Recent content by naqo
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Graduate Understanding the Complexity of Wavefunctions in Quantum Mechanics
Hi there, i have been studying a bit about QM, but ther's one fundamental question about the wavefunction i can't understand: why is the wavef. defined complex? I mean, couldn't one work from the beginning with a real wave? Thanks- naqo
- Thread
- Wavefunction
- Replies: 40
- Forum: Quantum Physics
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Undergrad What is the significance of the number e and Euler's formula?
aaaa sory about the symbols... just starting with latex... -
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Undergrad What is the significance of the number e and Euler's formula?
Okay, in order to understand a little more about the number e, one must analyze first the limit presented above, \lim_{n\rightarrow \infty}(1+\frac{1}{n})^n. In particular, one must prove that such limit exists, that means, that when you take n really big, the number doesn't go to infinity. To... -
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Undergrad Integral Test for Convergence/Divergence
Hi, i would proceed as follows: In the first case, one can easily see that the function f(n) is monotone decreasing and allways non-negative, so the integral test applies in this situation. Then, the series will converge only if the integral from 1 to infinity of x/(x^2+1) dx converges (sory... -
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Mainstream Science: What's the Mainstay?
\(1+frac{1}{n})^n- naqo
- Post #10
- Forum: Other Physics Topics
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Mainstream Science: What's the Mainstay?
\(1+frac{1}{n})^n=sum_{n=0}^{\n}\frac{\n!}{\(n-k)!\k!}- naqo
- Post #9
- Forum: Other Physics Topics
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Undergrad Converge/Diverge: Show Work & Tests Used
there was supposed to be a 1 in the numerator of the second sum... -
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Undergrad Converge/Diverge: Show Work & Tests Used
Hi, i suposse i would procceed in the following way: Note that due to the fact that cos(x) is always smaller or equal than 1, cos^2(x) has the same behavior. So, the sum that you are asking: \sum_{n=1}^{\infty}\frac{\cos^2n}{n\sqrt n} is smaller than \sum_{n=1}^{\infty}\frac{\1}{n\sqrt n}...