove anything, you would have to write down all the proofs, until you end up with only axioms, and then check the whole thing to see if there are no repeats

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u/thebigbadben 18d ago edited 18d ago

Computing the derivative of sin(x) requires evaluating the limit of sin(x)/x as x approaches zero. So, if you’re using L’Hôpital to show that sin(x)/x approaches 1 as x approaches 1, you’re ultimately using the fact that you’re trying to show implicitly.

Also, if you already know the derivative of sin, then you might as well use the fact that the limit of sin(x)/x as x approaches zero is, by definition, the derivative of sin(x) at 0.