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p-0-n-psh-a-bp-




Question Number 147828 by puissant last updated on 23/Jul/21
Σ_(p=0) ^n psh(a+bp)
$$\underset{\mathrm{p}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{psh}\left(\mathrm{a}+\mathrm{bp}\right) \\ $$
Answered by Olaf_Thorendsen last updated on 23/Jul/21
   Σ_(p=0) ^n ch(2a+2pb)=  ((ch(2a+nb)sh((n+1)b))/(shb))  (voir question precedente)  ⇒Σ_(p=0) ^n ch(a+pb)=  ((ch(a+(n/2)b)sh(((n+1)/2)b))/(sh(b/2)))  (∂/∂b)Σ_(p=0) ^n ch(a+pb) = Σ_(p=0) ^n psh(a+pb)      = (∂/∂b)(((ch(a+(n/2)b)sh(((n+1)/2)b))/(sh(b/2))))  ... calcul fastidieux !
$$ \\ $$$$\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ch}\left(\mathrm{2}{a}+\mathrm{2}{pb}\right)=\:\:\frac{\mathrm{ch}\left(\mathrm{2}{a}+{nb}\right)\mathrm{sh}\left(\left({n}+\mathrm{1}\right){b}\right)}{\mathrm{sh}{b}} \\ $$$$\left(\mathrm{voir}\:\mathrm{question}\:\mathrm{precedente}\right) \\ $$$$\Rightarrow\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ch}\left({a}+{pb}\right)=\:\:\frac{\mathrm{ch}\left({a}+\frac{{n}}{\mathrm{2}}{b}\right)\mathrm{sh}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}{b}\right)}{\mathrm{sh}\frac{{b}}{\mathrm{2}}} \\ $$$$\frac{\partial}{\partial{b}}\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{ch}\left({a}+{pb}\right)\:=\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}{p}\mathrm{sh}\left({a}+{pb}\right)\: \\ $$$$ \\ $$$$\:=\:\frac{\partial}{\partial{b}}\left(\frac{\mathrm{ch}\left({a}+\frac{{n}}{\mathrm{2}}{b}\right)\mathrm{sh}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}{b}\right)}{\mathrm{sh}\frac{{b}}{\mathrm{2}}}\right) \\ $$$$…\:\mathrm{calcul}\:\mathrm{fastidieux}\:! \\ $$
Commented by puissant last updated on 23/Jul/21
merci prof c′est trivial...
$$\mathrm{merci}\:\mathrm{prof}\:\mathrm{c}'\mathrm{est}\:\mathrm{trivial}… \\ $$

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