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nettmoney
Grup: membru
Mesaje: 15
20 Aug 2011, 12:50 |
a ori b se obtine scriind:
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nettmoney
Grup: membru
Mesaje: 15
20 Aug 2011, 12:51 |
$ a \cdot b $
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C.Telteu
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C.Telteu
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Este un site foarte bun pentru programarea in LaTex: codecogs.com/latex/eqneditor.php
Cel mai mare plus pe care eu il vad este, ca programatorul (pentru folosirea la majoritatea simbolurilor) nici nu are nevoe de cunostinte minime de programare de acest soft (un fel de programare vizuala).
Aceast fel de "programare" extrem de mult reduce timpul de lucru.
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(eroare: eq.0/35399)&n\int_{1}^{2}{\frac{dx}{x^2(1+x^n)}}=n\int_{1}^{2}{\frac{1+x^n-x^n}{x^2(1+x^n)}dx}=n(\int_{1}^{2}{\frac{1}{x^2}dx}-\int_{1}^{2}{\frac{x^{n-2}}{1+x^n}dx})=\frac{n}{2}-\int_{1}^{2}{\frac{nx^{n-1}}{x(1+x^n)}dx}=\frac{n}{2}-\int_{1}^{2}{\frac{1}{x}(ln(1+x^n))'dx}=\frac{n}{2}-\frac{ln(1+2^n)}{2}+ln2+\int_{1}^{2}{\frac{ln(1+x^n)}{x^2}dx}\geqslant\frac{1}{2}ln\frac{e^n}{1+2^n}\geqslant\frac{1}{2}ln(\frac{e}{2})^n\frac{1}{1+\frac{1}{2^n}}\rightarrow\infty&
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(eroare: eq.0/35400)n\int_{1}^{2}{\frac{dx}{x^2(1+x^n)}}=n\int_{1}^{2}{\frac{1+x^n-x^n}{x^2(1+x^n)}dx}=n(\int_{1}^{2}{\frac{1}{x^2}dx}-\int_{1}^{2}{\frac{x^{n-2}}{1+x^n}dx})=\frac{n}{2}-\int_{1}^{2}{\frac{nx^{n-1}}{x(1+x^n)}dx}=\frac{n}{2}-\int_{1}^{2}{\frac{1}{x}(ln(1+x^n))'dx}=\frac{n}{2}-\frac{ln(1+2^n)}{2}+ln2+\int_{1}^{2}{\frac{ln(1+x^n)}{x^2}dx}\geqslant\frac{1}{2}ln\frac{e^n}{1+2^n}\geqslant\frac{1}{2}ln(\frac{e}{2})^n\frac{1}{1+\frac{1}{2^n}}\rightarrow\infty
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