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Question Number 34678 Answers: 0 Comments: 0

prove that ∀ n≥3 (√n) <^n (√(n!))

$${prove}\:{that}\:\forall\:{n}\geqslant\mathrm{3}\:\:\:\:\:\sqrt{{n}}\:\:<^{{n}} \sqrt{{n}!} \\ $$$$ \\ $$

Question Number 34677 Answers: 0 Comments: 0

prove that Σ_(k=0) ^(n−1) [x +(k/n)] =[nx] ∀ n∈ ∈N^★

$${prove}\:{that}\:\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left[{x}\:+\frac{{k}}{{n}}\right]\:=\left[{nx}\right]\:\:\forall\:{n}\in\:\in{N}^{\bigstar} \\ $$

Question Number 34676 Answers: 0 Comments: 0

prove that Σ_(k=0) ^(2n−1) (((−1)^k )/(k+1)) =Σ_(k=n+1) ^(2n) (1/k)

$${prove}\:{that}\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\:=\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\:\frac{\mathrm{1}}{{k}} \\ $$

Question Number 34675 Answers: 0 Comments: 0

provethat e = Σ_(k=0) ^n (1/(k!)) +∫_0 ^1 (((1−t)^n )/(n!)) e^t dt .

$${provethat}\:{e}\:=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{k}!}\:\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left(\mathrm{1}−{t}\right)^{{n}} }{{n}!}\:{e}^{{t}} \:{dt}\:. \\ $$

Question Number 34674 Answers: 0 Comments: 0

find ∫_0 ^π ((x sinx)/(1+cos^2 x)) dx

$${find}\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{x}\:{sinx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:{dx} \\ $$

Question Number 34673 Answers: 0 Comments: 0

solve (((1+iz)/(1−iz)))^n = ((1+itanα)/(1−itanα)) with −(π/2)<α<(π/2)

$${solve}\:\left(\frac{\mathrm{1}+{iz}}{\mathrm{1}−{iz}}\right)^{{n}} \:=\:\frac{\mathrm{1}+{itan}\alpha}{\mathrm{1}−{itan}\alpha}\:\:{with}\:−\frac{\pi}{\mathrm{2}}<\alpha<\frac{\pi}{\mathrm{2}} \\ $$

Question Number 34672 Answers: 0 Comments: 0

prove that ∀n∈N ∣sin(nx)∣≤n∣sinx∣ .

$${prove}\:{that}\:\forall{n}\in{N}\:\:\:\mid{sin}\left({nx}\right)\mid\leqslant{n}\mid{sinx}\mid\:. \\ $$

Question Number 34671 Answers: 0 Comments: 0

calculste Σ_(n=1) ^∞ arctan((2/n^2 )).

$${calculste}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{arctan}\left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right). \\ $$

Question Number 34669 Answers: 0 Comments: 1

let P(x)=(1+x+ix^2 )^n −(1+x −ix^2 )^n 1) find the roots of P(x) 2) factorize inside C[x] P(x) 3) factorize indide R[x] P(x).

$${let}\:{P}\left({x}\right)=\left(\mathrm{1}+{x}+{ix}^{\mathrm{2}} \right)^{{n}} \:−\left(\mathrm{1}+{x}\:−{ix}^{\mathrm{2}} \right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{P}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{inside}\:{C}\left[{x}\right]\:{P}\left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{factorize}\:{indide}\:{R}\left[{x}\right]\:{P}\left({x}\right). \\ $$

Question Number 34668 Answers: 0 Comments: 0

find the roots of?p(x) = x^(2n) −2x^n cos(nθ) +1 2)?factorize p(x)

$${find}\:{the}\:{roots}\:{of}?{p}\left({x}\right)\:=\:{x}^{\mathrm{2}{n}} \:−\mathrm{2}{x}^{{n}} \:{cos}\left({n}\theta\right)\:+\mathrm{1} \\ $$$$\left.\mathrm{2}\right)?{factorize}\:{p}\left({x}\right)\: \\ $$

Question Number 34667 Answers: 0 Comments: 0

solve (x+1)^n = e^(2ina) then find the value of P_n = Π_(k=0) ^(n−1) sin(a +((kπ)/n))

$${solve}\:\:\left({x}+\mathrm{1}\right)^{{n}} \:=\:{e}^{\mathrm{2}{ina}} \:\:\:{then}\:{find}\:{the}\:{value}\:{of} \\ $$$${P}_{{n}} =\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left({a}\:+\frac{{k}\pi}{{n}}\right) \\ $$

Question Number 34666 Answers: 0 Comments: 0

simplify sin^2 ( ((arccosx)/2))

$${simplify}\:{sin}^{\mathrm{2}} \left(\:\frac{{arccosx}}{\mathrm{2}}\right) \\ $$

Question Number 34665 Answers: 0 Comments: 0

simplify sin (2arcsinx)

$${simplify}\:\:{sin}\:\left(\mathrm{2}{arcsinx}\right) \\ $$

Question Number 34664 Answers: 0 Comments: 0

simplify g(x)= arctan((1/(2x^2 ))) −arctan((x/(x+1))) +arctan(((x−1)/x))

$${simplify} \\ $$$${g}\left({x}\right)=\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right)\:−{arctan}\left(\frac{{x}}{{x}+\mathrm{1}}\right)\:+{arctan}\left(\frac{{x}−\mathrm{1}}{{x}}\right) \\ $$

Question Number 34663 Answers: 0 Comments: 0

simplify f(x)=arcsin((√(1−x^2 ))) −arctan((√((1−x)/(1+x))))

$${simplify}\: \\ $$$${f}\left({x}\right)={arcsin}\left(\sqrt{\left.\mathrm{1}−{x}^{\mathrm{2}} \right)}\:\:−{arctan}\left(\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\right)\right. \\ $$

Question Number 34662 Answers: 0 Comments: 0

calculate I(a) =∫_(1/a) ^a ((ln(x))/(1+x^2 )) dx with a>0 2) calculate ∫_0 ^(+∞) ((ln(x))/(1+x^2 )) dx .

$${calculate}\:{I}\left({a}\right)\:\:=\int_{\frac{\mathrm{1}}{{a}}} ^{{a}} \:\:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:\:{with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:. \\ $$

Question Number 34661 Answers: 0 Comments: 0

let f(x)= ∫_0 ^1 (e^(−(1+t^2 )x) /(1+t^2 )) dt find a simple form of f(x)

$${let}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{−\left(\mathrm{1}+{t}^{\mathrm{2}} \right){x}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:{find}\:{a}\:{simple}\:{form}\:{of} \\ $$$${f}\left({x}\right) \\ $$

Question Number 34653 Answers: 0 Comments: 0

Question Number 34647 Answers: 2 Comments: 0

if xsin^3 θ + ycos^3 θ=sinθcosθ and xsinθ −ycosθ=0 prove that x^2 + y^2 =1

$${if}\:\:\:{xsin}^{\mathrm{3}} \theta\:+\:{ycos}^{\mathrm{3}} \theta={sin}\theta{cos}\theta \\ $$$${and}\:{xsin}\theta\:−{ycos}\theta=\mathrm{0} \\ $$$${prove}\:{that}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} =\mathrm{1} \\ $$

Question Number 34646 Answers: 1 Comments: 0

show that 2tan^(−1) 2 + tan^(−1) 3= Π +tan^(−1) (1/3)

$${show}\:{that}\:\mathrm{2}{tan}^{−\mathrm{1}} \mathrm{2}\:+\:{tan}^{−\mathrm{1}} \mathrm{3}= \\ $$$$\Pi\:+{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}} \\ $$

Question Number 34645 Answers: 0 Comments: 0

show that 2tan^(−1) 2 + tan^(−1) 3= Π +tan^(−1) (1/3)

$${show}\:{that}\:\mathrm{2}{tan}^{−\mathrm{1}} \mathrm{2}\:+\:{tan}^{−\mathrm{1}} \mathrm{3}= \\ $$$$\Pi\:+{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}} \\ $$

Question Number 34639 Answers: 2 Comments: 0

Question Number 34635 Answers: 2 Comments: 4

calculate A(α) = ∫_0 ^1 ln(1+αix)dx 2) calculate ∫_0 ^1 ln(1+ix) dx (i^2 =−1)

$${calculate}\:{A}\left(\alpha\right)\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+\alpha{ix}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}+{ix}\right)\:{dx}\:\:\:\:\left({i}^{\mathrm{2}} \:=−\mathrm{1}\right) \\ $$

Question Number 34634 Answers: 1 Comments: 1

let f(x)=ln(1+ix) with ∣x∣<1 1) extract Re(f(x)) and Im(f(x)) 2) developp f(x) at integr serie.

$${let}\:{f}\left({x}\right)={ln}\left(\mathrm{1}+{ix}\right)\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{extract}\:{Re}\left({f}\left({x}\right)\right)\:{and}\:{Im}\left({f}\left({x}\right)\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\left({x}\right)\:{at}\:{integr}\:{serie}. \\ $$

Question Number 34633 Answers: 0 Comments: 0

let f(α) = ∫_(−∞) ^(+∞) ((arctan(1+αxi))/(1+x^2 ))dx find f(α) .

$${let}\:{f}\left(\alpha\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{arctan}\left(\mathrm{1}+\alpha{xi}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${find}\:{f}\left(\alpha\right)\:. \\ $$

Question Number 34632 Answers: 0 Comments: 0

let z∈C developp at integrserie f(z)=ln(1+z) with ∣z∣<1 . 2) give ln(2+i) at form of serie.

$${let}\:{z}\in{C}\:\:{developp}\:{at}\:{integrserie} \\ $$$${f}\left({z}\right)={ln}\left(\mathrm{1}+{z}\right)\:\:{with}\:\mid{z}\mid<\mathrm{1}\:. \\ $$$$\left.\mathrm{2}\right)\:{give}\:{ln}\left(\mathrm{2}+{i}\right)\:{at}\:{form}\:{of}\:{serie}. \\ $$

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