# X x n x

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is a specific positive integer known as a binomial coefficient.

(When an exponent is zero, the corresponding power expression is taken to be 1 and this multiplicative factor is often omitted from the term.

Binomial coefficients, as combinatorial quantities expressing the number of ways of selecting k objects out of n without replacement, were of interest to ancient Indian mathematicians.For security advisories please check our Security Page.Please notify us of any security issues by sending mail to [email protected] Org Foundation welcomes sponsorship (both cash and in-kind), and tries hard to put the donations of sponsors to transparent good use.Let us take it in the form $$F(x)=x^xg(x)$$ So it has to be: $$F'(x)=x^x((1 \ln(x))g(x) g'(x))=x^x$$ From there it is sufficient to take $g(x) \sim \frac$ So we can start our journey: $$F(x)=x^x(\frac f(g(x)))$$ If you calculate the derivative of this you have $$g'(x)f'(g(x))-\frac (\ln(x) 1)f(g(x))=0$$ For the purpose of cancellation if it best to take $$g'(x)=\ln(x) 1$$ meaning $$g(x)=x\ln(x)$$ Now we continue using the steps that are revealing the integral structure.$$F(x)=x^x(\frac f(x\ln(x)))$$ Take derivative once more and you have got $$f(x\ln(x))=\frac-f'(x\ln(x))$$ or $$F(x)=x^x(\frac \frac-f'(x\ln(x)))$$ We can then write: $$F(x)=x^x(\frac \sum_^f_n(x))$$ where $$\displaystyle f_=-\frac,\,f_0=\frac$$ From $x=0$ to

Binomial coefficients, as combinatorial quantities expressing the number of ways of selecting k objects out of n without replacement, were of interest to ancient Indian mathematicians.

Please notify us of any security issues by sending mail to [email protected] Org Foundation welcomes sponsorship (both cash and in-kind), and tries hard to put the donations of sponsors to transparent good use.

Let us take it in the form $$F(x)=x^xg(x)$$ So it has to be: $$F'(x)=x^x((1 \ln(x))g(x) g'(x))=x^x$$ From there it is sufficient to take $g(x) \sim \frac$ So we can start our journey: $$F(x)=x^x(\frac f(g(x)))$$ If you calculate the derivative of this you have $$g'(x)f'(g(x))-\frac (\ln(x) 1)f(g(x))=0$$ For the purpose of cancellation if it best to take $$g'(x)=\ln(x) 1$$ meaning $$g(x)=x\ln(x)$$ Now we continue using the steps that are revealing the integral structure.

$$F(x)=x^x(\frac f(x\ln(x)))$$ Take derivative once more and you have got $$f(x\ln(x))=\frac-f'(x\ln(x))$$ or $$F(x)=x^x(\frac \frac-f'(x\ln(x)))$$ We can then write: $$F(x)=x^x(\frac \sum_^f_n(x))$$ where $$\displaystyle f_=-\frac,\,f_0=\frac$$ From $x=0$ to $1$ you can use probably more suitably $F(x)=xg(\ln(x))$ The derivation is similar to the one given above. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

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Binomial coefficients, as combinatorial quantities expressing the number of ways of selecting k objects out of n without replacement, were of interest to ancient Indian mathematicians.For security advisories please check our Security Page.Please notify us of any security issues by sending mail to [email protected] Org Foundation welcomes sponsorship (both cash and in-kind), and tries hard to put the donations of sponsors to transparent good use.Let us take it in the form $$F(x)=x^xg(x)$$ So it has to be: $$F'(x)=x^x((1 \ln(x))g(x) g'(x))=x^x$$ From there it is sufficient to take $g(x) \sim \frac$ So we can start our journey: $$F(x)=x^x(\frac f(g(x)))$$ If you calculate the derivative of this you have $$g'(x)f'(g(x))-\frac (\ln(x) 1)f(g(x))=0$$ For the purpose of cancellation if it best to take $$g'(x)=\ln(x) 1$$ meaning $$g(x)=x\ln(x)$$ Now we continue using the steps that are revealing the integral structure.$$F(x)=x^x(\frac f(x\ln(x)))$$ Take derivative once more and you have got $$f(x\ln(x))=\frac-f'(x\ln(x))$$ or $$F(x)=x^x(\frac \frac-f'(x\ln(x)))$$ We can then write: $$F(x)=x^x(\frac \sum_^f_n(x))$$ where $$\displaystyle f_=-\frac,\,f_0=\frac$$ From $x=0$ to $1$ you can use probably more suitably $F(x)=xg(\ln(x))$ The derivation is similar to the one given above. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

$you can use probably more suitably$F(x)=xg(\ln(x))\$ The derivation is similar to the one given above. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

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$$\begin \int x^x \, dx&=\int u \, du\\[6pt] &=\frac\\[6pt] &=\dfrac\\[6pt] &=\frac \end$$ But it's certain that this isn't the correct way to evaluate that, and the answer must be wrong.

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