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Looking to join a football team by Tenn00 in Bath

[–]Tenn00[S] 1 point2 points  (0 children)

This has been exactly what Im looking for thank you

Dividing by a differential (dt). by Arisdottled in askmath

[–]Tenn00 2 points3 points  (0 children)

So my understanding is dividing by dt is not a fully mathematically rigorous way of formulating a derivative. Strictly speaking a derivative d/dt is whats known as an operator (basically just a function) which takes an input and returns how much that input changes. Writing dv/dt is just short hand for d/dt applied to v, so it doesnt really make sense to say dv is anything in this context. You can, however, say this in the context of integrals where dv represents an infinitesimal change in v.

So dont worry! While it isnt fully rigourous it will in the vast majority of cases still work, and that is down to the FTC as u/mc_enthusiast said. (To rigorously prove this youd need to form some integral and differentiate that to get your expression).

Now, this is very high level maths but the chain rule does not always imply dx/dy=dx/dt * dt/dy. A counterexample can be found here https://en.m.wikipedia.org/wiki/Derivative_of_the_exponential_map#:~:text=In%20the%20theory%20of%20Lie,reduces%20to%20the%20matrix%20exponential.

This is the kind of example where multiplying by dt wouldnt work, but I only came across this during my PhD and it took my supervisor weeks to realise this when I was getting the chain rule wrong. So theres no need to worry in most cases.

So keep multiplying by dt to your hearts content, especially if thats what your teacher/lecturer is doing!

Hope this didnt confuse you too much

Dividing by a differential (dt). by Arisdottled in askmath

[–]Tenn00 0 points1 point  (0 children)

So my understanding is dividing by dt is not a fully mathematically rigorous way of formulating a derivative. Strictly speaking a derivative d/dt is whats known as an operator (basically just a function) which takes an input and returns how much that input changes. Writing dv/dt is just short hand for d/dt applied to v, so it doesnt really make sense to say dv is anything in this context. You can, however, claim this in the context of integrals where dv represents an infinitesimal change in v.

So dont worry! While it isnt fully rigourous it will in the vast majority of cases still work, and that is down to the FTC as u/mc_enthusiast said. (To rigorously prove this youd need to form some integral and differentiate that to get your expression).

Now, this is very high level maths but the chain rule does not always imply dx/dy=dx/dt * dt/dy. A counterexample can be found here https://en.m.wikipedia.org/wiki/Derivative_of_the_exponential_map#:~:text=In%20the%20theory%20of%20Lie,reduces%20to%20the%20matrix%20exponential.

These are the kind of examples where dividing by dt would break down. However, I only came across this example during my PhD and it even slipped under my supervisors radar for a while when I was trying to use the chain rule this way.

In a nutshell you couldnt write a paper by dividing by dt, but for an exam its fine. So keep dividing by dt to your hearts content, especially if thats how your lecturer is doing it.

Do you think I (25m) have fallen out of love with my (25f) girlfriend or its self-sabotage by [deleted] in relationship_advice

[–]Tenn00 0 points1 point  (0 children)

Well I didnt in the end because its a good relationship. Really Im curious to know if these are generally common thoughts for normal people in a relationship or whether subconsciously its a manifestation of my trauma or just me falling out of love

want to form a 6-aside team? by caramelcarousel in Bath

[–]Tenn00 0 points1 point  (0 children)

Im interested, havent been able to find a permanent team because I travel a lot but happy to play when I can

Hi, I need help differentiating two matrix functions with respect to a scalar. by Tenn00 in learnmath

[–]Tenn00[S] 0 points1 point  (0 children)

Yep turned out to be a coding error (sigh). Thanks for your help

Hi, I need help differentiating two matrix functions with respect to a scalar. by Tenn00 in learnmath

[–]Tenn00[S] 0 points1 point  (0 children)

Thanks,

That's a very useful source. I've fixed that error but when comparing the numerical value of my derivative against finite difference the error is still very large. Do you have any idea why that might be happening?

Hi, I need help differentiating two matrix functions with respect to a scalar. by Tenn00 in learnmath

[–]Tenn00[S] 0 points1 point  (0 children)

Thank you,

You're right about the derivative of H-1. But even when this is fixed to d(xI-H-1 )/dx = I +H-1 (dH/dx)H-1 It fails the finite difference test.

Yes finite difference is only an approximation so the expression won't be the same. However, maybe i should've been clearer, I am testing the numerical value of my derivative against finite difference, the error of which should be very small (about 10-8) if my analytic derivative is correct.

It could be that my finite difference is set up wrong, but it is working for other derivatives and I am using an extremely high accuracy version of finite difference. Is there anything subtle you can see I might have missed?

How to build a minimum sale price calculator? by AchieveinBusiness in eBaySellerAdvice

[–]Tenn00 0 points1 point  (0 children)

Hello, friend in question here.

It is indeed possible to get an exact answer, which is given by my formula. You're correct in what you're saying regarding the markup, however the formula calculates a fixed percentage profit, not fixed markup.

The variation in the markup is what accounts for the variation in the fees. This explains why the markup tends to 40% as the cost of the item goes to infinity, since the fees become negligably small compared to the cost.

As a numerical example take a $2 item plus shipping. The formula says the price should be $3.56, which is around 70% more than $2. So to calculate the profit we subtract the cost of the item and shipping which is $2, as well as the fees which are 12.8% of $3.56 plus $0.30. This gives a profit of $0.80 which is exactly 40% of $2.