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Question Number 137668 Answers: 0 Comments: 2

.......advanced ... ... ... ... calculus.... prove that::::: ∗∗∗∗∗ 𝛀=Σ_(n=0) ^∞ ((Γ(n+1)Γ(x+1))/((n+1)Γ(n+x+2)))=ψ′(x+1) proof:: Ω=Σ_(n=0 ) ^∞ ((β (n+1,x+1))/(n+1))=Σ_(n=0) ^∞ {(1/((n+1)))∫_0 ^( 1) t^n .(1−t)^x dt} =∫_0 ^( 1) {(1−t)^x Σ_(n=0 ) ^∞ (t^n /(n+1))dt} =∫_0 ^( 1) {(1−t)^x Σ_(n=1) ^∞ (t^(n−1) /n)dt} =−∫_0 ^( 1) (((1−t)^x ln(1−t))/t)dt=−∫_0 ^( 1) ((t^x ln(t))/(1−t))dt = (∂/∂x)∫_0 ^( 1) ((1−t^x )/(1−t))dt=ψ′(1+x) .....✓✓ ....... 𝛀= ψ′(1+x) ......

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......{advanced}\:...\:...\:...\:...\:{calculus}.... \\ $$$$\:\:{prove}\:{that}:::::\:\ast\ast\ast\ast\ast \\ $$$$\:\:\:\:\boldsymbol{\Omega}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left({n}+\mathrm{1}\right)\Gamma\left({x}+\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)\Gamma\left({n}+{x}+\mathrm{2}\right)}=\psi'\left({x}+\mathrm{1}\right) \\ $$$${proof}:: \\ $$$$\:\:\:\:\Omega=\underset{{n}=\mathrm{0}\:} {\overset{\infty} {\sum}}\frac{\beta\:\left({n}+\mathrm{1},{x}+\mathrm{1}\right)}{{n}+\mathrm{1}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\int_{\mathrm{0}} ^{\:\mathrm{1}} {t}^{{n}} .\left(\mathrm{1}−{t}\right)^{{x}} {dt}\right\} \\ $$$$\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\left(\mathrm{1}−{t}\right)^{{x}} \underset{{n}=\mathrm{0}\:} {\overset{\infty} {\sum}}\frac{{t}^{{n}} }{{n}+\mathrm{1}}{dt}\right\} \\ $$$$\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\left(\mathrm{1}−{t}\right)^{{x}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{t}^{{n}−\mathrm{1}} }{{n}}{dt}\right\} \\ $$$$\:=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\left(\mathrm{1}−{t}\right)^{{x}} {ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt}=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{t}^{{x}} {ln}\left({t}\right)}{\mathrm{1}−{t}}{dt} \\ $$$$=\:\frac{\partial}{\partial{x}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{t}^{{x}} }{\mathrm{1}−{t}}{dt}=\psi'\left(\mathrm{1}+{x}\right)\:.....\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......\:\:\boldsymbol{\Omega}=\:\psi'\left(\mathrm{1}+{x}\right)\:...... \\ $$

Question Number 137667 Answers: 1 Comments: 0

y=log_2 (x^5 +4) find (dy/dx)

$${y}={log}_{\mathrm{2}} \left({x}^{\mathrm{5}} +\mathrm{4}\right)\:{find}\:\frac{{dy}}{{dx}} \\ $$

Question Number 137665 Answers: 0 Comments: 0

Question Number 137659 Answers: 0 Comments: 2

Question Number 137656 Answers: 1 Comments: 0

∫_0 ^∞ sin(x^p )dx

$$\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{{p}} \right){dx} \\ $$

Question Number 137652 Answers: 1 Comments: 0

Given that α,β,γ are the three roots of the equation (x−59)^3 +(2x−64)^3 =(3x−123)^3 , find the value of α+β+γ.

$$\mathrm{Given}\:\mathrm{that}\:\alpha,\beta,\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{three}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\left({x}−\mathrm{59}\right)^{\mathrm{3}} +\left(\mathrm{2}{x}−\mathrm{64}\right)^{\mathrm{3}} =\left(\mathrm{3}{x}−\mathrm{123}\right)^{\mathrm{3}} , \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha+\beta+\gamma. \\ $$

Question Number 137650 Answers: 2 Comments: 0

Question Number 137648 Answers: 1 Comments: 1

Question Number 137645 Answers: 1 Comments: 0

The sum of the squares of two consecutive odd numbers is 130. Find the larger number.

$${The}\:{sum}\:{of}\:{the}\:{squares}\:{of}\:{two}\:{consecutive} \\ $$$${odd}\:{numbers}\:{is}\:\mathrm{130}.\:{Find}\:{the}\:{larger}\:{number}. \\ $$

Question Number 137686 Answers: 0 Comments: 0

Question Number 137694 Answers: 1 Comments: 3

find ∫ (x^3 /((x−3)^3 (x(√2)+7)^4 ))dx

$${find}\:\int\:\:\:\frac{{x}^{\mathrm{3}} }{\left({x}−\mathrm{3}\right)^{\mathrm{3}} \left({x}\sqrt{\mathrm{2}}+\mathrm{7}\right)^{\mathrm{4}} }{dx} \\ $$

Question Number 137641 Answers: 1 Comments: 0

Question Number 137637 Answers: 0 Comments: 0

Prove or disprove ((cos(1+(1/( (√3))))𝛑)/1^2 )+((cos(1+(1/( (√3))))2π)/2^2 )+((cos(1+(1/( (√3))))3π)/3^2 )+...=0

$$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{disprove}} \\ $$$$\frac{{cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\boldsymbol{\pi}}{\mathrm{1}^{\mathrm{2}} }+\frac{{cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\mathrm{2}\pi}{\mathrm{2}^{\mathrm{2}} }+\frac{{cos}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\mathrm{3}\pi}{\mathrm{3}^{\mathrm{2}} }+...=\mathrm{0} \\ $$

Question Number 137635 Answers: 0 Comments: 1

Question Number 137634 Answers: 1 Comments: 0

x=2^(p ) and y=2^(q ) . Evaluate in terms of x and/ or y (i)2^(p+q) (ii) 2^(2q ) (iii) 2^(p−1)

$${x}=\mathrm{2}^{{p}\:} {and}\:{y}=\mathrm{2}^{{q}\:} .\:{Evaluate}\:{in}\:{terms}\:{of}\:{x}\:{and}/\:{or}\:{y}\: \\ $$$$\left({i}\right)\mathrm{2}^{{p}+{q}} \:\:\left({ii}\right)\:\mathrm{2}^{\mathrm{2}{q}\:} \:\:\:\left({iii}\right)\:\mathrm{2}^{{p}−\mathrm{1}} \\ $$

Question Number 137627 Answers: 1 Comments: 0

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Question Number 137624 Answers: 0 Comments: 0

(dy/dx) = ((6xy+y^2 )/(x+y))

$$\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}\:=\:\frac{\mathrm{6}\boldsymbol{\mathrm{xy}}+\boldsymbol{\mathrm{y}}^{\mathrm{2}} }{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}} \\ $$

Question Number 137618 Answers: 1 Comments: 2

Question Number 137615 Answers: 0 Comments: 1

∫_0 ^( (π/2)) ln(((1+tanx)/(1−tanx)))^(xcos(8x)) dx

$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{1}+{tanx}}{\mathrm{1}−{tanx}}\right)^{{xcos}\left(\mathrm{8}{x}\right)} {dx} \\ $$

Question Number 137613 Answers: 1 Comments: 0

Two particles of mass m and 4m are connected by a light inelastic string of length 3a, which passes through a small smooth fixed ring. The heavier particle hangs at rest at a distance 2a beneath the ring, while the lighter particle describes a horizontal circle with constant speed. Find (a) The distance of the plane of the circle below the ring. (b) The angular speed of the lighter particle.

$$\mathrm{Two}\:\mathrm{particles}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{and}\:\mathrm{4}{m}\:\mathrm{are}\:\mathrm{connected}\:\mathrm{by}\:\mathrm{a}\:\mathrm{light}\:\mathrm{inelastic} \\ $$$$\mathrm{string}\:\mathrm{of}\:\mathrm{length}\:\mathrm{3}{a},\:\mathrm{which}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{a}\:\mathrm{small}\:\mathrm{smooth}\:\mathrm{fixed}\:\mathrm{ring}. \\ $$$$\mathrm{The}\:\mathrm{heavier}\:\mathrm{particle}\:\mathrm{hangs}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{2}{a}\:\mathrm{beneath}\:\mathrm{the}\:\mathrm{ring}, \\ $$$$\mathrm{while}\:\mathrm{the}\:\mathrm{lighter}\:\mathrm{particle}\:\mathrm{describes}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{speed}. \\ $$$$\:\mathrm{Find} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{The}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{plane}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{below}\:\mathrm{the}\:\mathrm{ring}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{angular}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lighter}\:\mathrm{particle}. \\ $$

Question Number 137611 Answers: 1 Comments: 0

Question Number 137610 Answers: 0 Comments: 0

........ mathematical analysis (II).... prove that :: 𝛀=∫_0 ^( 1) (1/(1+x))ln(((x^2 +2x+1)/(1+x+x^2 )))=Σ_(n=1) ^∞ (1/(n^2 (((2n)),(( n)) )))=(π^2 /(18))..

$$\:\:\:\:\:\:\:\:\:\:........\:{mathematical}\:\:\:{analysis}\:\left({II}\right).... \\ $$$$\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}{ln}\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{\mathrm{1}+{x}+{x}^{\mathrm{2}} }\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix}}=\frac{\pi^{\mathrm{2}} }{\mathrm{18}}.. \\ $$

Question Number 137605 Answers: 0 Comments: 0

.....nice calculus... prove that:: Σ_(n=0) ^∞ tan^(−1) ((1/F_n )).tan^(−1) ((1/F_(n+1) ))=(π^2 /4) F_n is fibonacci sequence....

$$\:\:\:\:\:\:\:\:\:\:\:\:.....{nice}\:\:\:\:{calculus}... \\ $$$$\:\:\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{F}_{{n}} }\right).{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{F}_{{n}+\mathrm{1}} }\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:{F}_{{n}} \:{is}\:{fibonacci}\:{sequence}.... \\ $$

Question Number 137598 Answers: 0 Comments: 0

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