meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | 41 | |
| visits | member for | 10 months |
| seen | Oct 26 '12 at 6:50 | |
| stats | profile views | 260 |
|
Sep 20 |
revised |
Solving $(x+y) \exp(x+y) = x \exp(x)$ for $y$. exp -> \exp |
|
Sep 20 |
comment |
Proof of derivative of $e^x$ is $e^x$ without using chain rule @CameronBuie: Now better? |
|
Sep 20 |
revised |
Proof of derivative of $e^x$ is $e^x$ without using chain rule fixed incorrect calculation |
|
Sep 20 |
comment |
Proof of derivative of $e^x$ is $e^x$ without using chain rule @CameronBuie: Oops, you're right. Also, I now notice that I've forgotten the binomial coefficients in the sum. I'll try to fix it. |
|
Sep 20 |
comment |
Run away from lions in a cage Well, it is obvious that if your speed is $0$, you've got no chance to escape from the lions. It is also obvious that if you are much faster than the lions, you can always avoid them. Therefore there must exist a minimal speed $v$ for the problem as stated. Since the problem as given is well defined, there's no need to restate it. Of course, looking at more specific rules for the lions gives a lower bound to the solution (because if you are slower than that, we know a successful lion's strategy), and maybe that's the best we can hope for, but that doesn't invalidate the question as asked. |
|
Sep 20 |
comment |
What are imaginary numbers? That's up to now by far the best answer about the usefulness. +1 |
|
Sep 20 |
comment |
Find $f$ where $f'(x) = f(1+x)$ @Sasha: That is only the general solution if there are no solutions which are not of the form $\mathrm e^{\lambda x}$. For which there is no proof yet. |
|
Sep 20 |
comment |
Find $f$ where $f'(x) = f(1+x)$ OK, thinking again about it, the series does not give a recursive formula for $a_n$ (at least not immediately) because you get infinite sums. |
|
Sep 20 |
comment |
Find $f$ where $f'(x) = f(1+x)$ Chris's sister asked for the general form of $f$. This is just one solution. More solutions can be found by multiplying the function with a constant: $f(x)=a\mathrm e^{-W(-1)x}$. However is this already the most general form? |
|
Sep 20 |
comment |
Find $f$ where $f'(x) = f(1+x)$ Also you got $f^{(n)}(x) = f(x+n)$. Proof by induction: $f^{(0)}(x)=f(x)$ is trivial. $f^{(n+1)}(x) = (f^{(n)}(x))' = (f(x+n))' = f'(x+n) = f(x+n+1)$ where the second step was the induction assumption and the last step the defining equation. If you insert the power series from the previous comment and set $x=0$, you get $f(n) = n! a_n$. |
|
Sep 20 |
comment |
Find $f$ where $f'(x) = f(1+x)$ If you can write $f(x)$ as power series $f(x)=\sum_{n=0}^\infty a_n x^n$, this should give you a recursive formula for $a_n$. |
|
Sep 20 |
answered | What are imaginary numbers? |
|
Sep 20 |
comment |
What is $\lim\limits_{n\to\infty} \frac{n^d}{ {n+d \choose d} }$? The denominator in the product should be $n+j$. |
|
Sep 20 |
comment |
Proof of derivative of $e^x$ is $e^x$ without using chain rule You don't need to interchange limits, see my answer. |
|
Sep 20 |
answered | Proof of derivative of $e^x$ is $e^x$ without using chain rule |
|
Sep 20 |
comment |
is $\exp \left( - 1 \over x^2 \right)$ differentiable at $x=0$ Actually, you could define a derivative which also is well defined at isolated undefined points, provided that the function can be extended to an ordinarily differentiable function there: Just use $\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$ instead of the standard definition. Indeed, with that definition, you could even derive non-continuous functions, as long as they only differ from continuous ones in isolated points (this is similar to the Lebesgue integral which also doesn't care about localized changes, although there you've got even more freedom). |
|
Sep 20 |
answered | Proof: Symmetric and Positive Definite |
|
Sep 19 |
comment |
The norm of a diagonalizable matrix is its largest eigenvalue? @DirkCalloway: If the matrix is diagonalizable, you have the largest absolute value of the eigenvalues. Since a positive semidefinite matrix (I assume that's what you meant with SPD) has only positive eigenvalues, for this it is the same as the largest eigenvalue. I don't know what happens for non-diagonalizable matrices, though. |
|
Sep 19 |
revised |
Strictly increasing function fixed value of $f(0)$ (which doesn't change the argument) |
|
Sep 19 |
comment |
Strictly increasing function That function is not continuous at $x=-1$. |
