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 dont give any credit.
I'm not sure how to start off here. What does the S(a,b) mean...
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...
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...
Determine α>0 so that f'(0) exists
f_{\alpha }(x)=|x|^{\alpha }sin\left (\frac{1}{x} \right ) , \left [x\neq 0, f_{\alpha }(x)=0 \right ]
I derived the function in two cases, one where x<0 and one x>0, and saw that we get x in a denominator three times, As I understand it, it...
\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...