205 reputation
19
bio website sasq.programuj.com
location Earth, TX
age 31
visits member for 3 years
seen Apr 18 at 1:24

Apr
6
awarded  Revival
Apr
6
awarded  Yearling
Jan
10
comment Solving equations with high level exponents
Wolfram's "solutions" are not solutions at all. They're only restatement of the question.
Jan
6
comment compute limit (no l'Hospital rule)
Quick question: Why in the original function when we've got $0/0$ we had to go on with factoring, but when we came to the limit of $\frac{y-1}{y-1}$ times $-1/12$ (the result of the other limit), we could say that the $-1/12$ is the end result, disregarding that $\frac{y-1}{y-1}$ in the limit is still $0/0$?
Nov
8
awarded  Critic
Nov
8
comment Reducing a fraction, divisibility and indeterminate symbol
You've made a very important point here, and that's what I waited for. $g(a,b) = 3$ is defined everywhere. But $f(a,b) = \frac{3a+3b}{a+b}$ is defined everywhere except the line $a=-b$. So we can put the equal sign between these two functions only if we enforce additional condition upon $g(a,b)$. In other words, $g(a,b)$ is something more than $f(a,b)$, so we have to limit $g(a,b)$ to make them both equal. Is my stream of thought correct?
Nov
8
accepted Reducing a fraction, divisibility and indeterminate symbol
Nov
8
comment Reducing a fraction, divisibility and indeterminate symbol
Thanks for your answer. I added an update to my question to address that in somewhat more detail.
Nov
8
revised Reducing a fraction, divisibility and indeterminate symbol
Update about the difference between a/0 and 0/0.
Nov
8
asked Reducing a fraction, divisibility and indeterminate symbol
Nov
8
comment Simplifying this algebraic fraction
@mixedmath: You're right that factoring, by itself, is not necessarily a simplification. But it can be used further in a process of simplification, when there are some common factors which can be canceled. And how could we know that there are some until we factor? ;) Also, factoring polynomial expressions allows us to see their roots (solutions), while the non-factored form shows us nothing much.
Oct
31
awarded  Revival
Oct
30
revised Intuition for the Frobenius method
More explanations to answer the follow-up questions.
Oct
30
comment Intuition for the Frobenius method
Yes, it's a pity that in books & courses they rarely explain such things. So let me explain a little bit more. I edited my answer above and added some more explanations for your questions.
Oct
30
comment Frobenius series solution
RIAA would have a different viewpoint on that matter ;-) He would be better off by scrambling it into the bits of some cat photo ;-)
Oct
30
answered Solving ODEs: The Frobenius Method, worked examples
Oct
30
answered Intuition for the Frobenius method
Oct
21
comment Calculation of Bessel Functions
Are you sure your formula is correct? When I did some tests, it appears that $J_2$, $J_6$, $J_{10}$ etc. needs to change sign to match the original. Seems like a $(-1)^{2n}$ factor is needed somewhere. Odd ones work OK. Also, can you elaborate some more on how did you derive those formulas?
Oct
16
comment Solving ODEs: The Frobenius Method, worked examples
For your equation (2) after synchronizing indexes you've extracted $3a_0x^{r-1}$ while for n=1 it should be: $$3a_{n-1}x^{r+n-1} = 3a_{1-1}x^{r+1-1} = 3a_0x^r$$ So all the terms extracted outside the sums should be: $$r(r-1)a_0x^r - r(r-1)a_0x^{r-1} - r(r+1)a_1x^r + 3a_0x^r$$ Now you can collect them as: $$[r(r-1)a_0 - r(r+1)a_1 + 3a_0] x^r - r(r-1)a_0x^{r-1} \\ [(r(r-1)+ 3)a_0 - r(r+1)a_1] x^r - r(r-1)a_0x^{r-1}$$ and simplify a bit: $$[(r^2-r+3)a_0 - (r^2+r)a_1] x^r - r(r-1)a_0x^{r-1}$$ and you can get your indicial equation from this part of your equation.
Sep
26
awarded  Scholar