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$$du=tan(d\theta)$$

How to integrate this ?

I didn't try numerically but I thought of expanding the tangeant in series but then should for example $$\int d\theta^2$$ be understood as a double integration ?

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$$du=tan(d\theta)$$

How to integrate this ?

I didn't try numerically but I thought of expanding the tangeant in series but then should for example $$\int d\theta^2$$ be understood as a double integration ?

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You fell ? Did it hurt ?

Was there a context that can help us make it clear to you or did it get lost in the fall ?

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it depends whether we mean ##d(\theta^2)## or ##(d\theta)^2##No, ##d\theta^2 = 2\theta d\theta##.

And for me to answer the OP's question ##\int (d\theta)^n## is zero for all ##n\geq 2##

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Yes missed parentheses from my part.

The context is easy :

Let be the parametrization of a unit sphere ##\vec{r}(\theta,\phi)=(cos\theta cos\phi,cos\theta sin\phi,sin\theta),\theta\in [-\pi/2,\pi/2],\phi\in [0,2\pi] ## and the associated basis vectors ##\vec{e}_\theta=\frac{\partial\vec{r}}{\partial\theta}## and ##\vec{e}_\phi##

The metric is ##g=diag(1,cos(\theta)^2)##

I wanted to orthonormalize the metric via the associated quadratic form : ##\underbrace{(du\vec{e}_\theta+dv\vec{e}_\phi)}_{\vec{x}}g\vec{x}^T##

The ##du,dv## are infinitesimals on the tangent plane and I express them as a function of ##\theta,\phi##: ##du=tan(d\theta)##

##dv'=cos(\theta)^2tan(d\phi)##

Instead of linearizing I did what I wrote in the OP with successive integrations (##\int(d\theta)^n=\theta^n/n!## by fixing all the successive constants of integration to 0) and found

##u=\sum_{n=0}^\infty(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}\frac{\theta^{2n+1}}{(2n+2)!(2n+1)!}##

I wanted to compute the shape of u,v which if you want should be a map of the sphere.

Is it periodic and what's the asymptotic behaviour ?

Could anyone help me or give me references in webpages, it should be found in common textbooks about differential geometry or map projections I suppose.

What I found weird is that if I consider a path on the sphere ##\theta(t),\phi(t)## then the endpoint coordinate on the map could depend on the path itself ?!

Thanks.

The context is easy :

Let be the parametrization of a unit sphere ##\vec{r}(\theta,\phi)=(cos\theta cos\phi,cos\theta sin\phi,sin\theta),\theta\in [-\pi/2,\pi/2],\phi\in [0,2\pi] ## and the associated basis vectors ##\vec{e}_\theta=\frac{\partial\vec{r}}{\partial\theta}## and ##\vec{e}_\phi##

The metric is ##g=diag(1,cos(\theta)^2)##

I wanted to orthonormalize the metric via the associated quadratic form : ##\underbrace{(du\vec{e}_\theta+dv\vec{e}_\phi)}_{\vec{x}}g\vec{x}^T##

The ##du,dv## are infinitesimals on the tangent plane and I express them as a function of ##\theta,\phi##: ##du=tan(d\theta)##

##dv'=cos(\theta)^2tan(d\phi)##

Instead of linearizing I did what I wrote in the OP with successive integrations (##\int(d\theta)^n=\theta^n/n!## by fixing all the successive constants of integration to 0) and found

##u=\sum_{n=0}^\infty(-1)^n2^{2n+2}(2^{2n+2}-1)B_{2n+2}\frac{\theta^{2n+1}}{(2n+2)!(2n+1)!}##

I wanted to compute the shape of u,v which if you want should be a map of the sphere.

Is it periodic and what's the asymptotic behaviour ?

Could anyone help me or give me references in webpages, it should be found in common textbooks about differential geometry or map projections I suppose.

What I found weird is that if I consider a path on the sphere ##\theta(t),\phi(t)## then the endpoint coordinate on the map could depend on the path itself ?!

Thanks.

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I cannot anymore link geometry with algebra, it is as if the same geometric case could lead to several formulas.

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Yes, that is more-or-less correct. The expression ##\frac{d(\tan(x))}{dx}## is the derivative of the tangent function, with respect to its variable x.##du## is an infinitesimal while ##\frac{d(\tan(x))}{dx}## is a finite quantity if ##|x|<\frac{\pi}{2}##.

The substitution was ##u = \tan(x)## from which the differentials of the two sides would be as you have below. Note that we would usually write ##\tan^2(\theta)## rather than ##\tan(\theta)^2##.

However, I'm not sure this is relevant to your problem where you had ##du = \tan(d\theta)##. I would be interested in seeing the work that led up to this equation.

jk22 said:Did you mean ##du=d\tan(\theta=x)=(1+\tan(\theta)^2)d\theta##, so that we have not ##\tan(d\theta)##, which is meaningless ?

I cannot anymore link geometry with algebra, it is as if the same geometric case could lead to several formulas.

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In fact there is no work : I suppose a circle and the tangent line at coordinate ##\theta##, then an infinitesimal change ##d\theta## leads to a change ##du=\tan(d\theta)## on this line, with just basic trigonometrical geometry.However, I'm not sure this is relevant to your problem where you had ##du = \tan(d\theta)##. I would be interested in seeing the work that led up to this equation.

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