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2d
comment How many distinct roots $ax^5+bx^3+cx+d$ has
Over what domain are you looking for roots? The answer could be very different over $\mathbb{R}$ vs. $\mathbb{C}$...
Aug
27
answered Counting matchings, the modern way
Aug
27
comment Counting matchings, the modern way
There's a very clean description as $2^k\cdot(2k-1)!!$ that counts a smaller number of somethings (unordered pairings) and then multiplies by the $2^k$ ways of assigning a preferred partner per couple...
Aug
27
comment Let $T,S$ be linear transformations, $T:\mathbb R^4 \rightarrow \mathbb R^4$, such that $T^3+3T^2=4I, S=T^4+3T^3-4I$. Comment on S.
Well, you haven't proven that 4. is incorrect - in fact, you should be able to prove that any such $S$ is non-invertible. Also, and this is essential, $det(T-I)\neq 0$ is not the same thing as $T\neq I$ - there are many non-null matrices with zero determinants.
Aug
27
comment Let $T,S$ be linear transformations, $T:\mathbb R^4 \rightarrow \mathbb R^4$, such that $T^3+3T^2=4I, S=T^4+3T^3-4I$. Comment on S.
There are certainly other solutions for $T$ (and thus $S$), but just knowing that one solution makes those conditions false is enough to eliminate them as possible answers.
Aug
27
comment Let $T,S$ be linear transformations, $T:\mathbb R^4 \rightarrow \mathbb R^4$, such that $T^3+3T^2=4I, S=T^4+3T^3-4I$. Comment on S.
Diya: the point is that (as you noted yourself) $T=I$ satisfies the constraint, and if $T=I$ then $S=0$. The question is 'if all you know is <these givens>, what can you deduce?' You cannot possibly deduce that $S$ is invertible, or onto, or one-to-one, because there are a pair $T, S$ that satisfy the constraints for which those statements are false.
Aug
27
comment Let $T,S$ be linear transformations, $T:\mathbb R^4 \rightarrow \mathbb R^4$, such that $T^3+3T^2=4I, S=T^4+3T^3-4I$. Comment on S.
Given that you know that $S$ can be the zero matrix (if $T=I$, there are certainly several options that you can eliminate as correct possibilities.
Aug
27
comment Property for the natural numbers.
Property (1) looks like it's essentially omega-consistency in disguise; effectively it's saying that either there is some specific $i$ for which $f(i)$ is true, or we have $\forall k\neg f(k)$.
Aug
26
comment For every natural integer $N>3$ there are at least two distinct prime numbers $p$ and $q$ such that $\dfrac{p+q}{2}=N$ and $N-p=q-N$, $(p<q)$.
...both of those are equal to 17?
Aug
25
comment Is $\ln(x^{p(x)}) = p(x) \ln(x)$?
Also, while your solution works, you don't need to go through the exponential; you can simply take logarithms of both sides, since $\ln(a)=\ln(b)\implies a=b$.
Aug
25
comment The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$
The specific hole in this attempt is that adding the intervals on $x$ throws information away - you're no longer explicitly enforcing the condition that $\lceil\frac3x\rceil+\lceil\frac4x\rceil=5$.
Aug
24
revised Limit of $f_{n+1} = \sqrt{12 + f_n}$ with proof by contradiction
Fixed a matho
Aug
21
comment Express roots in polynomials of equation $x^3+x^2-2x-1=0$
Solving that quadratic doesn't seem to yield a polynomial expression; the discriminant is $b^2-4ac=(\alpha+1)^2-4(\alpha^2+\alpha-2)=-3\alpha^2-2\alpha+9$...
Aug
21
comment Can we take any decision about the quadrants, through which a line-segment passes, from its slope?
OTOH, if you're interested in a (doubly-infinite) line, then any line of positive slope will have portions in the first and third quadrants, and any line of negative slope will have portions in the second and fourth, and this isn't hard to see. (Lines of zero or infinite slope are rather special-casey.)
Aug
21
comment Can we take any decision about the quadrants, through which a line-segment passes, from its slope?
A line segment can be so short as to lie entirely in any quadrant regardless of its angle - imagine four separate line segments of length one with $(\pm 100, \pm 100)$ as their centers; then obviously the slope has no bearing on the quadrant.
Aug
20
comment $\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$
+1 for a slick solution - for clarity's sake, the top integral here is the difference between the two terms in the OP's equation (using $\log x-\log a=\log\frac xa$).
Aug
19
comment What's the order of growth of the 'double-and-rearrange' numbers?
@r.e.s. Yes, but the question asks (for convenience) about $\leq n$-digit integers (since the total number of these is $10^n$, possibly $-1$ if you don't count zero), so the total count of those numbers with all digits odd is $\sum_{i\leq n}5^i$.
Aug
19
comment Is it possible to construct a sequence that ends in 1000000000?
I got curious enough to ask my order-of-growth question: math.stackexchange.com/questions/1403089
Aug
19
asked What's the order of growth of the 'double-and-rearrange' numbers?
Aug
19
comment Finding closed form of finite product
I suggest posting up the precise problem you've been given and the work you've done on it so far; that's much more likely to get you a useful answer than trying to work things out from the little subproblem you've selected.