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<p>𝐼 π‘Žπ‘š π‘Ž π‘Ÿπ‘’π‘‘π‘–π‘Ÿπ‘’π‘‘ π‘šπ‘Žπ‘‘β„Žπ‘’π‘šπ‘Žπ‘‘π‘–π‘π‘  π‘‘π‘’π‘Žπ‘β„Žπ‘’π‘Ÿ. 𝐼 β„Žπ‘Žπ‘£π‘’ π‘‘π‘œπ‘›π‘’ π‘ π‘œπ‘šπ‘’ π‘€π‘œπ‘Ÿπ‘˜ π‘œπ‘› π‘Žπ‘π‘ π‘π‘œπ‘›π‘—π‘’π‘π‘‘π‘’π‘Ÿπ‘’.𝐼𝑑’𝑠 π‘Ÿπ‘’π‘Žπ‘™π‘™π‘¦ π‘ π‘’π‘šπ‘–π‘›π‘Žπ‘™ π‘‘β„Žπ‘Žπ‘‘β€™π‘  π‘€β„Žπ‘¦ 𝐼 π‘€π‘Žπ‘›π‘‘ π‘‘π‘œ π‘‘π‘’π‘šπ‘œπ‘›π‘ π‘‘π‘Ÿπ‘Žπ‘‘π‘’ 𝑖𝑛 π‘‘β„Žπ‘’ π‘π‘œπ‘›π‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ π‘œπ‘“ π‘π‘œπ‘›π‘”π‘Ÿπ‘’π‘ π‘  π‘œπ‘“ π‘šπ‘Žπ‘‘β„Žπ‘’π‘šπ‘Žπ‘‘π‘–π‘π‘–π‘Žπ‘›π‘  π‘œπ‘›π‘™π‘¦. 10.17605/OSF.IO/YJR86</p>

Posts

  • Post #3142165

    https://math.stackexchange.com/questions/733754/visually-stunning-math-concepts-which-are-easy-to-explain

  • Post #3142164

    You wanted the proof of \[(m+q)(n+q)=2mn+q^2\] \[(m+q)(n+q)&amp;lt;2mn+\frac{mn}{k+2}\] \[(m+q)(n+q)&amp;lt;mn(2+\frac{1}{3})\] \[\left(\frac{m+q}{m}\right)\cdot \left(\frac{n+q}{n}\right)&amp;lt;\frac{7}{3}\] \[2\left(\frac{m+q}{m}\right)\cdot \left(\frac{n+q}{n}\right)&amp;lt;\frac{14}{3}\] \[\left\{\left(\frac{m+q}{m}\right)+ \left(\frac{n+q}{n}\right)\right\}^2=3^2\] \[\left(\frac{m+q}{m}\right)^2+ 2\left(\frac{m+q}{m}\right)\cdot \left(\frac{n+q}{n}\right)+\left(\frac{n+q}{n}\right)^2=3^2\...

  • Post #3142163

    \[(π‘˜+2)π‘˜Ξ΅^β€²+(π‘˜+1)Ξ΅=1 \] DOI: https://doi.org/10.17605/OSF.IO/KR2MQ

  • Post #3142162

    \subsection{Numerical Verification of the Expression} The following table shows the numerical verification of the expression \[ \frac{5}{3} \approx \frac{m+q}{m} \approx \frac{6 + (k^2 + 1)\epsilon&amp;#39; + k\epsilon}{4} \] on several abc triples with different values of \(k\). \begin{table}[h] \centering \begin{tabular}{|c|l|c|c|c|c|} \hline \(k\) &amp;amp; Triple (example) &amp;amp; Actual \(\frac{m+q}{m}\) &amp;amp; Right side \(\frac{6 + (k^2+1)\epsilon&amp;#39; + k\epsilon}{4}\) &amp;am...

  • Post #3142161

    If you want to know truth, learn mathematics.

  • Post #3142160

    \[(1+\epsilon)^2\leqslant q&amp;lt;\left(\frac{(2)\cdot(14)\cdot\left\{2k+1+(k-1)\epsilon\right\}}{9(k+2)\cdot(2+(1-k)\epsilon^{\prime})}\right)\leqslant \left(\frac{5}{3}\right)\] \[\qquad \because(1+\epsilon)&amp;lt; \left(\frac{14}{9}\right)\approx q \leqslant \left(\frac{5}{3}\right)\]

  • Post #3142159

    \[(1+\epsilon)^2\leqslant q&amp;lt;\left(\frac{(2)\cdot(14)\cdot\left\{2k+1+(k-1)\epsilon\right\}}{9(k+2)\cdot(2+(1-k)\epsilon^{\prime})}\right)\leqslant \frac{5}{3}\] \[ \because(1+\epsilon)&amp;lt; \left(\frac{14}{9}\right)\approx q \leqslant \left(\frac{5}{3}\right)\] \[\textit{Verification by ChatGpt Let us verify the corrected inequality}\] \[\boxed{ n&amp;lt; \frac{28}{9\bigl(2+(1-k)\epsilon&amp;#39;\bigr)} \bigl(2k+1+(k-1)\epsilon\bigr). }\] carefully on actual abc triples. πŸ”· Triple 1 $...