meta user
network profile
| bio | website | mathematica.stackexchange.com |
|---|---|---|
| location | United States | |
| age | 36 | |
| visits | member for | 2 years, 8 months |
| seen | 6 hours ago | |
| stats | profile views | 166 |
The gravitar was made with Mathematica, and inspired by this.
The contents of my posts are my own opinion, and do not reflect the opinions of my employer.
|
Sep 29 |
comment |
differentiation of Polynomials I think you've misinterpreted the OPs equation, the $3 x$ only divides the second term. |
|
Sep 28 |
comment |
Confused about implicit differentiation @JordanCarlyon, $dy^{20}/dy = 20 y^{19}$. |
|
Sep 27 |
comment |
What is the result of $\infty - 1$? +1, for "Cat - 1." |
|
Sep 27 |
awarded | Nice Answer |
|
Sep 27 |
revised |
Pure maths vs applied added additional thoughts. |
|
Sep 26 |
revised |
Pure maths vs applied added clarifying statement |
|
Sep 26 |
comment |
Pure maths vs applied If your interested in the use of Lie algebras in quantum, the book by Lipkin is a good read, and it's a Dover reprint! Essentially it shows you how to extract a lot of information from the Hamiltonian by just knowing the operator algebra. |
|
Sep 25 |
comment |
Do real matrices always have real eigenvalues? @commenter. No, the "iff" is absolutely correct. Please review the fundamental theorem of algebra and the definition of the characteristic polynomial. |
|
Sep 25 |
comment |
Do real matrices always have real eigenvalues? @commenter, no. The equation $A v = \lambda v$ has a non-trivial solution iff $\det(A - \lambda I) = 0$ which is identical to finding the roots of a polynomial in $\lambda$. We know the roots of a polynomial with real coefficients may have complex roots. If we try to restrict ourselves to only real roots, we're not guaranteed to find all of them. |
|
Sep 25 |
awarded | Critic |
|
Sep 25 |
comment |
Do real matrices always have real eigenvalues? No, it doesn't. Finding the eigenvalues of a matrix is identical to finding the zeroes of a polynomial, and polynomials with real coefficients can have complex zeroes. Consider $(a - \lambda)(b - \lambda) - c d = 0$ which has complex roots if $(a+b)^2 < 4 (ab - cd)$. |
|
Sep 24 |
revised |
Solving an equation with trig functions and two different angles added LaTeX markup |
|
Sep 24 |
suggested | suggested edit on Solving an equation with trig functions and two different angles |
|
Sep 24 |
answered | Pure maths vs applied |
|
Sep 22 |
revised |
derivatives using chain rule deleted 2 characters in body |
|
Sep 22 |
answered | derivatives using chain rule |
|
Sep 22 |
comment |
Derivative using product rule and chain rule As a second point, I'm not a fan of Newtonian notation to indicate a derivative, e.g. $f'(x)$, and I prefer Leibniz's, e.g. $df/dx$, because it is difficult to tell which variable to use when calculating a derivative like, $\cos(a^3 + x^3)'$. In first semester calculus, this is easier as it is likely to be $x$. But, Leibniz's notation leaves less room for ambiguity, and that can be a life saver when doing these. However, professors insist on using Newton's form, so you have to learn it, also. But, I'd change it to the other form when working these. |
|
Sep 22 |
awarded | Citizen Patrol |
|
Sep 22 |
comment |
Derivative using product rule and chain rule This is just a relatively straightforward application of your previous question. I'd suggest looking at Jonas' answer again. |
|
Sep 12 |
awarded | Analytical |
