Question and Answers Forum

All Questions Topic List

AllQuestion and Answers: Page 1564

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

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

Question Number 53881 Answers: 0 Comments: 2

Question Number 53894 Answers: 2 Comments: 0

∫_(π/6) ^(5π/6) (√(4−4 sin^2 t)) dt =

$$\underset{\pi/\mathrm{6}} {\overset{\mathrm{5}\pi/\mathrm{6}} {\int}}\sqrt{\mathrm{4}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} {t}}\:{dt}\:= \\ $$

Question Number 53867 Answers: 1 Comments: 0

Question Number 53852 Answers: 0 Comments: 0

G_(μν) = R_(μν) − (1/2) Rg_(μν) + 𝚲g_(μν) Wich theory of modern physic belongs this equation? and what does it mean?

$${G}_{\mu\nu} =\:{R}_{\mu\nu} −\:\frac{\mathrm{1}}{\mathrm{2}}\:{Rg}_{\mu\nu} \:+\:\boldsymbol{\Lambda}{g}_{\mu\nu} \\ $$$$\mathrm{Wich}\:\mathrm{theory}\:\mathrm{of}\:\mathrm{modern}\:\mathrm{physic}\:\mathrm{belongs}\:\mathrm{this}\:\mathrm{equation}? \\ $$$$\mathrm{and}\:\mathrm{what}\:\mathrm{does}\:\mathrm{it}\:\mathrm{mean}? \\ $$

Question Number 53843 Answers: 1 Comments: 0

Question Number 53842 Answers: 0 Comments: 0

Find all the real valued f satisfying f[2x + f(2y)] + f[f(y)] = 4x + 8y for all real numbers x and y.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{real}\:\mathrm{valued}\:\:\mathrm{f}\:\:\mathrm{satisfying}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\left[\mathrm{2x}\:+\:\mathrm{f}\left(\mathrm{2y}\right)\right]\:+\:\mathrm{f}\left[\mathrm{f}\left(\mathrm{y}\right)\right]\:\:=\:\:\mathrm{4x}\:+\:\mathrm{8y} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{real}\:\mathrm{numbers}\:\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}.\: \\ $$

Pg 1559 Pg 1560 Pg 1561 Pg 1562 Pg 1563 Pg 1564 Pg 1565 Pg 1566 Pg 1567 Pg 1568

Terms of Service

Contact: info@tinkutara.com