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bio website ephorie.de
location Germany
age 44
visits member for 4 years, 6 months
seen 7 hours ago

just an amateur fascinated by math


May
15
awarded  Necromancer
May
15
comment Example for finite dimensional analog of integral transforms
I tried this with Mathematica and the vector $\{1,2,3,4,5\}$. The result after multiplying with $B(z)$ and transforming back should be $\{1,1,1,1,1\}$ because these are the first differences, right? (I am not sure about the last one, but anyway). Interestingly after multiplying with $B(z)$ I get $\{1,2,3,4,5\}$ and transforming this back gives the discrete delta function times $\{1,2,3,4,5\}$: wolframalpha.com/input/… - what's wrong?!?
May
15
comment Example for finite dimensional analog of integral transforms
This is exactly what I was looking for! Thank you very much indeed! :-)
May
15
accepted Example for finite dimensional analog of integral transforms
May
14
comment Example for finite dimensional analog of integral transforms
@MattL. Ok, I tried this with vectors but there are still things that don't work out :-( If you gave me an example with the z-transform on vectors where multiplication/division becomes differentiation/integration I would happily accept your answer :-) Thank you
May
14
reviewed Approve Why does $\sin(0)$ exist?
May
14
comment Discrete Laplace Tranform.
I asked a similar question here: math.stackexchange.com/questions/793550/… - what I find strange though is that while you can also read that the Laplace transform is the continuous analog of the dot product for infinite dimensional vectors (=functions) you still calculate infinitely many terms in the discrete version (and not just over the available dimensions of your vectors as with the dot product) - perhpas you could contribute :-)
May
14
comment Discrete Laplace Tranform.
I asked a similar question here: math.stackexchange.com/questions/793550/… - what I find strange though is that while you can also read that the Laplace transform is the continuous analog of the dot product for infinite dimensional vectors (=functions) you still calculate infinitely many terms in the discrete version (and not just over the available dimensions of your vectors as with the dot product) - perhpas you could contribute :-)
May
14
comment Example for finite dimensional analog of integral transforms
@MattL.: Thank you, what still bothers me with this finite version is that the sum is over infinitely many terms because the dot product is only over the available dimensions.
May
13
reviewed Approve Trigonometric substitution
May
13
comment Example for finite dimensional analog of integral transforms
@MattL.: I had a quick look at it and it looks promising - so first thank you! Could the $z$ also be real valued like e.g. $2$?
May
13
comment Example for finite dimensional analog of integral transforms
@MattL.: To be honest with you I don't know because I don't know the Z-transform well enough. But perhaps you could give an example that contains the elements I asked for in my question?
May
13
asked Example for finite dimensional analog of integral transforms
May
9
awarded  Notable Question
Apr
28
awarded  Nice Question
Apr
28
revised Intuition for complex eigenvalues
edited title
Mar
9
accepted Intuition and counterexamples for higher-order derivative test
Mar
8
comment Intuition and counterexamples for higher-order derivative test
I was just talking about your example: In our axiomatic system it is not defined at point $0$ because you are dividing by zero there! I am not talking about nature but only about things happening within the realms of pure maths. Yet could you give me a reference for the number of smooth functions and formulas accessible? Thank you again.
Mar
8
comment Intuition and counterexamples for higher-order derivative test
The intuition is appealing - thank you and +1! Considering your example: I disagree: It does not have a minimum at zero - it has literally nothing at zero because it is not defined there, neither are any of its derivatives (see also comments to my question).
Mar
8
comment Intuition and counterexamples for higher-order derivative test
Is it possible to find a counterexample where you don't have a separate case at the extreme value (as with this example at $0$)? This feels a little bit like cheating because the "original function" is not defined at this point.