One ##3##-dimensional representation of ##SO(3)## is ##3\times 3## rotation matrices parametrized by three Euler angles. The representation acts on ##3##-dimensional vectors and rotates them in ##\mathbb{R}^3 ##. That makes sense. What doesn't make sense is the interpretation of, say, the...
Problem
This is a conceptual problem from my self-study. I'm trying to learn the basics of group theory but this business of representations is a problem. I want to know how to interpret representations of a group in different dimensions.
Relevant Example
Take SO(3) for example; it's the...
Homework Statement
I want to estimate f(x)=\ln (\frac{1}{1+x}) on the interval (1/10,1) with the error on the approximation being no more than 0.1.
Homework Equations
http://en.wikipedia.org/wiki/Taylor's_theorem#Example
The Attempt at a Solution
Following the example from Wikipedia, I...
If I'm understanding correctly, R is the area under the graph of f(x) = 4cos(x) between 0 and π/2, which you've determined correctly to be 4.
This is true, but it doesn't really help you.
This is what you want. Note that you don't actually need \int_{a}^{b}4cos(x)dx=4/3, since the other...
Yes.
The "coolness" of the final answer would suggest that there might be a simple geometrical argument, but I can't think of one. This method seems like a lot of work (because you've been working on the problem for a long time), but it really isn't. Noting that the integral transfers fairly...
The formula for the surface area of z = f(x, y) above some region D is:
SA=\underset{D}{{\iint}} \sqrt{1+(f_x)^2+(f_y)^2}\: dA
(Alternatively, write down a surface integral and parametrize the sphere - but it will end up being the same thing.) The question now is: what is the f(x, y) to...
Exactly.
Alternatively, you could note that since the left side is positive, the right side must also be positive, so we have to take the plus sign since the square root function is never negative.
Okay, let's back up for a second. You presumably found that:
y= \pm t \sqrt{te^c-1}
I'll show you a "cool" trick. Since e is a constant, e^C must also be a constant, so you can just replace it with C.
y= \pm t \sqrt{Ct-1}
You sub in y(1) = 1:
1 = \pm \sqrt{C-1}
Now do you...
Dude, ln(a) + ln(b) = ln(ab). You should have ln|t e^{c}| = ln|u^2 + 1|. Also, don't forget the plus-or-minus sign when you take the square root of both sides. You'll need check which sign you want by which one works with the initial condition.
First of all, 4u / (2u^2 + 2) = 2u / (u^2 + 1), which simplifies the anti-derivative slightly to ln|u^2+1|. You can't make the assumption that C=0; you have an initial condition that must be obeyed!
Note that C = ln(e^C); then you can combine logarithms with the laws of logarithm addition and...
It's not showing that the integral and sum are equivalent that's my problem, it's coming up with the integral in the first place :wink:. Let me try to rephrase this:
Given a sum, I want to know how to write down a computable integral whose value is the value of that sum.
Well what I'm really asking is how to convert the sum to an integral. Like If I just saw \sum_{n=1}^{\infty }\frac{1}{n^2} somewhere, I would start with this:
\sum_{n=1}^{\infty} (xy)^{n-1}=\frac{1}{1-xy}
But how did I know to start with that? It appears to have no relation to...
Homework Statement
Check out matt grime's post in this thread (it's the last one):
https://www.physicsforums.com/showthread.php?p=470773#post470773"
How exactly did he know that the sum could be represented as that double integral? Also, is there a method of converting sums like that...
Okay, wait a minute. You're expanding this:
\sum_{i=0}^{n} \frac{6}{n}[(1 + \frac{6i}{n})^2 - 4 (1 + \frac{6i}{n}) + 2]
It should give you this:
\sum_{i=0}^{n} [\frac{216 i^2}{n^3} - \frac{72 i}{n^2} - \frac{6}{n}]
You have to find this sum first before taking the limit. To do this...
I'd love to, but that's not the way it works around here. The best I (or anyone else) is allowed to do is guide you towards the solution as per the forum rules. You should post the method you used to compute the sum, or perhaps some steps from your simplification, and then we can probably spot...
The value of the integral is 30.
http://www.wolframalpha.com/input/?i=integrate+x^2+-+4x+%2B+2+from+1+to+7
You must have gone wrong somewhere in the computation of the sum:
\sum_{i=0}^{n} \frac{6}{n}[(1 + \frac{6i}{n})^2 - 4 (1 + \frac{6i}{n}) + 2]
The Wronskian method didn't give me zero. You should find that:
W(e^{x}, x e^{x}) = e^{2x}
Then the integrals are:
Y_{p}(x) = -e^{x} \int \frac{x e^{x}}{(e^{x}+1)^2}dx + xe^{x} \int \frac{e^{x}}{(e^{x}+1)^2}dx
That looks good! Now that you have those, you can invert them term-by-term since the inverse Laplace transform is linear, so \mathcal{L}^{-1}(af(t)+bg(t)) = a\mathcal{L}^{-1}(f(t)) + b\mathcal{L}^{-1}(g(t)) for constants a and b.
Well using partial fractions is definitely on track. But remember that when you have a quadratic term in the denominator, you have to have a linear term in the numerator (plus, it's squared, so there are going to be two.) The form for the expansion should be:
\frac{1}{(s)(s^2+4)^2} =...
Actually, I remember reading something relevant to this in J. Nearing's Mathematical Tools for Physics. Consider:
\int cos(ax)e^{bx}dx + i \int sin(ax)e^{bx}dx
Then by adding integrands:
\int e^{bx}[cos(ax) + i \, sin(ax)]dx
Here we make the substitution cos(ax) + i \, sin(ax) =...
The inverse Laplace transform of a product isn't the product of the inverse Laplace transforms in general. You can go for the convolution theorem (might be a mess), or use partial fractions to decompose that, which requires you to use a few other inverse identities once you get the decomposition.
You're welcome, chexmix!
I would highly suggest learning substitutions though. They're pretty easy to understand and it removes the guessing aspect :smile: