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bio website motls.blogspot.com
location Czech Republic
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visits member for 3 years, 4 months
seen Aug 4 at 12:45

Hi, I am a string theorist and a publicist.


May
19
awarded  Supporter
May
19
comment How can I solve this non-linear differential equation?
Dear @Hannesh, it's not only legal but mandatory to allow all integration constants throughout the calculation being arbitrary complex numbers. Solving equations - algebraic or differential - in the reals isn't simpler than in complex numbers. Quite on the contrary, it's more complicated because you must solve it using all possible complex values of the parameters, and at the very end, you must do an extra job of filtering out the solutions that are not real. See the exchanges right under your question.
May
19
comment How can I solve this non-linear differential equation?
Exactly, this is the right compact form of the solution. tanh is sinh/cosh so its derivative is $(\cosh^2 t - \sinh^2 t)/\cosh^2 t = 1/\cosh^2 t$ which is equal to $1-\tanh^2 t$, indeed.
May
19
comment How can I solve this non-linear differential equation?
Dear @Ross, I don't think your comment is right. "Working in the reals" only means that you must filter the solutions at the very end to make sure that they're real if this is what you were asked about. However, all intermediate steps can and should use complex numbers, otherwise you're missing some solutions. This is, in fact, why complex numbers were first used. Some cubic equations have all 3 roots being real, but you still need complex numbers in intermediate steps (of the Cardan formula) to calculate these roots. This situation is completely analogous.
May
19
comment How can I solve this non-linear differential equation?
Yes, you may set $k$ to a negative number because $c$ can be complex. Any solution with any complex values of the integration constants is OK, and this particular choice even ends up with a real $y$ for real $t$. A related comment: the absolute value as a part of the logarithm's argument is counterproductive because it is not a holomorphic function. I think it's a good idea to avoid all such symbols that only work on the real axis - and create a wrong discontinuity when the argument of the absolute value vanishes.
May
19
answered Trig equation help please
May
19
comment How is the uniform boundedness principle compatible with this seemingly weak convergent sequence?
Of course that I am allowed to use $x$ even if it is not in your arbitrarily chosen basis. The weak convergence is defined so that the inner products with all vectors in the Hilbert space converge to the right value, see en.wikipedia.org/wiki/Weak_convergence_(Hilbert_space) - it would be very counterproductive to make a definition of a vector's property (or the property of their sequence) that depends on the choice of basis. Important properties don't depend on the choice of bases. @Jonas, thanks for interpreting, you understood me well.
May
19
revised Collecting coupons which arrive as a Poisson process
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May
19
answered Collecting coupons which arrive as a Poisson process
May
19
answered Is $\frac{m-1}{x}$ an unbiased estimator of $\theta$ for given pdf?
May
19
comment How is the uniform boundedness principle compatible with this seemingly weak convergent sequence?
If I understand you well: on the contrary: for the weak convergence it is enough to show the convergence of the inner product with any $\phi_i$ because that's how the weak convergence is defined. What I tried to argue is that you did not prove the convergence of the inner products to the "right" inner product and you could not because this convergence of inner products is not true.
May
19
answered Complex Exponential Expansion
May
19
revised How is the uniform boundedness principle compatible with this seemingly weak convergent sequence?
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May
19
answered How is the uniform boundedness principle compatible with this seemingly weak convergent sequence?
May
19
answered How to prove the inequality $\Theta(x,y)\le \Theta(x,z)+\Theta(z,y)$?
May
18
answered Complex Exponents
May
18
answered Deduce plus and minus with Cross Product in 3th and 4th Maxwell equations
May
18
awarded  Commentator
May
18
comment Global conformally flat coordinates in 2d spacetimes
Dear @Dionigi, the Lorentzian-signature "disks" defined as any contractible manifolds can't be mapped to each other. Imagine that the shape is described by an equation involving $x^+,x^-$, the light-like coordinates. The conformal transformations in this case are separate reparametrizations of $x^+$ and of $x^-$. This is clearly not enough to relate all contractible shapes. For example, a null boundary of a "diamond" will always stay null under conformal transformations, and manifolds with piecewise spacelike and then timelike boundaries etc. will always have these pieces.
May
18
awarded  Nice Answer