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Question Number 54013    Answers: 1   Comments: 0

2^x =−4.solve for x

$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} =−\mathrm{4}.\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}} \\ $$

Question Number 54011    Answers: 1   Comments: 2

calculate f(a) =∫ (dx/((√(1+ax))−(√(1−ax)))) with a>0 . 2) calculate U_n =∫_(−(1/(na))) ^(1/(na)) (dx/((√(1+ax))−(√(1−ax)))) with n from N and n>1 find lim_(n→+∞) U_n and study the convergence of Σ U_n

$${calculate}\:{f}\left({a}\right)\:=\int\:\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}+{ax}}−\sqrt{\mathrm{1}−{ax}}}\:\:{with}\:{a}>\mathrm{0}\:. \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:{U}_{{n}} =\int_{−\frac{\mathrm{1}}{{na}}} ^{\frac{\mathrm{1}}{{na}}} \:\:\frac{{dx}}{\sqrt{\mathrm{1}+{ax}}−\sqrt{\mathrm{1}−{ax}}}\:\:{with}\:{n}\:{from}\:{N}\:{and}\:{n}>\mathrm{1} \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \:\:\:{and}\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{U}_{{n}} \\ $$

Question Number 53998    Answers: 0   Comments: 1

In the next link, Barry Barrish, who won the physic nobel prize in 2017; gave an exclusive interview to IFT-Instituto de Fisica Teo^ rica-(part 1):

$$\mathrm{In}\:\mathrm{the}\:\mathrm{next}\:\mathrm{link},\:\mathrm{Barry}\:\mathrm{Barrish},\:\mathrm{who}\:\mathrm{won}\:\mathrm{the}\:\mathrm{physic}\:\mathrm{nobel}\:\mathrm{prize}\:\mathrm{in}\:\mathrm{2017}; \\ $$$$\mathrm{gave}\:\mathrm{an}\:\mathrm{exclusive}\:\mathrm{interview}\:\mathrm{to}\:\mathrm{IFT}-{Instituto}\:{de}\:{Fisica}\:{Te}\acute {{o}rica}-\left(\mathrm{part}\:\mathrm{1}\right): \\ $$

Question Number 53967    Answers: 2   Comments: 1

1)calculate A_t =∫_0 ^∞ e^(−xt) sinxdx with x>0 2) by using Fubuni theorem find the value of ∫_0 ^∞ ((sinx)/x)dx .

$$\left.\mathrm{1}\right){calculate}\:{A}_{{t}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} \:{sinxdx}\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{by}\:{using}\:{Fubuni}\:{theorem}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{x}}{dx}\:. \\ $$

Question Number 53966    Answers: 0   Comments: 1

let f(x) =xsinx ,2π periodic even developp f at Fourier serie .

$${let}\:{f}\left({x}\right)\:={xsinx}\:,\mathrm{2}\pi\:{periodic}\:{even} \\ $$$${developp}\:{f}\:{at}\:{Fourier}\:{serie}\:. \\ $$

Question Number 53965    Answers: 1   Comments: 0

solve (x+1)y^(′′) +(2−3x^2 )y^′ =xsin(x) let y^′ =z so (e) ⇔(x+1)z^′ +(2−3x^2 )z =xsinx(e) let first find z (he) →(x+1)z^′ +(2−3x^2 )z =0 ⇒(x+1)z^′ =(3x^2 −2)z ⇒(z^′ /z) =((3x^2 −2)/(x+1)) ⇒ ln∣z∣ = ∫ ((3x^2 −2)/(x+1)) dx +c =∫ ((3(x^2 −1)+1)/(x+1))+c =∫3(x−1)dx +∫ (dx/(x+1)) +c =3((x^2 /2)−x)+ln∣x+1∣ +c ⇒z =K∣x+1∣e^((3/2)x^2 −3x) solution on]−1,+∞[ ⇒ z(x) =K(x+1) e^((3/2)x^2 −3x) mvc method give z^′ =K^′ (x+1)e^((3/2)x^2 −3x) +K{ e^((3/2)x^2 −3x) +(x+1)(3x−3) e^((3/2)x^2 −3x) } ={ K^′ (x+1) + K +3K(x+1)(x−1) } e^((3/2)x^2 −3x) (e) ⇒{(x+1)^2 K^′ +K(x+1) +3K(x+1)^2 (x−1)} e^((3/2)x^2 −3x) (2−3x^2 )K(x+1)e^((3/2)x^2 −3x) =xsinx ⇒ (x+1)K^′ +K +3K(x^2 −1) +K(2−3x^2 ) =((xsinx)/(x+1)) ⇒ (x+1)K^′ +K −K =((xsinx)/(x+1)) ⇒K^′ =((xsinx)/((x+1)^2 )) ⇒ K(x) =∫ ((xsinx)/((x+1)^2 ))dx +c_0 ⇒z(x) =(x+1)e^((3/2)x^2 −3x) { ∫ ((xsinx)/((x+1)^2 ))dx +c_0 } =c_0 (x+1) e^((3/2)x^2 −3x) +(x+1)e^((3/2)x^2 −3x) ∫_. ^x ((tsint)/((t+1)^2 )) dt we have y^′ (x)=z(x) ⇒y(x) =∫ z(x)dx +λ ⇒ y(x) =∫ c_0 (x+1)e^((3/2)x^2 −3x) dx +∫ {(x+1)e^((3/2)x^2 −3x) ∫_. ^x ((tsint)/((t+1)^2 ))dt}dx +λ .

$${solve}\:\:\left({x}+\mathrm{1}\right){y}^{''} \:+\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){y}^{'} \:={xsin}\left({x}\right) \\ $$$${let}\:{y}^{'} ={z}\:\:\:{so}\:\left({e}\right)\:\Leftrightarrow\left({x}+\mathrm{1}\right){z}^{'} \:+\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){z}\:={xsinx}\left({e}\right)\:{let}\:{first}\:{find}\:{z} \\ $$$$\left({he}\right)\:\rightarrow\left({x}+\mathrm{1}\right){z}^{'} \:+\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){z}\:=\mathrm{0}\:\Rightarrow\left({x}+\mathrm{1}\right){z}^{'} \:=\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\right){z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$${ln}\mid{z}\mid\:=\:\int\:\:\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}}{{x}+\mathrm{1}}\:{dx}\:+{c}\:\:=\int\:\frac{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}}{{x}+\mathrm{1}}+{c}\:=\int\mathrm{3}\left({x}−\mathrm{1}\right){dx}\:+\int\:\frac{{dx}}{{x}+\mathrm{1}}\:+{c} \\ $$$$\left.=\mathrm{3}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right)+{ln}\mid{x}+\mathrm{1}\mid\:+{c}\:\Rightarrow{z}\:={K}\mid{x}+\mathrm{1}\mid{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:{solution}\:{on}\right]−\mathrm{1},+\infty\left[\:\Rightarrow\right. \\ $$$${z}\left({x}\right)\:={K}\left({x}+\mathrm{1}\right)\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:\:{mvc}\:{method}\:{give} \\ $$$${z}^{'} ={K}^{'} \left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:+{K}\left\{\:\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:+\left({x}+\mathrm{1}\right)\left(\mathrm{3}{x}−\mathrm{3}\right)\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \right\} \\ $$$$=\left\{\:{K}^{'} \left({x}+\mathrm{1}\right)\:+\:{K}\:+\mathrm{3}{K}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\:\right\}\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \\ $$$$\left({e}\right)\:\Rightarrow\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} {K}^{'} \:+{K}\left({x}+\mathrm{1}\right)\:+\mathrm{3}{K}\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{1}\right)\right\}\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \\ $$$$\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){K}\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:={xsinx}\:\Rightarrow \\ $$$$\left({x}+\mathrm{1}\right){K}^{'} \:+{K}\:+\mathrm{3}{K}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:+{K}\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right)\:=\frac{{xsinx}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$$\left({x}+\mathrm{1}\right){K}^{'} \:+{K}\:−{K}\:=\frac{{xsinx}}{{x}+\mathrm{1}}\:\Rightarrow{K}^{'} \:=\frac{{xsinx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\int\:\:\:\frac{{xsinx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:+{c}_{\mathrm{0}} \:\Rightarrow{z}\left({x}\right)\:=\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \left\{\:\int\:\:\frac{{xsinx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:+{c}_{\mathrm{0}} \right\} \\ $$$$={c}_{\mathrm{0}} \left({x}+\mathrm{1}\right)\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:+\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\int_{.} ^{{x}} \:\:\:\frac{{tsint}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:{dt} \\ $$$${we}\:{have}\:{y}^{'} \left({x}\right)={z}\left({x}\right)\:\Rightarrow{y}\left({x}\right)\:=\int\:{z}\left({x}\right){dx}\:+\lambda\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\int\:{c}_{\mathrm{0}} \left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} {dx}\:+\int\:\:\left\{\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\int_{.} ^{{x}} \:\:\frac{{tsint}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}\right\}{dx}\:+\lambda\:. \\ $$

Question Number 53956    Answers: 0   Comments: 1

give∫_0 ^1 e^(−x) ln(1−x)dx at form of serie

$${give}\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} {ln}\left(\mathrm{1}−{x}\right){dx}\:\:{at}\:{form}\:{of}\:{serie} \\ $$

Question Number 53957    Answers: 0   Comments: 1

let f(x) =arctan(1+2x) 1) calculate f^((n)) (x) then f^((n)) (0) 2) developp f at integr serie . we have f^′ (x)=(2/(1+(1+2x)^2 )) ⇒ f^((n)) (x) =2 {(1/((2x+1)^2 +1))}^((n−1)) with n>0 let W(x)=(1/((2x+1)^2 +1)) ⇒W(x) =(1/((2x+1+i)(2x+1−i))) =(1/(4(x+((1+i)/2))(x+((1−i)/2)))) =(1/(4(x +(1/(√2)) e^((iπ)/4) )( x +(1/(√2)) e^(−((iπ)/4)) ))) but ((1/(x +(1/(√2))e^(−((iπ)/4)) )) −(1/(x+(1/(√2))e^((iπ)/4) ))) =(1/(√2)) (((2isin((π/4))))/((x+(1/(√2))e^(−((iπ)/4)) )(x+(1/(√2))e^((iπ)/4) ))) ⇒ W(x) =(1/(4i)){ (1/(x+(1/(√2))e^(−((iπ)/4)) )) −(1/(x+(1/(√2))e^((iπ)/4) ))} ⇒ W^((n−1)) (x)=(1/(4i)){ (((−1)^(n−1) (n−1)!)/((x+(1/(√2))e^(−((iπ)/4)) )^n )) −(((−1)^(n−1) (n−1)!)/((x+(1/(√2))e^((iπ)/4) )^n ))} ⇒ f^((n)) (x)=(((−1)^(n−1) (n−1)!)/(2i)){ (1/((x+(1/(√2))e^(−((iπ)/4)) )^n )) −(1/((x+(1/(√2))e^((iπ)/4) )^n ))} and f^((n)) (0) =(((−1)^(n−1) (n−1)!)/(2i)){ (1/((√2))^(−n) e^(−i((nπ)/4)) )) −(1/(((√2))^(−n) e^(i((nπ)/4)) ))} =(((−1)^(n−1) (n−1)!)/(2i)){ ((√2))^n e^((inπ)/4) − ((√2))^n e^(−((inπ)/4)) } =(((−1)^(n−1) (n−1)!)/(2i)) ((√2))^n 2i sin(((nπ)/4)) =(−1)^(n−1) (n−1)! ((√2))^n sin(((nπ)/4))

$${let}\:{f}\left({x}\right)\:={arctan}\left(\mathrm{1}+\mathrm{2}{x}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{then}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)=\frac{\mathrm{2}}{\mathrm{1}+\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }\:\Rightarrow\:{f}^{\left({n}\right)} \left({x}\right)\:=\mathrm{2}\:\left\{\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\right\}^{\left({n}−\mathrm{1}\right)} \:\:{with}\:{n}>\mathrm{0} \\ $$$${let}\:{W}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow{W}\left({x}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}+{i}\right)\left(\mathrm{2}{x}+\mathrm{1}−{i}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\left({x}+\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}\left({x}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\:{x}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:{but} \\ $$$$\left(\frac{\mathrm{1}}{{x}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:−\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\right)\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\frac{\left(\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right)}{\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}\:\Rightarrow \\ $$$${W}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:\:\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:−\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\right\}\:\Rightarrow \\ $$$${W}^{\left({n}−\mathrm{1}\right)} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\:\:\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\:−\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\right\}\:{and} \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\:\frac{\mathrm{1}}{\left.\sqrt{\mathrm{2}}\right)^{−{n}} \:{e}^{−{i}\frac{{n}\pi}{\mathrm{4}}} }\:−\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}\right)^{−{n}} \:{e}^{{i}\frac{{n}\pi}{\mathrm{4}}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{e}^{\frac{{in}\pi}{\mathrm{4}}} \:−\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{e}^{−\frac{{in}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:\mathrm{2}{i}\:{sin}\left(\frac{{n}\pi}{\mathrm{4}}\right)\:=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\:\left(\sqrt{\mathrm{2}}\right)^{{n}} {sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$

Question Number 53950    Answers: 0   Comments: 1

calculate ∫_(1/3) ^(1/2) Γ(x)Γ(1−x)dx with Γ(x) =∫_0 ^∞ t^(x−1) e^(−t) dt with x>0 .

$$\:{calculate}\:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right){dx}\:\:\:{with}\:\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} {dt}\:\:\:\:{with}\:{x}>\mathrm{0}\:. \\ $$

Question Number 53958    Answers: 0   Comments: 0

find the value of ∫_0 ^1 ln(x)ln(1−x^2 )dx

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$

Question Number 53963    Answers: 0   Comments: 1

let f(x) = x∣x∣ , 2π periodic odd developp f at fourier serie .

$${let}\:{f}\left({x}\right)\:=\:{x}\mid{x}\mid\:\:\:,\:\mathrm{2}\pi\:{periodic}\:\:{odd}\: \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie}\:. \\ $$

Question Number 53962    Answers: 1   Comments: 3

Question Number 53961    Answers: 0   Comments: 2

let ϕ(x) =((arctan(2x))/(1−x^2 )) 1) calculate ϕ^((n)) (x) 2) calculate ϕ^((n)) (0) anddevelpp ϕ at integr serie

$${let}\:\varphi\left({x}\right)\:=\frac{{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\varphi^{\left({n}\right)} \left({x}\right)\: \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\varphi^{\left({n}\right)} \left(\mathrm{0}\right)\:{anddevelpp}\:\varphi\:{at}\:{integr}\:{serie} \\ $$

Question Number 53960    Answers: 0   Comments: 0

let f(x) =arctan(x^2 ) developp f at i serie. the Q . is developp f at integr serie.

$${let}\:{f}\left({x}\right)\:={arctan}\left({x}^{\mathrm{2}} \right)\:{developp}\:{f}\:{at}\:{i}\:{serie}. \\ $$$${the}\:{Q}\:.\:{is}\:{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$

Question Number 53947    Answers: 0   Comments: 0

let U_n =Σ_(k=1) ^n (1/(k[(√k)])) find a equivalent of U_n when n→+∞

$${let}\:{U}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}\left[\sqrt{{k}}\right]} \\ $$$${find}\:{a}\:{equivalent}\:{of}\:{U}_{{n}} \:{when}\:{n}\rightarrow+\infty \\ $$

Question Number 53935    Answers: 1   Comments: 0

(√8^(x−2) )×(4^(2x−3) )^(1/(x+1)) =(2^(5x+5) )^(1/6)

$$\sqrt{\mathrm{8}^{{x}−\mathrm{2}} }×\sqrt[{{x}+\mathrm{1}}]{\mathrm{4}^{\mathrm{2}{x}−\mathrm{3}} }=\sqrt[{\mathrm{6}}]{\mathrm{2}^{\mathrm{5}{x}+\mathrm{5}} } \\ $$$$ \\ $$

Question Number 53931    Answers: 0   Comments: 1

∫x!dx

$$\int{x}!{dx} \\ $$

Question Number 53924    Answers: 2   Comments: 0

If the non−zero numbers x, y, z are in AP, and tan^(−1) x, tan^(−1) y, tan^(−1) z are also in AP, then

$$\mathrm{If}\:\mathrm{the}\:\mathrm{non}−\mathrm{zero}\:\mathrm{numbers}\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP}, \\ $$$$\mathrm{and}\:\mathrm{tan}^{−\mathrm{1}} {x},\:\mathrm{tan}^{−\mathrm{1}} {y},\:\mathrm{tan}^{−\mathrm{1}} {z}\:\mathrm{are}\:\mathrm{also}\:\mathrm{in}\:\mathrm{AP}, \\ $$$$\mathrm{then} \\ $$

Question Number 53919    Answers: 1   Comments: 0

Question Number 53912    Answers: 3   Comments: 0

{ ((2^x −2^y =1)),((4^x −4^y =(5/3))) :}

$$\begin{cases}{\mathrm{2}^{{x}} −\mathrm{2}^{{y}} =\mathrm{1}}\\{\mathrm{4}^{{x}} −\mathrm{4}^{{y}} =\frac{\mathrm{5}}{\mathrm{3}}}\end{cases} \\ $$

Question Number 53910    Answers: 1   Comments: 12

Question Number 53909    Answers: 1   Comments: 0

a angle is 14° more than its complement then its measure

$${a}\:{angle}\:{is}\:\mathrm{14}°\:{more}\:{than}\:{its}\:{complement}\:\:{then}\:{its}\:{measure} \\ $$

Question Number 53907    Answers: 1   Comments: 0

Question Number 53902    Answers: 1   Comments: 3

How many zeros and how many ones are there in the numbers from 1 to 9999? Example: in the number 1010 there are 2 zeros and 2 ones.

$${How}\:{many}\:{zeros}\:{and}\:{how}\:{many}\:{ones} \\ $$$${are}\:{there}\:{in}\:{the}\:{numbers}\:{from}\:\mathrm{1}\:{to} \\ $$$$\mathrm{9999}? \\ $$$${Example}:\:{in}\:{the}\:{number}\:\mathrm{1010}\:{there} \\ $$$${are}\:\mathrm{2}\:{zeros}\:{and}\:\mathrm{2}\:{ones}. \\ $$

Question Number 53901    Answers: 1   Comments: 0

Question Number 53890    Answers: 1   Comments: 0

pls. help me solve this if y=((x−2)/(2x−1)) show that (2x−1)(d^2 y/dx^2 )+4(dy/dx)=0

$$\mathrm{pls}.\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{this} \\ $$$$\mathrm{if}\:\mathrm{y}=\frac{\mathrm{x}−\mathrm{2}}{\mathrm{2x}−\mathrm{1}}\:\mathrm{show}\:\mathrm{that}\:\left(\mathrm{2x}−\mathrm{1}\right)\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }+\mathrm{4}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{0} \\ $$

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