Oh okay, so you are able to do contour integration, but for the first book(and possibly only book if you do retire) you are going to be using real methods. If you do a second book it will be using complex methods?
@Committingtoachallenge Integrals, series and limits only.
Well, in the math community I met more haters than in any other part of life. If I wanna ever continue my work, I can do it alone, away from any kind of community.
I think my main issue with those classes is that the professor makes huge logical leaps in between steps and I feel like it's nigh impossible for someone like me to fill them in on my own
@ᴇʏᴇs Hmm, originality in a subject like that where the proofs have been simplified a huge lot isn't usually a good thing. The standard proofs tend to be standard for a reason.
@Chris'ssis It is cleaner to simply move the entire contour a finite amount. Furthermore, being close to an essential singularity makes it hard to perform the integral over the indent.
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@robjohn @DanielFischer with some cleverness one can finish this one ALSO by real methods $$\int_0^{\infty} \cos\left(x\frac{x^2-\pi^2}{x^2-e^2}\right)\frac{1}{1+x^2} \ dx=\frac{\pi}{2}\exp\left(-\frac{\pi^2+1}{e^2+1}\right)$$
In my notes there is the following:
$$U_{\xi \eta}(\xi, \eta)=0 \Rightarrow \frac{\partial}{\partial{\eta}} \left (\frac{\partial}{\partial{\xi}} U \right )=0 \\ \Rightarrow U_{\xi}(\xi, \eta)=a(\xi) \\ \Rightarrow \frac{\partial}{\partial{\xi}}\left (U(\xi, \eta)\right )=\frac{\partial}{\parti...
@Owatch What can you add to the first integral to simplify the numerator and denominator? Remember to apply the opposite operation to the second integral to maintain equivalence
Not quite, but you're almost there. Recall that I asked you to apply the opposite operation to the second integral, but notice that there's already a factor of -1 in front.
Okay, so I added by three on one side. And made the denominator a match so I can add without modifying the three accordingly. Then I must undo this operation. Since I am subtracting, I cannot subtract from the other side, because I will have subtracted twice? I must add the same to the other side so that when I subtract them they neutralise each other?
If the previous version was hard, then what to say about the alternating version, that is $$ \sum_{n=1}^{\infty} (-1)^{n+1}H_n\left( \sum_{k=n+1}^{\infty}\frac{1}{k^3}\right)$$? :-)
Let's use the same terms, just ignoring the integrals @Owatch. $\frac{3t-2}{t+1}=\frac{3t}{t+1}-\frac{2}{t+1}=\frac{3t}{t+1}+\frac3{t+1}-\frac{3}{t+1}-\frac{2}{t+1}$
@TimDavids If you look at the lines $y = c\cdot x$ for $c\neq 0$, you have a cusp at the origin for that slice of the graph. The graphs of differentiable functions have no cusps.
@DanielFischer Okay. Would it be right to state that a function is not differentiable at a point if the tangent lines at a point do not lie in the same plane?
@DanielFischer I am referring to the tangent line of the curve $C$ which is the curve of intersection of $y=x$ and $f(x,y)$. So that has a vertical tangent line as I showed in the post but $f_{x}(0,0)$ and $f_{y}(0,0)$ are both equal to zero, so the tangent lines at the origin are not in the same plane. What do you think? Tangent line being line which touches the curve at only on place.
@TimDavids Yes, if $f$ were differentiable, the directional derivatives would be the appropriate linear combination of the partials, and hence the tangent lines would sweep out a plane. That doesn't happen here, so $f$ cannot be differentiable. But where you have no differentiability, the concept of tangent line becomes a little intricate. The "touches at only one point" thing is bad, consider a straight line for your curve. Oops, the "only one point" completely screws things up.
@DanielFischer Hello!!! Could I ask you something? If $u(x,t)=A(t-x)+B(t+\frac{x}{2})$ and we have the initial value $u(x,0)=f(x)$, do if we have $u(x,0)=f(x) \iff A(-x)+B(\frac{x}{2})=f(x)$ or $u(x,0)=f(x) \Rightarrow A(-x)+B(\frac{x}{2})=f(x)$ ?
@TimDavids Mostly such that the tangent line exists if and only if the curve is differentiable (with non-zero derivative) at the point. Sometimes it is useful to forget about the orientation and say that a tangent line also exists at cusps - like the $C$ in the question has - by splitting the curve and looking at one-sided derivatives (using a suitable parametrisation of the parts of the curve).
@evinda If you have the equation - that is, $u, A, B$ are determined - then you have $\iff$. If $u$ and $f$ are given and you look for $A$ and $B$, then $\Rightarrow$.
@DanielFischer Something else, with similar ideas. If you consider the function $g(x,y) = \{ \frac{y^{2}(x-y)}{x^{2}+y^{2}}~~\text{ if }(x,y) \neq (0,0) \text{ and }~~0~\text{ if } (x,y) = (0,0) \}$. We then have that $g_{x}(0,0)= 0$ and $g_{y}(0,0) = -1$. Would you expect the tangent plane (assuming it exists) at $(0,0,0)$ to be $y + z = 0$?
@teadawg1337 Could you explain how this changed to the line after that you wrote on chat after? I see you multiplied by denominators (cross) but why did you do that again? Is it just another way to write it?
@DanielFischer One more thing I want to confirm. For the function in the post $f(x,y) = x^{\frac{1}{3}}y^{\frac{1}{3}}$, would you say that the level curves are just hyperbolas of the form $xy = c^{3}$?
@DanielFischer For my original question with regard to the post, how do you see that there is a cusp at the origin? How do you define the branches of the cusp?
@TimDavids If you slice along $x = y$, you have $z = \lvert x\rvert^{2/3}$, and we know that $\lvert x\rvert^\alpha$ has a cusp at $0$ for $0 < \alpha < 1$.
Guys, first peano axioms says : for all a , ( Sa =/= 0 ) , do you guys view it as specifying the property ( Sx =/= 0 ) holds for ALL numbers in the universe of the model, or do you guys view it as specifying that the property \forall a ( Xa =/= Y ) holds for a specific tuple ( S,0 ) ?
hello, if i have $x_{n+1}= f(x_n)$ where f is continuous ,if $x_n$ has a convergent sub sequence $x_{\varphi(n)}$ , is $ x_{\varphi(n)+1}$ converge to the same limite ?
I have a 16 GB Apple Mini iPad using iOS 8.1.2 and am unable to install the bookmark found here that renders LaTeX in the Math StackExchange chat room. I can't find a browser with a bookmark bar to drag the given link to.
Other users in the room have also been unable to render LaTeX with their ...
In my notes there is the following:
$$U_{\xi \eta}(\xi, \eta)=0 \Rightarrow \frac{\partial}{\partial{\eta}} \left (\frac{\partial}{\partial{\xi}} U \right )=0 \\ \Rightarrow U_{\xi}(\xi, \eta)=a(\xi) \\ \Rightarrow \frac{\partial}{\partial{\xi}}\left (U(\xi, \eta)\right )=\frac{\partial}{\parti...
