To show that the set $\{3x-x^5, 4x+x^3, 5x-x^5-x^6\}$ in $\mathbb{Q}[x]$ is linearly independent, do we have to find the $a,b,c,d$ such that $a(3x-x^5)+ b(4x+x^3)+c(5x-x^5-x^6)=0$ ? And these coefficients are in $\mathbb{Q}$ or $\mathbb{Q}[x]$ ?
So, $(3a+4b+5c)x+bx^3-(a+c)x^5-cx^6=0 \Rightarrow 3a+4b+5c=0 , b=0, a+c=0, c=0$ and therefore, $c=a=b=0$, and so the set is linearly independent, right? @Null
So, from that set we have the vectors $(0,3,0,0,0,-1,0), (0,4,0,1,0,0,0), (0,5,0,0,0,-1,-1)$ and we add each vector that represent the monomials: $(1,0,0,0,0,0,0), (0,0,1,0,0,0,0), (0,0,0,0,1,0,0)$ Now I have 6 vectors... Which one is left?
The degree 4 is (0,0,0,0,1,0,0), the degree 2 is (0,0,1,0,0,0,0) and the degree 0 is (1,0,0,0,0,0,0). The degree 1 can we written as a linear combination of the set?
For example, the degree 6 cannot be represented by the set: $(3a+4b+5c)x+bx^3-(a+c)x^5-cx^6=x^6 \Rightarrow c=-1, a+c=0, b=0, 3a+4b+5c=0$ This system does not have a solution since replacing $c=-1, a=1, b=0$ at $3a+4b+5c=0$ we get $3-5=0$.
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I have also an other question.... I want to calculate the q-adic expansion for q=3 and q=4. I have found that for q=3 we get $\frac{1}{5}=2+0\cdot 3+1\cdot 3^2+2\cdot 3^3+1\cdot 3^4+0\cdot 3^5+\ldots$, so $}\frac{1}{5}=2,01210121\ldots$.
It is the same method for q=4, which is not a prime, right?
@BalarkaSen People say "It's just a theory" to discredit the theory of evolution. Of course, this makes no sense in the context of Ramsey theory (well, it doesn't make sense in the context of the theory of evolution since "theory" doesn't mean "guess" in that context either, but whatever).
I want to find the $4$-adic expansion of $\frac{1}{5}$. I have done the following:
$\displaystyle{\frac{1}{5}\equiv a_0\pmod 4 \Rightarrow a_0\equiv \frac{1}{5}\pmod 4 \Rightarrow 5a_0\equiv 1\pmod 4}$. The only residue of division by $4$ that solves this is $a_0=1$.
Then we have $\displaystyle{\frac{1}{5}-1\equiv 4a_1 \pmod {4^2} \Rightarrow -\frac{4}{5}\equiv 4a_1\pmod {4^2}\Rightarrow -\frac{1}{5}\equiv a_1\pmod {4}\Rightarrow -1\equiv 5a_1\pmod {4}\Rightarrow 5a_1\equiv -1 \pmod 4 \Rightarrow 5a_1\equiv 3\pmod 4 \Rightarrow a_1\equiv 3\pmod 4}$. The only residue of division by $4$ th…
But I got, at the calculation of $a_4$, $4a_4\equiv 10 \pmod {4^2}$. Here the result of the multiplication is not zero... So, we don' see in this case the property that it is not an integral domain... @DHMO
@DHMO For $q=-\frac{1}{4}$ we get $\sum \left (-\frac{1}{4}\right ) ^k=\frac{1}{1+\frac{1}{4}}=\frac{4}{5}$, and so $\frac{1}{5}=\frac{1}{4}\sum \left (-\frac{1}{4}\right ) ^k$, right?
A q-adic expansion is when we write the number as $x=\sum a_i q^i$. I have seen also this form $x=a_0, a_1a_2...$ . Is this just an other form of the q-adic expansion?
@DHMO Why is it 0 and not 1? Do we not have $\dfrac15 = \dfrac14 - \dfrac1{16} + \dfrac1{64} - \dfrac{256} + \cdots=4^{-1}-4^{-2}+4^{-3}-4^{-4}+...=4^{-1}+3\cdot 4^{-2}+4^{-3}+3\cdot 4^{-4}+...$ ?
@DHMO I see... So, when we had 3-adic instead of 4-adic, I got $\frac{1}{5}=2\cdot 3^0+0\cdot 3+1\cdot 3^2+2\cdot 3^3+1\cdot 3^4+0\cdot 3^5+\ldots$ do we write this as ...121012102 ?
In this book they found the same coefficients as I did: https://books.google.gr/books?id=cn07Oz6nUNAC&pg=PA13&lpg=PA13&dq=3-adic+expansion+of+1/5&source=bl&ots=Umzw58694N&sig=WShJEuzoe19V6S4q0kzhZoK-7ik&hl=el&sa=X&ved=0ahUKEwjBnKjm5eXQAhUDXBoKHUZhCFYQ6AEIYTAI#v=onepage&q=3-adic%20expansion%20of%201%2F5&f=false
Let $\bar{x}^{(1)},...,\bar{x}^{(n)}$ be linearly independent solutions of $\bar{x}' = \bar{P}(t)\bar{x}$ where $\bar{P}$ is continuous on $\alpha < t < \beta$. Show that any solution $\bar{x} = \bar{z}(t)$ can be written in the form $$\bar{z}(t) = c_1\bar{x}^{(1)}(t)+...+c_n\bar{x}^{(n)}(t)$$for suitable constants.
" Can you make friends with only a compass and a straightedge?" Gauss gave a neusis construction, but it involved "going outside" and "talking to people" which can't be done in the usual axiom system
I got this unanswered question with at bounty of 150 that ends in 5 hours. Any answer which even just has a useful reference will be awarded the bounty. math.stackexchange.com/questions/2035989/…
hey quesiton... would a question that asked "Based on the current electoral college system in the US what is the most candidates that could tie for first place and what is the number of votes they would have" be on topic here?
Could I ask if there is a way to figure that out besides brute force?
Hey guys I have a quick question: Let $(X,d)$ be a metric space and let $(\tilde(X),d)$ be its completion. There exists an isometric embedding $I:X\to\tilde(X)$, but does it holds that the closure of $I(X)$ is $\tilde(X)$, where the closure is with respect to the topology on $\tilde(X)$?
Let's say you have a real variable $t$, and for simplicity we'll use $t \ge 0$. Let's say you have two functions $x=f\left({t}\right)$ and $y = g\left({t}\right)$ and both $f$ and $g$ are well-behaved
follow so far?
I'll illustrate with an example. Give me two continuously differentiable functions
Hey I have a tiny doubt, If I have a series $ a_n $ that converges, that means $ a_n = 0 < 1 $, so If I want to see if $ \dfrac {a_n} {b_n} $ converges, then I can compare it with $ \dfrac {1} {b_n} $ because $ a_n < 1 $ right ?
And if $ \dfrac {1} {b_n} $ converges so does $ \dfrac {a_n} {b_n} $
the problems is, can I use that in everything that tends to 0 ?
