Recent content by WigneRacah

Try the substitution: y=f(x).EDIT: I suppose f is a "regular" function.

Welcome! :smile:

Try with this. Find the product \prod_{k=1}^{\infty} \left( 1 + \frac{2}{k^2 + 7} \right).

We have I = \int_0^1 \ln x \ln (1-x) dx = \lim_{a\rightarrow 1} \int_0^a \ln x \ln (1-x) dx. Integrating by part I= \lim_{a \rightarrow 1}( - \ln (1-a) + \int_0^a \frac{x \ln x}{1-x} - \int_0^a \frac{x}{1-x}). By performing such a limit we have I= 2 + \int_0^1 \frac{\ln x}{1-x} dx...

Since e^{iz^2} is an analytic function on \mathbb{C} we have: \int_{\Gamma_R} e^{iz^2} d z = 0 where \Gamma_R is the "pizza-slice" countour given in the link. Now \int_{\Gamma_R} e^{iz^2} d z = \int_0^R e^{i t^2} d t + \int_0^{\frac{\pi}{4}} e^{i R^2 e^{i 2 \theta}} i R e^{i \theta} d...

In general, it would be good to destinguish between "exact" and "closed" forms.

Quite easy: it is a diverging series ... :rolleyes:

Let \epsilon be a small positive real number. We have to prove that there exists a correspondent \delta such as | e^x - e | < \epsilon for all 1-\delta < x < 1+\delta and x \neq 1. We have | e^x - e | < \epsilon \Rightarrow -\epsilon < e^x - e < \epsilon \Rightarrow e -...

The equation you wrote doesn't describe a hyperbola but two straight lines intersecting in the point (-1, 0). In fact you can write (x+1)^2- 4 y^2 = 0 \Rightarrow (x +1 - 2 y) (x +1 + 2 y) =0 which implies x+1 -2y=0 or x + 1 + 2 y =0 (the stright line equations). If the equation was...

No, you don't. For the angular velocity, you only need initial position, initial w and final position.

good luck! :wink:

We don't get that! :cry: I think you have some problems with the fundamental theorems of calculus or with logarithmic properties. \ln r - \ln r_0 = \ln \frac{r}{r_0} \ln \omega - \ln \omega_0 = \ln \frac{\omega}{\omega_0}

Very nice! :approve:

The integrand function on the left hand side of your equation is wrong ... From the second equation in post n. 8 you get: 2 \int_{r_0}^r \frac{d \rho}{\rho} = - \int_{\omega_0}^{\omega} \frac{d \omega}{\omega} .

Perfect! In fact the general solution for the given equation has the form y = C x^3 - \frac{k}{3}.