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Question Number 213648 Answers: 0 Comments: 0

a_m =Σ_(h=m) ^∞ ((1/h))^2 =(π^2 /6)−Σ_(h=1) ^m ((1/h))^2 a_m ≈𝛙^((1)) (m+1) 𝛙^((n)) (z)=((d^(n+1) )/dz^(n+1) )ln(𝚪(z)) , C\Z_(≤0) aka polygamma function 1. lim_(m→∞) m𝛙^((1)) (m+1)=1. and... 2. lim_(m→∞) m^2 𝛙^((1)) (m+1)=∞ div m^2 𝛙^((1)) (m+1)≈m−𝛜 lim_(m→∞) m^2 𝛙^((1)) (m+1)≈lim_(m→∞) (m−𝛜)=∞

$${a}_{{m}} =\underset{{h}={m}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{h}}\right)^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\underset{{h}=\mathrm{1}} {\overset{{m}} {\sum}}\:\left(\frac{\mathrm{1}}{{h}}\right)^{\mathrm{2}} \\ $$$${a}_{{m}} \approx\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right) \\ $$$$\boldsymbol{\psi}^{\left({n}\right)} \left({z}\right)=\frac{\mathrm{d}^{{n}+\mathrm{1}} \:\:}{\mathrm{d}{z}^{{n}+\mathrm{1}} }\mathrm{ln}\left(\boldsymbol{\Gamma}\left({z}\right)\right)\:,\:\mathbb{C}\backslash\mathbb{Z}_{\leq\mathrm{0}} \\ $$$$\mathrm{aka}\:\:\mathrm{polygamma}\:\mathrm{function} \\ $$$$\mathrm{1}.\:\:\underset{{m}\rightarrow\infty} {\mathrm{lim}}\:{m}\boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right)=\mathrm{1}. \\ $$$$\mathrm{and}... \\ $$$$\mathrm{2}.\:\:\:\underset{{m}\rightarrow\infty} {\mathrm{lim}}{m}^{\mathrm{2}} \boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right)=\infty\:\mathrm{div} \\ $$$${m}^{\mathrm{2}} \boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right)\approx{m}−\boldsymbol{\varepsilon} \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\:{m}^{\mathrm{2}} \boldsymbol{\psi}^{\left(\mathrm{1}\right)} \left({m}+\mathrm{1}\right)\approx\underset{{m}\rightarrow\infty} {\mathrm{lim}}\:\left({m}−\boldsymbol{\varepsilon}\right)=\infty \\ $$

Question Number 213643 Answers: 3 Comments: 0

Find the maximum value of 3sin^2 x−8cosx+5=?

$$\:{Find}\:{the}\:{maximum}\:{value}\:{of} \\ $$$$\:\mathrm{3}{sin}^{\mathrm{2}} {x}−\mathrm{8}{cosx}+\mathrm{5}=? \\ $$

Question Number 213642 Answers: 1 Comments: 0

Question Number 213641 Answers: 1 Comments: 0

Question Number 213639 Answers: 1 Comments: 0

Question Number 213636 Answers: 2 Comments: 0

x^2 − x + 1 = 0 find: x^(10) = ?

$$\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{find}:\:\:\:\mathrm{x}^{\mathrm{10}} \:=\:? \\ $$

Question Number 213632 Answers: 0 Comments: 0

Question Number 213628 Answers: 1 Comments: 0

Question Number 213626 Answers: 1 Comments: 3

Question Number 213624 Answers: 1 Comments: 0

x^3 + 5x − 42 = (x − 3)∙P(x) Find: P(3) = ?

$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{5x}\:−\:\mathrm{42}\:=\:\left(\mathrm{x}\:−\:\mathrm{3}\right)\centerdot\mathrm{P}\left(\mathrm{x}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{P}\left(\mathrm{3}\right)\:=\:? \\ $$

Question Number 213621 Answers: 1 Comments: 0

a,b>0 (1/a) + (1/b) = 2 and a^2 + b^2 = 12 a + b = ?

$$\mathrm{a},\mathrm{b}>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:=\:\mathrm{2}\:\:\:\mathrm{and}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{12} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:=\:? \\ $$

Question Number 213615 Answers: 1 Comments: 0

Question Number 213614 Answers: 0 Comments: 1

Question Number 213613 Answers: 1 Comments: 0

Question Number 213607 Answers: 0 Comments: 0

Give a,b,c > 0 such that ab+bc+ca=abc Prove that : a+b+c≥4((a/(bc))+(b/(ca))+(c/(ab)))+5

$${Give}\:{a},{b},{c}\:>\:\mathrm{0}\:{such}\:{that} \\ $$$${ab}+{bc}+{ca}={abc} \\ $$$${Prove}\:{that}\:: \\ $$$${a}+{b}+{c}\geqslant\mathrm{4}\left(\frac{{a}}{{bc}}+\frac{{b}}{{ca}}+\frac{{c}}{{ab}}\right)+\mathrm{5} \\ $$

Question Number 213606 Answers: 2 Comments: 0

△ABC. 2a+b=2c. Find the minimum of (3/(sin C)) + (1/(tan A)).

$$\bigtriangleup{ABC}.\:\mathrm{2}{a}+{b}=\mathrm{2}{c}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of} \\ $$$$\frac{\mathrm{3}}{\mathrm{sin}\:{C}}\:+\:\frac{\mathrm{1}}{\mathrm{tan}\:{A}}. \\ $$

Question Number 213604 Answers: 0 Comments: 0

show that ∫_C e^z^3 dz=0 where C is any simple closed contour. Evaluate the integral ∫_( C_1 ) f(z)dz , ∫_( C_2 ) f(z)dz where f(z)=(y−x)−3x^2 i C_3 =OA ; z(y)=x+iy=iy , (0≤y≤1) C_1 =AB ; z(x)=x+iy=x+i , (0≤x≤1) C_2 =OB ; z(x)=x+iy=x+ix , (0≤x≤1) Let′s C be the quadrant z=2e^(iθ) ,0≤θ≤(π/2) show that ∣∫_( C) ((z^ +4)/(z^3 −1)) dz∣≤((6π)/7) Let C be any simple closed contour described in the positive sense in the z plane and write g(z)=∫_( C) ((s^3 +2s)/((z−2s)^3 )) ds show that g(z)=6πiz when z is inside C show that g(z)=0 when z is outside C

$$\mathrm{show}\:\mathrm{that}\:\:\int_{\boldsymbol{\mathcal{C}}} \:{e}^{{z}^{\mathrm{3}} } \:\mathrm{d}{z}=\mathrm{0} \\ $$$$\mathrm{where}\:\boldsymbol{\mathcal{C}}\:\mathrm{is}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}. \\ $$$$\:\:\:\: \\ $$$$\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\int_{\:{C}_{\mathrm{1}} } {f}\left({z}\right)\mathrm{d}{z}\:,\:\int_{\:{C}_{\mathrm{2}} } {f}\left({z}\right)\mathrm{d}{z} \\ $$$$\mathrm{where}\:{f}\left({z}\right)=\left({y}−{x}\right)−\mathrm{3}{x}^{\mathrm{2}} \boldsymbol{{i}} \\ $$$${C}_{\mathrm{3}} ={OA}\:;\:{z}\left({y}\right)={x}+\boldsymbol{{i}}{y}=\boldsymbol{{i}}{y}\:,\:\left(\mathrm{0}\leq{y}\leq\mathrm{1}\right) \\ $$$${C}_{\mathrm{1}} ={AB}\:;\:{z}\left({x}\right)={x}+\boldsymbol{{i}}{y}={x}+\boldsymbol{{i}}\:,\:\left(\mathrm{0}\leq{x}\leq\mathrm{1}\right) \\ $$$${C}_{\mathrm{2}} ={OB}\:;\:{z}\left({x}\right)={x}+\boldsymbol{{i}}{y}={x}+\boldsymbol{{i}}{x}\:,\:\left(\mathrm{0}\leq{x}\leq\mathrm{1}\right) \\ $$$$\: \\ $$$$\mathrm{Let}'\mathrm{s}\:{C}\:\mathrm{be}\:\mathrm{the}\:\mathrm{quadrant} \\ $$$${z}=\mathrm{2}{e}^{\boldsymbol{{i}}\theta} \:,\mathrm{0}\leq\theta\leq\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{show}\:\mathrm{that}\: \\ $$$$\mid\int_{\:{C}} \:\frac{{z}^{\:} +\mathrm{4}}{{z}^{\mathrm{3}} −\mathrm{1}}\:\mathrm{d}{z}\mid\leq\frac{\mathrm{6}\pi}{\mathrm{7}} \\ $$$$\: \\ $$$$\mathrm{Let}\:{C}\:\mathrm{be}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour} \\ $$$$\mathrm{described}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{sense}\:\mathrm{in}\:\mathrm{the}\:{z}\:\mathrm{plane} \\ $$$$\mathrm{and}\:\mathrm{write} \\ $$$$\mathrm{g}\left({z}\right)=\int_{\:{C}} \:\:\frac{{s}^{\mathrm{3}} +\mathrm{2}{s}}{\left({z}−\mathrm{2}{s}\right)^{\mathrm{3}} }\:\mathrm{d}{s} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{g}\left({z}\right)=\mathrm{6}\pi\boldsymbol{{i}}{z}\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{inside}\:{C}\: \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{g}\left({z}\right)=\mathrm{0}\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{outside}\:{C} \\ $$

Question Number 213589 Answers: 2 Comments: 0

Find: (((3 - (3/4))∙(3 - (3/5))∙(3 - (1/2))∙(3 - (3/7))∙...∙(3 - (1/6)))/(27^5 )) = ?

$$\mathrm{Find}: \\ $$$$\frac{\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{4}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{5}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{1}}{\mathrm{2}}\right)\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{3}}{\mathrm{7}}\right)\centerdot...\centerdot\left(\mathrm{3}\:-\:\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{27}^{\mathrm{5}} }\:=\:? \\ $$

Question Number 213580 Answers: 0 Comments: 2