Recent content by Alv95

@ Vela I have found an oblique asymptote: y = \frac{x}{2} - \frac{1}{4}

ok :wink: I have got: \frac{1}{y^2-1} = \frac{1}{2(y-1)}-\frac{1}{2(y+1)} therefore: \frac{1}{y+1} = \frac{1}{y-1}-\frac{2}{y^2-1} My function is then: \frac{1}{e^u+1} = \frac{1}{e^u-1}-\frac{2}{e^{2u}-1}

I don't know yet what Bernoulli numbers are (we don't do them in my school) :smile: Thanks anyways for your help :smile: ... From what I have understood from Wikipedia (:rolleyes:) I should rewrite my function: \frac{1}{1+e^u} as a "generating function" so that it will then generate the...

... The problem is now that I get a different expansion depending on how far I go in the approximation ... those \left( \frac{\cdots \, \, - 1}{2}\right)^n change the first term of the expansion ... EDIT: Sorry :smile: I should replace \left( \frac{e^u - 1}{2}\right) with his expansion...

Thank you very much Office_Shredder :smile: I had thought of adding/subtracting 1 but not of factorising the 2 :redface: Thank you again :wink:

I have some problems finding Taylor's expansion at infinity of f(x) = \frac{x}{1+e^{\frac{1}{x}}} I tried to find Taylor's expansion at 0 of : g(u) = \frac{1}{u} \cdot \frac{1}{1+e^u} \hspace{10 mm} \mbox{ where } \hspace{10 mm} u = 1/x in order to then use the known expansion of...

Thanks, I will :)

Well, I prefer to be sure about what I write especially if it is on a new topic that I have just learned at school ;) By seeking help and advices on the internet I hope to improve and strenghten my knowledge and thus be more confident. I didn't expect it to be a problem.

Thanks ;)

So it's not a group? :) The text of the homework that the teacher gave us seemed to imply that it was :)

I have to find if the set [1;+∞[ x [1;+∞[ with the operation (x;y)°(v;w) = (x+v-1; yw) is a group I have already proven Closure, associativity and Identity but I have some problems with invertibility :) The neutral element that I have found is (1;1) I did (x;y)°(x1;y1)= (1;1) and I have...

Thank you very much tiny-tim for your precious Latex tips and your suggestions :) Here is a little resumé of all this thread: The resultant Electrical field vector (Etot) on a test charge due to two charges of equal magnitude but opposite sign, equidistant from the y-axis and lying on the...

The equation for the angle ψ of the Etot vector is: \psi = \arctan(\frac{\frac{x+a}{[(x+a)^2+y^2]^(\frac{3}{2})}-\frac{x-a}{[(x+-a)^2+y^2]^(\frac{3}{2})}}{\frac{y}{[(x+a)^2+y^2]^(\frac{3}{2})}-\frac{y}{[(x-a)^2+y^2]^(\frac{3}{2})}}) I showed my equations yesterday in class (I was the only one...

It is right! :) For what concerns its direction I think to know how to do it but now I have to go to school :) I will post it later ;)

Eureka! :) I have eventually found the final total expanded equation. Thanks for your help tiny-tim! :D It works with the data that the teacher has given me! I will try with other ones :) Etot =\sqrt{[\frac{kQ}{(x+a)^2+y^2}\frac{x+a}{\sqrt{(x+a)^2+y^2...