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Pietro Majer
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Put $t=1-\sqrt{s}\in[0,1/2)$ so the equation writes $$ \Big(1-\frac p2\Big)\, t^p+ \frac p2\, t^{p-1}=2^{-\frac p2}$$

With another explicit change of variables the form $t=\alpha u^\beta$, for suitable $\alpha$ and $\beta$, andNow if we put $y=y(p)$, the$u:=t^{p-1}$ the equation can be put intakes the form $$u+u^q=y$$$$u+\Big( \frac2p -1\Big)\,u^q =\frac {2^{1-\frac p 2 }} p$$ with $q=q(p)>1$$q=\frac p{p-1} >1$, thatthat can be solved by series (see e.g. here) (this way one covers an interval $1.57<p\le2$ if I'm not wrong. To cover the other values of $p$, close to $1$, one needs to put the equation in other forms).

Put $t=1-\sqrt{s}\in[0,1/2)$ so the equation writes $$ \Big(1-\frac p2\Big)\, t^p+ \frac p2\, t^{p-1}=2^{-\frac p2}$$

With another explicit change of variables the form $t=\alpha u^\beta$, for suitable $\alpha$ and $\beta$, and $y=y(p)$, the equation can be put in the form $$u+u^q=y$$ with $q=q(p)>1$, that can be solved by series (see e.g. here).

Put $t=1-\sqrt{s}\in[0,1/2)$ so the equation writes $$ \Big(1-\frac p2\Big)\, t^p+ \frac p2\, t^{p-1}=2^{-\frac p2}$$

Now if we put $u:=t^{p-1}$ the equation takes the form $$u+\Big( \frac2p -1\Big)\,u^q =\frac {2^{1-\frac p 2 }} p$$ with $q=\frac p{p-1} >1$, that can be solved by series (see e.g. here) (this way one covers an interval $1.57<p\le2$ if I'm not wrong. To cover the other values of $p$, close to $1$, one needs to put the equation in other forms).

Source Link
Pietro Majer
  • 60.5k
  • 4
  • 122
  • 269

Put $t=1-\sqrt{s}\in[0,1/2)$ so the equation writes $$ \Big(1-\frac p2\Big)\, t^p+ \frac p2\, t^{p-1}=2^{-\frac p2}$$

With another explicit change of variables the form $t=\alpha u^\beta$, for suitable $\alpha$ and $\beta$, and $y=y(p)$, the equation can be put in the form $$u+u^q=y$$ with $q=q(p)>1$, that can be solved by series (see e.g. here).