question Maths for purchasing domains in bulk

[Zain110]

i want to understand maths . Like if i buy 100 plus domains as a bull purchase and if it costs me around $2 each , does it mean that 100 domains will cost me $200 for 1 year ?

 

[SCP Foundation]

it's a hard question i would say

if we modelise the problem with a 2-variables function, and we then use the Gauss composition for the quadratic forms of that function, we could then consider the derivativity of that function on the R² set, which would allow us to calculate the corresponding hertzian matrix, we could then state that depending on the signature of the quadratic form and the positivity or the negativity of the trace of the hertzian matrix (considering that the determinant would obviously be positive, as the original function is certainly at least C1 on R²), we would be able to find the critical points (where both of the 1st partial derivatives are 0) whose value would correspond to the complex form of the amount you should pay a year for those domains

consider we have a signature of ( 100*sqrt(3) , 100 ), we then could state that the corresponding complex we found is z = 100*sqrt(3) + 100i whose modulus is |z|= 200

we can state with this that if you buy 100 domains at 2$ each then it would cost you 200$ the 1st year (if there are no discounts in the bulk purchase and if we don't take into account the renewal fee for the years after the 1st year, because then we would need a C² or C³ 2-variable function for modelisation)

if someone thinks i'm wrong please correct me

 

[Armageddon]

 

[NickB]

Yes.......unless you register a domain/s at a promotional price......then it will cost more to renew

That $2 figure seems to be a promo, make sure to check the renewal pricing

 

[forge]

if someone thinks i'm wrong please correct me

That's far too simplistic imo.

 

[pb]

i want to understand maths . Like if i buy 100 plus domains as a bull purchase and if it costs me around $2 each , does it mean that 100 domains will cost me $200 for 1 year ?

If by "buy" you mean "register" (free domains), then yes.
If by "buy" you mean actually "buy" (already registered domains), then no, you will only own them until their expiration date.

 

[karmaco]

I have a math question too. Where are you going to get domains for $2 a year? Make sure you check the renewal prices. They are likely sky high.

Cheap isn’t always better if the registrar sucks.

200 Dot coms are going to cost you at least 2K a year to renew and 365 days goes very fast in domain land.

 

[VadimK]

it's a hard question i would say

if we modelise the problem with a 2-variables function, and we then use the Gauss composition for the quadratic forms of that function, we could then consider the derivativity of that function on the R² set, which would allow us to calculate the corresponding hertzian matrix, we could then state that depending on the signature of the quadratic form and the positivity or the negativity of the trace of the hertzian matrix (considering that the determinant would obviously be positive, as the original function is certainly at least C1 on R²), we would be able to find the critical points (where both of the 1st partial derivatives are 0) whose value would correspond to the complex form of the amount you should pay a year for those domains

consider we have a signature of ( 100*sqrt(3) , 100 ), we then could state that the corresponding complex we found is z = 100*sqrt(3) + 100i whose modulus is |z|= 200

we can state with this that if you buy 100 domains at 2$ each then it would cost you 200$ the 1st year (if there are no discounts in the bulk purchase and if we don't take into account the renewal fee for the years after the 1st year, because then we would need a C² or C³ 2-variable function for modelisation)

if someone thinks i'm wrong please correct me

Hmmmm.... that's very interesting... I would say to that if XX is a non-locally convex space, it is easy to construct a sequence of simple functions sn:[0,1]→X,sn(t)=∑m=1nχAm,n(t)xm,n,sn:[0,1]→X,sn(t)=∑m=1nχAm,n(t)xm,n,
where (Am,n)nm=1(Am,n)m=1n is a partition of the interval [0, 1] for each n∈Nn∈N, and χχ denotes the characteristic function, such as: sup1≤m≤n∥xm,n∥→0,∑m=1nμ(Am,n)xm,n↛0,sup1≤m≤n‖xm,n‖→0,∑m=1nμ(Am,n)xm,n↛0,
as n goes to infinity, where μμ denotes the Lebesgue measure. Therefore, Bochner–Lebesgue integration cannot be extended to non-locally convex spaces. On the other hand, the definition of the Riemann integral extends verbatim for functions defined on an interval [a, b] with values in an F-space XX. However, it has some problems in the non-locally convex setting. For example, Mazur and Orlicz proved that the F-space XX is non-locally convex if and only if there is a continuous function f:[0,1]→Xf:[0,1]→X which is not Riemann integrable. But the main drawback is that the Riemann integral operator IRIR, acting from the set of XX-valued simple functions S([a,b],X)S([a,b],X) to XX by IR(∑j=1nxjχ[tj−1,tj))=∑j=1n(tj−tj−1)xj,IR(∑j=1nxjχ[tj−1,tj))=∑j=1n(tj−tj−1)xj,
isn't it continuous when XX is not locally convex, what do you think??

 

[Ostrados]

Tricky question I had to ask ChatGPT:

Omg ChatGPT did its best to make the answer looks complicated

 

[SCP Foundation]

Hmmmm.... that's very interesting... I would say to that if XX is a non-locally convex space, it is easy to construct a sequence of simple functions sn:[0,1]→X,sn(t)=∑m=1nχAm,n(t)xm,n,sn:[0,1]→X,sn(t)=∑m=1nχAm,n(t)xm,n,
where (Am,n)nm=1(Am,n)m=1n is a partition of the interval [0, 1] for each n∈Nn∈N, and χχ denotes the characteristic function, such as: sup1≤m≤n∥xm,n∥→0,∑m=1nμ(Am,n)xm,n↛0,sup1≤m≤n‖xm,n‖→0,∑m=1nμ(Am,n)xm,n↛0,
as n goes to infinity, where μμ denotes the Lebesgue measure. Therefore, Bochner–Lebesgue integration cannot be extended to non-locally convex spaces. On the other hand, the definition of the Riemann integral extends verbatim for functions defined on an interval [a, b] with values in an F-space XX. However, it has some problems in the non-locally convex setting. For example, Mazur and Orlicz proved that the F-space XX is non-locally convex if and only if there is a continuous function f:[0,1]→Xf:[0,1]→X which is not Riemann integrable. But the main drawback is that the Riemann integral operator IRIR, acting from the set of XX-valued simple functions S([a,b],X)S([a,b],X) to XX by IR(∑j=1nxjχ[tj−1,tj))=∑j=1n(tj−tj−1)xj,IR(∑j=1nxjχ[tj−1,tj))=∑j=1n(tj−tj−1)xj,
isn't it continuous when XX is not locally convex, what do you think??

well, your reasoning is good, and i would complete it with my theorical development

 

[Ritik bhardwaj]

Hmmmm.... that's very interesting... I would say to that if XX is a non-locally convex space, it is easy to construct a sequence of simple functions sn:[0,1]→X,sn(t)=∑m=1nχAm,n(t)xm,n,sn:[0,1]→X,sn(t)=∑m=1nχAm,n(t)xm,n,
where (Am,n)nm=1(Am,n)m=1n is a partition of the interval [0, 1] for each n∈Nn∈N, and χχ denotes the characteristic function, such as: sup1≤m≤n∥xm,n∥→0,∑m=1nμ(Am,n)xm,n↛0,sup1≤m≤n‖xm,n‖→0,∑m=1nμ(Am,n)xm,n↛0,
as n goes to infinity, where μμ denotes the Lebesgue measure. Therefore, Bochner–Lebesgue integration cannot be extended to non-locally convex spaces. On the other hand, the definition of the Riemann integral extends verbatim for functions defined on an interval [a, b] with values in an F-space XX. However, it has some problems in the non-locally convex setting. For example, Mazur and Orlicz proved that the F-space XX is non-locally convex if and only if there is a continuous function f:[0,1]→Xf:[0,1]→X which is not Riemann integrable. But the main drawback is that the Riemann integral operator IRIR, acting from the set of XX-valued simple functions S([a,b],X)S([a,b],X) to XX by IR(∑j=1nxjχ[tj−1,tj))=∑j=1n(tj−tj−1)xj,IR(∑j=1nxjχ[tj−1,tj))=∑j=1n(tj−tj−1)xj,
isn't it continuous when XX is not locally convex, what do you think??

You can also try to apply tensor vector equations in anti-D-dimensional space for spatial dimensions in vector space. Instead of summation, you can prefer integral for continuous convex space.