Embedded in {$ \sqrt{x^2 + 2xy + y^2} $} text |
Embedded in \sqrt{x^2 + 2xy + y^2} text
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Centered {$$ \sqrt{x^2 + 2xy + y^2} $$} |
Centered \sqrt{x^2 + 2xy + y^2}
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{$$ {1\over 2} \quad {n+1\over 3} \quad {n+1 \choose 3} \quad \sum_{n=1}^3
Z_n^2 $$} |
{1\over 2} \quad {n+1\over 3} \quad {n+1 \choose 3} \quad \sum_{n=1}^3 Z_n^2
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{$$ {\textstyle \sum x_n}\quad {\sum x_n} \quad \sum_{n=1}^m x_n^2 $$} |
{\textstyle \sum x_n}\quad {\sum x_n} \quad \sum_{n=1}^m x_n^2
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For some reason the built-in demonstration markup does not interpret correctly the expression under the integral droping dt one line. However, no everything is lost. Outside the demonstration markup the integral comes out right: \int_0^\infty f(t)\,dt . You can see the corresponding LaTex code by clicking on that expression.
{$$\sqrt{2} = \sup \left\{ x\in\mathbb{Q}: x^{2}<2\right\} $$} |
\sqrt{2} = \sup \left\{ x\in\mathbb{Q}: x^{2}<2\right\}
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Would you guess it:
{$$ 2 = \sqrt{2+\sqrt{2+\sqrt{2+ ....}}} $$} |
2 = \sqrt{2+\sqrt{2+\sqrt{2+ ....}}}
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And this is how you get the Taylor series:
{$$ f(u_1) = f(u_0) + hf'(u_0)+{h^2\over 2!}f''(u_0) + \cdots +
{h^{n+1}\over (n+1)!}f^{(n+1)}(u_0) + o(h^{n+1}). $$} |
f(u_1) = f(u_0) + hf'(u_0)+{h^2\over 2!}f''(u_0) + \cdots + {h^{n+1}\over (n+1)!}f^(n+1)(u_0) + o(h^{n+1}).
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