Recent content by bhoom

I got it now, thanks. My problem were how to express c_v and c_p.

Homework Statement The temperature of one mole of ideal gas is 360K. The gas is allowed to expand adiabatically to the double volume. Then it is compressed isothermally to original volume. The specified amount of heat is measured at 1304J. Determine C_p / C_v, and specify the type of ideal gas...

Homework Statement Prove that if gcd(a,b)=1 then N\ S(a,b) is a finite set. Homework Equations The Attempt at a Solution I'm new to set theory and this question is from a voluntary course that don't give any credit. I'm not sure how to start off here. What does the S(a,b)...

I think i got it earlier today. I am the phone atm, so I can't supply any details, but I ended up comparing (1+ak)(a+1) \geq (1+ka+a)

Ok, I will go from my base case, n=1 is true. Now I assume n=k is true. From that assumption I show that n=k+1 is true. (a+1)^(k+1) ≥ 1+a(k+1) (a+1)*(a+1)^k ≥ 1+ak+a (a+1) ≥ (1+ak+a)/(a+1)^k (a+1) ≥ (1+ak+a)(a+1)^-k or perhaps (a+1)^(k+1) ≥ 1+a(k+1) (a+1)*(a+1)^k ≥ 1+ak+a...

Ok, I'm off to sleep now but I will do another take on it tomorrow. The brackets were to show that 0<1<k<n<n+1. Thanks so far though.

Ahh, I had completely forgotten that I could simplify on both sides of the ≥. Thanks both of you.

In words: The original problem, if I set up a base(n=0 already tested and proven) n=1[n=n-(n-1)], I get the statement to be true. Then I will try to see what happens if n=k[n=n-(n-k)], 1<k<n. Then I will get (a+1)^k ≥ 1+ak. If this is true, then k-1 will also be true. n=k-1[n=n-(n-(k+1))...

If I set n=k+1 I can write the expression like this(a+1)^k(a+1)\geq 1+(k+1)a (a+1)^k(a+1)\geq 1+ak+a Since we assumed that (1 + a)^k ≥ 1 + ka, can I then look at (a+1)^k(a+1)-(a+1)^k\geq 1+ak+a-(1+ak) a(a+1)^k\geq a Since k is at least 0, which gives me a ≥ a. This whole thing feels very...

Homework Statement a> -1, (1+a)^n \geq 1+na Homework Equations a>-1 a+1> 0 The Attempt at a Solution If I let n=0, and then n=1 i get that 1≥1, and a+1≥ a+1. Add to each side n+1 (1+a)^{n+1}+(1+a)^{n} \geq 1+na+(1+a(n+1)) Then...? Perhaps subtract the original expressions on...

Ahh yes. Thanks!

\frac{-2x^2-x-3}{x^3+2x^2-x-2}=\frac{-2x^2-x-3}{(x-1)(x+1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2} Multiply by common denominator gives: -2x^2-x-3=A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)\Leftrightarrow -2x^2-x-3=Ax^2+A3x+A2+Bx^2+Bx-B2+Cx^2-C1 System of equations gives -2=A+B+C -1=A+B...

If I use Wien's law of displacement(λmax=b/T) I get the temperature to be ~6,000K. That is because the λmax of the visual spectrum is ~4,800Å. However, the actual temperature of Sirius A is ~10,000K and so according to λmax=b/T the λmax should be ~2,800Å I read something like that on a site...

Hello, I'v been taking an interest to astronomical spectroscopy lately and I have some questions about that. In order to get myself going I wanted to get as much information about Sirius A and Sirius B using the distance, obtained by using parallax(I didn't do that but imagine I did) , and...