Menu Close

let-f-x-0-pi-2-d-x-cos-2-with-x-gt-0-1-calculate-f-x-and-f-x-2-find-f-n-x-and-f-n-0-3-developp-f-at-integr-serie-

Tinku Tara 0 Comments
Pin




Question Number 40619 by math khazana by abdo last updated on 25/Jul/18
let f(x)=∫_0 ^(π/2) (dθ/(x +cos^2 θ)) with x>0 . 1) calculate f(x) and f^′ (x) 2) find f^((n)) (x) and f^((n)) (0) 3) developp f at integr serie.
$${let}\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{d}\theta}{{x}\:\:+{cos}^{\mathrm{2}} \theta}\:\:{with}\:{x}>\mathrm{0}\:. \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right)\:{and}\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by abdo mathsup 649 cc last updated on 27/Jul/18
1) we have proved that f(x)=((π(√2))/(2(√x))) ⇒ f^′ (x)=((π(√2))/2)(x^(−(1/2)) )^′ =−(1/2) ((π(√2))/2) x^(−(3/2)) =−((π(√2))/(4x(√x))) 2) f^((n)) (x)=((π(√2))/2) (x^(−(1/2)) )^((n)) let find (x^p )^((n)) with p ∈Q (x^p )^((1)) =px^(p−1) , (x^p )^((2)) =p(p−1)x^(p−2) ⇒ (x^p )^((n)) =p(p−1)...(p−n+1)x^(p−n) ⇒ (x^(−(1/2)) )^((n)) =(−(1/2))(−(3/2))...(−(1/2)−n+1)x^(−(1/2)−n) =(−(1/2))(−(3/2))...(((−1−2n+2)/2))x^(−(1/2)−n) =(−(1/2))(−(3/2))....(−((2n−1)/2))(1/(x^n (√x))) ⇒ f^((n)) (x)=(π/( (√2))) (−(1/2))(−(3/2))...(−((2n−1)/2)) (1/(x^n (√x))) f^((n)) (1)=(π/( (√2)))(−(1/2))(−(3/2))...(−((2n−1)/2)) 3) f(x) =Σ_(n=0) ^∞ ((f^((n)) (1))/(n!))(x−1)^n = Σ_(n=0) ^∞ (π/(n!(√2)))(−(1/2))(−(3/2))...(−((2n−1)/2))(x−1)^n .
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{proved}\:{that}\:{f}\left({x}\right)=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{{x}}}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\left({x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{'} \:=−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:{x}^{−\frac{\mathrm{3}}{\mathrm{2}}} =−\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}{x}\sqrt{{x}}} \\ $$$$\left.\mathrm{2}\right)\:{f}^{\left({n}\right)} \left({x}\right)=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\left({x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\left({n}\right)} \:\:\:{let}\:{find} \\ $$$$\left({x}^{{p}} \right)^{\left({n}\right)} \:{with}\:{p}\:\in{Q}\: \\ $$$$\left({x}^{{p}} \right)^{\left(\mathrm{1}\right)} ={px}^{{p}−\mathrm{1}} \:,\:\:\left({x}^{{p}} \right)^{\left(\mathrm{2}\right)} ={p}\left({p}−\mathrm{1}\right){x}^{{p}−\mathrm{2}} \Rightarrow \\ $$$$\left({x}^{{p}} \right)^{\left({n}\right)} \:={p}\left({p}−\mathrm{1}\right)…\left({p}−{n}+\mathrm{1}\right){x}^{{p}−{n}} \:\Rightarrow \\ $$$$\left({x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\left({n}\right)} =\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)…\left(−\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right){x}^{−\frac{\mathrm{1}}{\mathrm{2}}−{n}} \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)…\left(\frac{−\mathrm{1}−\mathrm{2}{n}+\mathrm{2}}{\mathrm{2}}\right){x}^{−\frac{\mathrm{1}}{\mathrm{2}}−{n}} \\ $$$$=\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)….\left(−\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)\frac{\mathrm{1}}{{x}^{{n}} \sqrt{{x}}}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)…\left(−\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)\:\frac{\mathrm{1}}{{x}^{{n}} \sqrt{{x}}} \\ $$$${f}^{\left({n}\right)} \left(\mathrm{1}\right)=\frac{\pi}{\:\sqrt{\mathrm{2}}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)…\left(−\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{3}\right)\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{1}\right)}{{n}!}\left({x}−\mathrm{1}\right)^{{n}} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\pi}{{n}!\sqrt{\mathrm{2}}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)…\left(−\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)\left({x}−\mathrm{1}\right)^{{n}} . \\ $$
Commented by abdo mathsup 649 cc last updated on 27/Jul/18
2) the Q is find f^((n)) (x) and f^((n)) (1) 3) the Q is developp f at integr serie at v(1).
$$\left.\mathrm{2}\right)\:{the}\:{Q}\:{is}\:{find}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right)\:{the}\:{Q}\:{is}\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:{at}\:{v}\left(\mathrm{1}\right). \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *