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Question Number 142236    Answers: 2   Comments: 0

f is an endomorphism of V such that f○f=−Id_V . 1. Show that f is an isomorphism of V and express f^(−1) in function of f. 2. show that 0^→ is the one invariant vector by f. 3. Given u^→ ≠0^→ and u^→ ∈ V. a. Show that (u^→ ; f(u^→ )) is a base of V. b. Write the matrix of f in base (u^→ ; f(u^→ )).

$$\mathrm{f}\:\mathrm{is}\:\mathrm{an}\:\mathrm{endomorphism}\:\mathrm{of}\:\mathrm{V}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{f}\circ\mathrm{f}=−\mathrm{Id}_{\mathrm{V}} \:. \\ $$$$\mathrm{1}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{f}\:\mathrm{is}\:\mathrm{an}\:\mathrm{isomorphism}\:\mathrm{of} \\ $$$$\mathrm{V}\:\mathrm{and}\:\mathrm{express}\:\mathrm{f}^{−\mathrm{1}} \:\mathrm{in}\:\mathrm{function}\:\mathrm{of}\:\mathrm{f}. \\ $$$$\mathrm{2}.\:\mathrm{show}\:\mathrm{that}\:\overset{\rightarrow} {\mathrm{0}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{one}\:\mathrm{invariant} \\ $$$$\mathrm{vector}\:\mathrm{by}\:\mathrm{f}. \\ $$$$\mathrm{3}.\:\mathrm{Given}\:\overset{\rightarrow} {\mathrm{u}}\neq\overset{\rightarrow} {\mathrm{0}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{u}}\:\in\:\mathrm{V}. \\ $$$$\mathrm{a}.\:\mathrm{Show}\:\mathrm{that}\:\left(\overset{\rightarrow} {\mathrm{u}};\:\mathrm{f}\left(\overset{\rightarrow} {\mathrm{u}}\right)\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{base}\:\mathrm{of}\:\mathrm{V}. \\ $$$$\mathrm{b}.\:\mathrm{Write}\:\mathrm{the}\:\mathrm{matrix}\:\mathrm{of}\:\mathrm{f}\:\mathrm{in}\:\mathrm{base} \\ $$$$\left(\overset{\rightarrow} {\mathrm{u}};\:\mathrm{f}\left(\overset{\rightarrow} {\mathrm{u}}\right)\right). \\ $$

Question Number 141694    Answers: 0   Comments: 0

log((((√5)+1)/(10))9e^γ )=((ζ(2))/2)(((1^2 +9^2 )/(10^2 )))−((ζ(3))/3) (((1^3 +9^3 )/(10^3 )) )+((ζ(4))/4)(((1^4 +9^4 )/(10^4 )))−... γ=Euler Mascheroni Constant

$${log}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{10}}\mathrm{9}{e}^{\gamma} \right)=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}\left(\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }\right)−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{3}}\:\left(\frac{\mathrm{1}^{\mathrm{3}} +\mathrm{9}^{\mathrm{3}} }{\mathrm{10}^{\mathrm{3}} }\:\right)+\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{4}}\left(\frac{\mathrm{1}^{\mathrm{4}} +\mathrm{9}^{\mathrm{4}} }{\mathrm{10}^{\mathrm{4}} }\right)−... \\ $$$$\gamma={Euler}\:{Mascheroni}\:{Constant} \\ $$

Question Number 141693    Answers: 0   Comments: 0

n∈N^+ a_5 =a_(13) =0 b_(n+1) −b_n =2 b_n =a_(n+1) −a_n ⇒ a_1 =?

$${n}\in{N}^{+} \\ $$$${a}_{\mathrm{5}} ={a}_{\mathrm{13}} =\mathrm{0} \\ $$$${b}_{{n}+\mathrm{1}} −{b}_{{n}} =\mathrm{2} \\ $$$${b}_{{n}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} \:\:\:\Rightarrow\:\:{a}_{\mathrm{1}} =? \\ $$$$ \\ $$

Question Number 141692    Answers: 1   Comments: 0

Daniel and Bruno are playing with perfect cube Daniel is the first player if he obtains 1 or 2 he wins the game and the party stopping or else Bruno plays and if he have {3.4.6} Bruno won and the game stopping Determine the probability that Daniel winand the probability that Bruno win

$${Daniel}\:{and}\:{Bruno}\:{are}\:{playing}\:{with}\:{perfect}\:{cube} \\ $$$${Daniel}\:{is}\:{the}\:{first}\:{player}\:{if}\:{he}\:{obtains}\:\mathrm{1}\:{or}\:\mathrm{2} \\ $$$${he}\:{wins}\:{the}\:{game}\:{and}\:{the}\:{party}\:{stopping} \\ $$$${or}\:{else}\:{Bruno}\:{plays}\:{and}\:{if}\:{he}\:{have}\:\left\{\mathrm{3}.\mathrm{4}.\mathrm{6}\right\}\:{Bruno}\:{won}\:{and}\:{the}\:{game}\:{stopping} \\ $$$${Determine}\:{the}\:{probability}\:{that}\:{Daniel}\:{winand}\:{the}\:{probability}\:{that}\:{Bruno}\:{win} \\ $$$$ \\ $$

Question Number 141691    Answers: 1   Comments: 0

....Calculus(I).... 𝛗:=∫_(1/(2 )) ^( 1) (1/(x^2 (1+x^4 )^(3/4) ))dx=???

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{Calculus}\left({I}\right).... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\frac{\mathrm{1}}{\mathrm{2}\:}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} }{dx}=??? \\ $$

Question Number 141685    Answers: 1   Comments: 0

......nice ... ... ... calculus..... If lim_(x→0) ((tan(x))/x) = 1 , prove that: lim(1/x)((1/x)−(1/(tan(x))))=(1/3)

$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:......{nice}\:...\:...\:...\:{calculus}..... \\ $$$$\:\:\mathrm{I}{f}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{tan}\left({x}\right)}{{x}}\:=\:\mathrm{1}\:,\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:{lim}\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{tan}\left({x}\right)}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Question Number 141681    Answers: 1   Comments: 0

On the Argand Diagram, the variable point Z represents a complex number z. Find the equation of the locus of a point Z which moves such that ∣((z−1)/(z+2))∣=2

$$\mathrm{On}\:\mathrm{the}\:\mathrm{Argand}\:\mathrm{Diagram},\:\mathrm{the}\:\mathrm{variable}\:\mathrm{point} \\ $$$$\mathrm{Z}\:\mathrm{represents}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\:{z}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{a}\:\mathrm{point} \\ $$$$\mathrm{Z}\:\mathrm{which}\:\mathrm{moves}\:\mathrm{such}\:\mathrm{that}\:\mid\frac{{z}−\mathrm{1}}{{z}+\mathrm{2}}\mid=\mathrm{2} \\ $$

Question Number 143167    Answers: 2   Comments: 0

∫arctan((√((√x)+1)))dx=??? propose′ par Rodrigue

$$\int\boldsymbol{{arctan}}\left(\sqrt{\sqrt{\boldsymbol{{x}}}+\mathrm{1}}\right)\boldsymbol{{dx}}=??? \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{propose}}'\:\boldsymbol{{par}}\:\boldsymbol{{Rodrigue}} \\ $$

Question Number 141672    Answers: 1   Comments: 0

Solve the equation x^4 −2x^3 −5x^2 +10x−3=0

$$\:\:{Solve}\:{the}\:{equation}\: \\ $$$$\:\:{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{3}=\mathrm{0} \\ $$

Question Number 141668    Answers: 1   Comments: 0

.......Advanced ...★ ...★ ... Calculus....... if Ω =Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) then prove that :: (1/2) = e^(Ω−1) proof :: method (1): ψ (1+x )= −γ+Σ_(n=2) ^∞ (−1)^n ζ(n)x^(n−1) ( Maclaurin series for ψ(x+1) ) x:=(1/2) ⇒ ψ ((3/2) )=−γ + 2Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) (∗ ) we know that :: ψ(1+x)=(1/x)+ψ(x) ( ∗ ) ⇛ ψ ((3/2))=2+ψ((1/2))=−γ+2Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) (∗) ⇛ 2−γ−ln(4)=−γ+2Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) ln((e/2))= Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) =Ω (1/2) = e^(Ω −1) ....✓ ...m.n.july.1970...

$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:.......{Advanced}\:...\bigstar\:...\bigstar\:...\:{Calculus}....... \\ $$$$\:\:\:\:\:\:\:\:\:{if}\:\:\:\:\Omega\:=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} }\:{then}\:{prove} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{that}\:::\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:{e}^{\Omega−\mathrm{1}} \:\: \\ $$$$\:\:\:\:{proof}\::: \\ $$$$\:\:\:\:{method}\:\left(\mathrm{1}\right): \\ $$$$\:\:\:\:\:\psi\:\left(\mathrm{1}+{x}\:\right)=\:−\gamma+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right){x}^{{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\left(\:{Maclaurin}\:{series}\:{for}\:\psi\left({x}+\mathrm{1}\right)\:\right) \\ $$$$\:\:\:\:{x}:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\psi\:\left(\frac{\mathrm{3}}{\mathrm{2}}\:\right)=−\gamma\:+\:\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} }\:\:\left(\ast\:\right) \\ $$$$\:\:\:\:{we}\:{know}\:{that}\:::\:\psi\left(\mathrm{1}+{x}\right)=\frac{\mathrm{1}}{{x}}+\psi\left({x}\right) \\ $$$$\:\:\:\:\:\:\left(\:\ast\:\right)\:\:\Rrightarrow\:\psi\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{2}+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\gamma+\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} } \\ $$$$\:\:\:\:\:\:\:\left(\ast\right)\:\:\:\:\:\Rrightarrow\:\:\:\:\:\:\:\:\:\mathrm{2}−\gamma−{ln}\left(\mathrm{4}\right)=−\gamma+\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{{e}}{\mathrm{2}}\right)=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} }\:=\Omega \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:{e}^{\Omega\:−\mathrm{1}} \:\:\:....\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.{july}.\mathrm{1970}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 141665    Answers: 0   Comments: 1

Question Number 141663    Answers: 0   Comments: 1

_( −−−−−−−−−−−−−−−−−−−) x^2 (x−12)(x−15)=k(x−16) find x in terms of k (>0). ^(−−−−−−−−−−−−−−−−−−−)

$$\:\underset{\:\:−−−−−−−−−−−−−−−−−−−} {\:} \\ $$$$\:\:{x}^{\mathrm{2}} \left({x}−\mathrm{12}\right)\left({x}−\mathrm{15}\right)={k}\left({x}−\mathrm{16}\right) \\ $$$$\:\:\:\:{find}\:{x}\:{in}\:{terms}\:{of}\:{k}\:\left(>\mathrm{0}\right). \\ $$$$\:\:\overset{−−−−−−−−−−−−−−−−−−−} {\:} \\ $$

Question Number 141658    Answers: 1   Comments: 0

Question Number 141654    Answers: 0   Comments: 2

Solve for real numbers 5∙(((1−z))^(1/5) + ((1+z))^(1/5) = 2+4∙(((1−z))^(1/4) + ((1+z))^(1/4) )

$${Solve}\:{for}\:{real}\:{numbers} \\ $$$$\mathrm{5}\centerdot\left(\sqrt[{\mathrm{5}}]{\mathrm{1}−{z}}\:+\:\sqrt[{\mathrm{5}}]{\mathrm{1}+{z}}\:=\:\mathrm{2}+\mathrm{4}\centerdot\left(\sqrt[{\mathrm{4}}]{\mathrm{1}−{z}}\:+\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{z}}\right)\right. \\ $$

Question Number 141653    Answers: 0   Comments: 0

what is condition to have log( I +A)=Σ a_n A^n and determine the sequence (a_n ) A ∈ M_n (C)

$${what}\:{is}\:{condition}\:{to}\:{have} \\ $$$${log}\left(\:{I}\:+{A}\right)=\Sigma\:{a}_{{n}} {A}^{{n}} \\ $$$${and}\:{determine}\:{the}\:{sequence}\:\left({a}_{{n}} \right) \\ $$$${A}\:\in\:{M}_{{n}} \left({C}\right) \\ $$

Question Number 141652    Answers: 0   Comments: 0

A = (((1 2)),((−1 1)) ) find e^(A ) and e^(tA) find ch(A) and sh(A)

$${A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${find}\:{e}^{{A}\:} \:{and}\:{e}^{{tA}} \\ $$$${find}\:{ch}\left({A}\right)\:{and}\:{sh}\left({A}\right) \\ $$

Question Number 141651    Answers: 0   Comments: 2

Question Number 141649    Answers: 1   Comments: 0

.......advanced calculus...... prove that−:: φ:=∫_0 ^( ∞) ((cos(2πx^2 ))/(cosh^2 (πx)))dx=(1/4) ✓

$$\:\:\:\:\:\:\:\:\:.......{advanced}\:\:{calculus}...... \\ $$$$\:\:\:\:{prove}\:\:{that}−:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\phi:=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{cos}\left(\mathrm{2}\pi{x}^{\mathrm{2}} \right)}{{cosh}^{\mathrm{2}} \left(\pi{x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\checkmark \\ $$

Question Number 141937    Answers: 0   Comments: 2

Write and graph the equation of the graph of y=sin(πx) It is stretched up by a factor of 5 and shifted (1/2) unit to the right Help me please

$${Write}\:{and}\:{graph}\:{the}\:{equation}\:{of}\:{the}\:{graph}\:{of}\:{y}={sin}\left(\pi{x}\right) \\ $$$${It}\:{is}\:{stretched}\:{up}\:{by}\:{a}\:{factor}\:{of}\:\mathrm{5}\:{and}\:{shifted}\:\frac{\mathrm{1}}{\mathrm{2}}\:{unit}\:{to}\:{the}\:{right} \\ $$$${Help}\:{me}\:{please} \\ $$$$ \\ $$

Question Number 141645    Answers: 0   Comments: 0

Question Number 141643    Answers: 1   Comments: 0

Question Number 141642    Answers: 0   Comments: 0

Question Number 141640    Answers: 1   Comments: 0

Question Number 141635    Answers: 0   Comments: 0

solve the differential equation (PDE), z((∂z/∂x)−(∂z/∂y))=z^2 +(x+y)^2 .

$${solve}\:{the}\:{differential}\:{equation}\:\left({PDE}\right), \\ $$$${z}\left(\frac{\partial{z}}{\partial{x}}−\frac{\partial{z}}{\partial{y}}\right)={z}^{\mathrm{2}} +\left({x}+{y}\right)^{\mathrm{2}} . \\ $$

Question Number 141633    Answers: 0   Comments: 0

1<a<b ,prove that : b^n = Σ_(k=0) ^n (−1)^k C_n ^k a^((ln(Σ_(p=0) ^(n−k) C_(n−k) ^p a^(n−p) b^p ))/(ln(a)))

$$\mathrm{1}<\mathrm{a}<\mathrm{b}\:,\mathrm{prove}\:\mathrm{that}\:: \\ $$$${b}^{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \mathrm{C}_{{n}} ^{{k}} \:{a}^{\frac{{ln}\left(\underset{{p}=\mathrm{0}} {\overset{{n}−{k}} {\sum}}\mathrm{C}_{{n}−{k}} ^{{p}} {a}^{{n}−{p}} {b}^{{p}} \right)}{{ln}\left({a}\right)}} \\ $$

Question Number 141632    Answers: 0   Comments: 0

Let f(x)=((sin(x))/x) , prove that : Σ_(n=0) ^∞ [ f(nπ+α)+f(nπ−α) ]= 1+f(α)

$$\mathrm{Let}\:{f}\left({x}\right)=\frac{{sin}\left({x}\right)}{{x}}\:,\:\mathrm{prove}\:\mathrm{that}\:: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\:{f}\left({n}\pi+\alpha\right)+{f}\left({n}\pi−\alpha\right)\:\right]=\:\mathrm{1}+{f}\left(\alpha\right) \\ $$

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