right. restoring my earlier statement: $y=f(x)$ and $y=g(x)$ having a triple point of intersection amounts to $f(x)-g(x)$ having a zero of multiplicity 3
Let C be the curve y = x3 (where x takes all real values). The tangent at A meets the curve again at B. Prove that the gradient at B is 4 times the gradient at A.
to get a regular intersection, you require $f(a)-(ma+b)=0$. to get a double intersection, you further require that the first derivative also vanishes i.e. $f'(a)-m=0$. and when you solve that for $m,b$ you get the usual equation of the tangent line
so say the tangent line at $P(x_0,y_0)$ to $y = x^3$ intersects the cubic in the other point $B(x_1,y_1)$, why does the tangent have a double point of intersection?
and that means that, in the first case, $f(x_0)-g(x_0)=0$ and $f'(x_0)-g'(x_0)=0$. and that means it's a double root of $f(x)-g(x)$ i.e. the tangent line will be a double point of intersection.
do you mean, it's possible to draw a tangent line to a cubic and have the second point of intersection also have the same slope there?
@Julian Its a terribly important skill for a mathematician to realize when you are wasting your time. I'll tell you I have wasted more time than I am willing to admit.
we don't. in fact, there has to be at least one (though it may not be real)
by that i mean that, if you draw a tangent line to $x^3$, you'll get a double point of intersection at the point of tangency and a single point of intersection somewhere else
because, well, a cubic polynomial has 3 roots (possibly complex)
and if you've got a root of multiplicity 2, the last one is multiplicity 1.
@Julian You are studying things developed to study things you don't know anything about. I think this is almost certainly a mistake of wasting your time (everyone makes similar mistakes). If you want to succeed in mathematics, you should try and recognize when you are wasting your time. I'm not sure I have much else to say on this.
well, if i draw a typical tangent line it's just multiplicity 2. but if i draw it at the origin, it'll be multiplicity 5.
and there's a simple way to see that it can't be higher. to get multiplicity 2, you need an intersection where the slopes also match. to get multiplicity 3, you also need the second derivatives to match.
but the second derivative of a straight line is zero everywhere, and the only place where $y=x^5$ has zero second derivative is at zero.
so the only possible way to get a higher multiplicity is to pick the origin as the tangent point.
my book says that since the tangent line $y=0$ to $y=x^3$ has a triple intersection point as opposed to the double intersection points at the tangencies, it is reasonable to say it meets the curve again at $(0,0)$
on the other hand, if you do a different quintic polynomial then you certainly can have zero second derivative somewhere else. so $x^5$ isn't typical in that regard.
not sure i follow what it's getting at there, aside from the following: if you move your point of tangency to the origin, then the single point of intersection moves to collide with the double point
I have this kind of inequalities where I want to determine whether it is transitive or not, the C terms are cut sets of a graph. How is transitivity usually proved in graph theory?
I am stuck to transitivity consideration with graphs
where $x^a,x^b$ are boolean monomials such as $x_1x_3$ and $x_5$, the cut sets contains $C=\{x^{C_i}=0\mid C_i\in C\}$, $C=\{C_1,C_2,\ldots,C_n\}$, $C_{\alpha/\beta}=\{x^{C_i}\mid \alpha\in C_i,\beta\not\in C_i, C_i \in C\}$ and $C_{\beta/\a...
@robjohn Ok... I will think about it again. In order to show that E(t,x) is integrable does it suffice to fix t and show that E(t,x) is continuous for the fixed t and then show that it is also continuous for a fixed x?
I just want to confirm something. Consider the function $f(x)= \sqrt{x}$, am I right in stating that this function is continuous on $[0,\infty)$ but not on $\mathbb{R}$ (since it is not defined for negative real numbers).
Can someone reason with me? I have this question: imgur.com/KSGzerX and if we simplify both sides i find that Quantity A = 1 and Quantity B = -1 which makes Quantity A greater than quantity B, am I right though? Cause they say that it can't be determined.
Hey @DanielFischer I have a question. I want to show that a function f(x,t) of two variables is integrable. Does it suffice to fix t and show that for a fixed t f is continuous? And then that for a fixed x f is continuous?
@ypercubeᵀᴹ $f(x)$ is continuous at $a$ iff $\lim\limits_{x \to a}f(x) = f(a)$. So are you saying that it is continuous even in $\mathbb{R}$ since it is clearly continuous on $(0,\infty)$ and at $0$ we have $\lim\limits_{x \to 0}f(x) = \lim\limits_{x \to 0^+}f(x) = f(0)$ and anywhere else doesn't matter since it is not defined there.
@Evinda I don't know any special statement, but then again I don't know much measure theory. It is however also wrong that for all $t$ the function $f(x,t)$ should be integrable wrt to $x$ and for all $x$ it should be integrable wrt to $t$. Consider as a counterexample to that $\exp(-|t+x|)$ on $\mathbb R^2$ which is integrable for all $t$ wrt to $x$ (and converse), but is not integrable on $\mathbb R^2$.
@MatsGranvik an operator is a linear map between two vector spaces. As such a $2\times2$ matrix can represent the action of an operator on the basis of a $2$ dimensional vector space. An operator $A$ is bounded if there is a positive real number $C$ so that $\|A x\|≤C\|x\|$ for all $x$ in the vector space. If there is no such number then the operator is unbounded.
The norm is part of the data of a normed vector space, the definition of boundedness only makes sense in the case of a normed vector space.
An example of an unbounded operator would be on the vector space given by finite linear combinations of $\{e_n \mid n \in \mathbb N\}$ with norm $\|\sum_n a_n e_n\| := \sup_n |a_n|$, then the operator that sends $\sum_n a_n e_n$ to $\sum_n a_n$ in $\mathbb C$ is unbounded.
@Oh yes, right!!! @robjohn So it should be like that: $\frac{\partial{E}}{\partial{t}}=\left\{\begin{matrix} \frac{\frac{|x|^2}{4 t^2} e^{-\frac{|x|^2}{4t}}(2 \sqrt{\pi t})^n- e^{-\frac{|x|^2}{4t}} (2 \sqrt{\pi})^n \frac{n}{2} t^{\frac{n-2}{2}}}{((2 \sqrt{\pi t})^n)^2} & , t >0\\ \\ 0 &, t \leq 0 \end{matrix}\right.$
does anyone know how to add new lines in your code when you are typing an answer? Like it says to do a line break just do two spaces at the end I try that and no line break happens for me.
I think this would prevent the trouble I have on here to with my text/math stuff getting all wrapped up into each other.
so hey @robjohn when I'm typing a answer in that latex stuff, how do I put new lines in... I read I just need to put two spaces and it moves to a new line but that doesn't work.
@Tobias: He's definitely putting plenty of effort in. And his posts generally include his thoughts and specific things that confuse/bother him. I have no complaints.
I am stuck to transitivity consideration with graphs
where $x^a,x^b$ are boolean monomials such as $x_1x_3$ and $x_5$, the cut sets contains $C=\{x^{C_i}=0\mid C_i\in C\}$, $C=\{C_1,C_2,\ldots,C_n\}$, $C_{\alpha/\beta}=\{x^{C_i}\mid \alpha\in C_i,\beta\not\in C_i, C_i \in C\}$ and $C_{\beta/\a...
