- #176
Gib Z
Homework Helper
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O well since I guess I got the last one, I'll just ask again what post 169 asked: How can one find the surface area of a ring?
Because |G| is odd, so too is o(g) for each g in G. Consequently, o(g^2)=o(g). This means that <g^2>=<g>, and do this for each g in G. [Alternatively, let o(g)=2k-1 so that g^(2k-1)=1 => (g^k)^2 = g.]StatusX said:Here's one:
Let G be a group of odd order. Prove that for every element g in G, there is exactly one element h in G such that h2=g.
Hmm. Are there any restrictions on x, i.e. does it have to be a nonnegative integer? And can we use knowledge gathered from evaluating at the first value to pick the second one?rs1n said:I didn't read this entire thread, so this problem may have already been posted. It was a problem that got tossed around by graduate students, and I thought it was cute.
Suppose [tex]p(x)[/tex] is a polynomial of degree d with nonnegative integer coefficients. You are allowed to evaluate [tex]p(x)[/tex] at exactly TWO values of x. What two values do you use in order to completely determine the coefficients of [tex]p(x)[/tex]?
morphism said:Hmm. Are there any restrictions on x, i.e. does it have to be a nonnegative integer? And can we use knowledge gathered from evaluating at the first value to pick the second one?
My initial guess was +/- 1, but that won't be sufficient most of the time. So maybe +/- 1/d, but this also doesn't seem to be very helpful for large degrees. It seems impossible to do it with two values picked at the same time!
Has something to do with the dth root of something like, I dunno, 2. Or maybe a better one to do is this: first evaluate p(1), that's the sum of all the coefficients. Then pick a prime P > p(1), and evaluate p(P). Then if you consider p(P) mod P², p(P) mod P³, ..., p(P) mod Pd or something you can recover the coefficients.rs1n said:I didn't read this entire thread, so this problem may have already been posted. It was a problem that got tossed around by graduate students, and I thought it was cute.
Suppose [tex]p(x)[/tex] is a polynomial of degree d with nonnegative integer coefficients. You are allowed to evaluate [tex]p(x)[/tex] at exactly TWO values of x. What two values do you use in order to completely determine the coefficients of [tex]p(x)[/tex]?
AKG said:Or maybe a better one to do is this: first evaluate p(1), that's the sum of all the coefficients. Then pick a prime P > p(1), and evaluate p(P). Then if you consider p(P) mod P², p(P) mod P³, ..., p(P) mod Pd or something you can recover the coefficients.
ThirstyDog said:I will have a guess and I think it goes something like this...
We start with
[tex]
\int_{- \infty}^{\infty}{e^{-x^2}} dx = \sqrt{\pi}
[/tex]
We consider
[tex]
\int_{- \infty}^{\infty}{e^{-tx^2}} dx
= \int_{- \infty}^{\infty}{e^{-u^2}} \frac{d(t^{-0.5}u)}{du}du
= \int_{- \infty}^{\infty}{e^{-u^2}} t^{-0.5} du
= t^{-0.5} \sqrt{\pi}
[/tex]
This leads to
[tex]
\frac{d}{dt} \int_{- \infty}^{\infty}{e^{-tx^2}} dx
= \frac{d}{dt} t^{-0.5} \sqrt{\pi}
= \frac{-\sqrt{\pi}}{2} t^{-\frac{3}{2}}
[/tex]
Ignea_unda said:Okay new to this, and with my luck probably wrong but...
inside the integral we have exp[-tx^2] = exp[t]exp[-x^2]
We can then pull exp[t] out of the integral and evaluate the integral
Finally we take the derivative and multiply the two together getting sqrt(pi)*exp[t]
StatusX said:Here's a problem:
You find yourself in the middle of a forest, with no idea how you got there. You find a note on a nearby tree that says the edge of the forest is 1 mile away. Of course, you have no idea what direction it's in. Assuming you know the forest is a half-plane (ie, from above, it looks like the set of points {(x,y) | y>0}), what path should you take to minimize the distance you need to travel to find the edge?
(Specifically, what path minimizes the distance traveled in the worst case scenario. It might also be interesting to find the path that minimizes the expected distance, although this sounds harder).
Hint:You can do better than the obvious path of length 2pi + 1.
maze said:edit: nm
To make 100! divisible by 1249, we need to find the highest power of 12 that is a factor of 100!. This can be found by dividing 100 by 12, which gives us 8. This means that 128 is the highest power of 12 that is a factor of 100!. To make it divisible by 1249, we need to multiply 1241 to 100!.
Yes, we can make 100! divisible by 1249 without changing its value by multiplying it with 1241. This ensures that the value of 100! remains the same, but it becomes divisible by 1249.
Making 100! divisible by 1249 is important because it allows us to easily perform calculations involving large numbers. By making it divisible by 1249, we can break down the calculation into smaller, more manageable parts.
1249 is the highest power of 12 that is a factor of 100!. This means that it is the smallest number that can divide 100! without leaving a remainder. It is significant because it allows us to make 100! divisible by a large number, making calculations easier.
Yes, there is a general rule for making a factorial divisible by a large number. We need to find the highest power of the number that is a factor of the factorial, and then multiply it to the factorial. This will ensure that the factorial is divisible by the large number without changing its value.