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Question Number 60621 by fjdjdcjv last updated on 22/May/19
if π is rational then there  exists a I_n =(v^(2n) /(n!))∫_0 ^π x^n (x−π)^n sin(x)dx  can someone give a easier way to expaned this
$${if}\:\pi\:{is}\:{rational}\:{then}\:{there} \\ $$$${exists}\:{a}\:{I}_{{n}} =\frac{{v}^{\mathrm{2}{n}} }{{n}!}\underset{\mathrm{0}} {\overset{\pi} {\int}}{x}^{{n}} \left({x}−\pi\right)^{{n}} {sin}\left({x}\right){dx} \\ $$$${can}\:{someone}\:{give}\:{a}\:{easier}\:{way}\:{to}\:{expaned}\:{this} \\ $$
Commented by MJS last updated on 23/May/19
I don′t understand. we can solve the integral  for any given n∈N even if π∉Q. and what is v?
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\mathrm{for}\:\mathrm{any}\:\mathrm{given}\:{n}\in\mathbb{N}\:\mathrm{even}\:\mathrm{if}\:\pi\notin\mathbb{Q}.\:\mathrm{and}\:\mathrm{what}\:\mathrm{is}\:{v}? \\ $$
Commented by fjdjdcjv last updated on 23/May/19
π=(u/v)
$$\pi=\frac{{u}}{{v}} \\ $$
Commented by MJS last updated on 23/May/19
ok. still the integral exists, quite apart from  the nature of π  another question: sin x with which period,  2π or 2(u/v)?  is it (v^(2n) /(n!))∫_0 ^(u/v) x^n (x−(u/v))sin ((vπx)/u) dx?  or shall we calculate I_n  using the usual π  and then show that π∉Q?
$$\mathrm{ok}.\:\mathrm{still}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{exists},\:\mathrm{quite}\:\mathrm{apart}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\pi \\ $$$$\mathrm{another}\:\mathrm{question}:\:\mathrm{sin}\:{x}\:\mathrm{with}\:\mathrm{which}\:\mathrm{period}, \\ $$$$\mathrm{2}\pi\:\mathrm{or}\:\mathrm{2}\frac{{u}}{{v}}? \\ $$$$\mathrm{is}\:\mathrm{it}\:\frac{{v}^{\mathrm{2}{n}} }{{n}!}\underset{\mathrm{0}} {\overset{\frac{{u}}{{v}}} {\int}}{x}^{{n}} \left({x}−\frac{{u}}{{v}}\right)\mathrm{sin}\:\frac{{v}\pi{x}}{{u}}\:{dx}? \\ $$$$\mathrm{or}\:\mathrm{shall}\:\mathrm{we}\:\mathrm{calculate}\:{I}_{{n}} \:\mathrm{using}\:\mathrm{the}\:\mathrm{usual}\:\pi \\ $$$$\mathrm{and}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\pi\notin\mathbb{Q}? \\ $$
Commented by MJS last updated on 23/May/19
sorry this is quite confusing me
$$\mathrm{sorry}\:\mathrm{this}\:\mathrm{is}\:\mathrm{quite}\:\mathrm{confusing}\:\mathrm{me} \\ $$
Commented by MJS last updated on 23/May/19
I_1 =−4v^2   I_2 =2v^4 (−π^2 +12)  I_3 =24v^6 (π^2 −10)  I_4 =2v^8 (π^4 −180π^2 +1680)  I_5 =60v^(10) (−π^4 +112π^2 −1008)  ...  all calculated with the usual π  if π=(u/v)  I_1 =−4v^2   I_2 =−2(u^2 −12v^2 )v^2   I_3 =24(u^2 −10v^2 )v^4   I_4 =2(u^4 −180u^2 v^2 +1680v^4 )v^4   I_5 =−60(u^4 −112u^2 v^2 +1008v^4 )v^6   ...  but what do we know now?
$${I}_{\mathrm{1}} =−\mathrm{4}{v}^{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\mathrm{2}{v}^{\mathrm{4}} \left(−\pi^{\mathrm{2}} +\mathrm{12}\right) \\ $$$${I}_{\mathrm{3}} =\mathrm{24}{v}^{\mathrm{6}} \left(\pi^{\mathrm{2}} −\mathrm{10}\right) \\ $$$${I}_{\mathrm{4}} =\mathrm{2}{v}^{\mathrm{8}} \left(\pi^{\mathrm{4}} −\mathrm{180}\pi^{\mathrm{2}} +\mathrm{1680}\right) \\ $$$${I}_{\mathrm{5}} =\mathrm{60}{v}^{\mathrm{10}} \left(−\pi^{\mathrm{4}} +\mathrm{112}\pi^{\mathrm{2}} −\mathrm{1008}\right) \\ $$$$… \\ $$$$\mathrm{all}\:\mathrm{calculated}\:\mathrm{with}\:\mathrm{the}\:\mathrm{usual}\:\pi \\ $$$$\mathrm{if}\:\pi=\frac{{u}}{{v}} \\ $$$${I}_{\mathrm{1}} =−\mathrm{4}{v}^{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =−\mathrm{2}\left({u}^{\mathrm{2}} −\mathrm{12}{v}^{\mathrm{2}} \right){v}^{\mathrm{2}} \\ $$$${I}_{\mathrm{3}} =\mathrm{24}\left({u}^{\mathrm{2}} −\mathrm{10}{v}^{\mathrm{2}} \right){v}^{\mathrm{4}} \\ $$$${I}_{\mathrm{4}} =\mathrm{2}\left({u}^{\mathrm{4}} −\mathrm{180}{u}^{\mathrm{2}} {v}^{\mathrm{2}} +\mathrm{1680}{v}^{\mathrm{4}} \right){v}^{\mathrm{4}} \\ $$$${I}_{\mathrm{5}} =−\mathrm{60}\left({u}^{\mathrm{4}} −\mathrm{112}{u}^{\mathrm{2}} {v}^{\mathrm{2}} +\mathrm{1008}{v}^{\mathrm{4}} \right){v}^{\mathrm{6}} \\ $$$$… \\ $$$$\mathrm{but}\:\mathrm{what}\:\mathrm{do}\:\mathrm{we}\:\mathrm{know}\:\mathrm{now}? \\ $$

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