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Dr Sardonicus · 2020-10-22, 11:37

Yes, substituting x for 7! does make things much easier to handle. The obvious regrouping of the first part of the expression gives

$2x^{\frac{1}{4}}\cdot$x^{\frac{1}{2}}\;+\;x^{\frac{-1}{2}}$^{\frac{1}{2}}$

The "obvious" multiplication then gives $2\sqrt{x+1}$ .

I note that things can go wrong for complex values of "x" that aren't positive real numbers.

2020-10-22, 11:37	#6
Dr Sardonicus Feb 2017 Nowhere 2²·13·83 Posts	Yes, substituting x for 7! does make things much easier to handle. The obvious regrouping of the first part of the expression gives $2x^{\frac{1}{4}}\cdot\(x^{\frac{1}{2}}\;+\;x^{\frac{-1}{2}}\)^{\frac{1}{2}}$ The "obvious" multiplication then gives $2\sqrt{x+1}$ . I note that things can go wrong for complex values of "x" that aren't positive real numbers.