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Question Number 113637    Answers: 0   Comments: 0

Montrer que pour 0<z<1 on a Γ(z)Γ(1−z)=(π/(sin(πz)))

$${Montrer}\:{que}\:{pour}\:\mathrm{0}<{z}<\mathrm{1}\:{on}\:{a} \\ $$$$\Gamma\left({z}\right)\Gamma\left(\mathrm{1}−{z}\right)=\frac{\pi}{{sin}\left(\pi{z}\right)} \\ $$

Question Number 113499    Answers: 0   Comments: 0

Question Number 113490    Answers: 0   Comments: 1

Question Number 113456    Answers: 0   Comments: 0

Question Number 113455    Answers: 1   Comments: 1

Question Number 113354    Answers: 0   Comments: 3

( ((n),(0) )/2)−( ((n),(1) )/3)+( ((n),(2) )/4)−( ((n),(3) )/5)+.....n

$$\frac{\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}}{\mathrm{2}}−\frac{\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}}{\mathrm{3}}+\frac{\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}}{\mathrm{4}}−\frac{\begin{pmatrix}{{n}}\\{\mathrm{3}}\end{pmatrix}}{\mathrm{5}}+.....{n} \\ $$

Question Number 113336    Answers: 0   Comments: 0

If 1, a^2 ,a^3 ,...,a^(n−1) are the roots nth of unity , prove that : (1+a)(1+a^2 )(1+a^3 )...(1+a^(n−1) ) = n−2⌊(n/2)⌋

$${If}\:\mathrm{1},\:{a}^{\mathrm{2}} ,{a}^{\mathrm{3}} \:,...,{a}^{{n}−\mathrm{1}} \:{are}\:{the}\:{roots}\: \\ $$$${nth}\:{of}\:{unity}\:,\: \\ $$$${prove}\:{that}\::\:\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{a}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{3}} \right)...\left(\mathrm{1}+{a}^{{n}−\mathrm{1}} \right) \\ $$$$=\:{n}−\mathrm{2}\lfloor\frac{{n}}{\mathrm{2}}\rfloor \\ $$$$ \\ $$

Question Number 113190    Answers: 1   Comments: 3

∫_0 ^1 ((logx)/(x−1))dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{logx}}{{x}−\mathrm{1}}{dx} \\ $$

Question Number 113185    Answers: 1   Comments: 0

Question Number 112838    Answers: 1   Comments: 0

What is the sum of all the solutions of the equation ∣2x+8∣^2 −∣9x+36∣−9=0

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{solutions} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mid\mathrm{2x}+\mathrm{8}\mid^{\mathrm{2}} −\mid\mathrm{9x}+\mathrm{36}\mid−\mathrm{9}=\mathrm{0} \\ $$

Question Number 112809    Answers: 1   Comments: 0

Question Number 112707    Answers: 0   Comments: 0

Question Number 112454    Answers: 1   Comments: 1

(1) find the locus ∣z−z_1 ∣ = 2 meets the positive real axis (2)On a single Argand diagram, sketch the loci → { ((∣z−z_1 ∣=2)),((arg(z−z_2 )=(π/4))) :}

$$\left(\mathrm{1}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{locus}\:\mid\mathrm{z}−\mathrm{z}_{\mathrm{1}} \mid\:=\:\mathrm{2}\:\mathrm{meets} \\ $$$$\mathrm{the}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{axis} \\ $$$$\left(\mathrm{2}\right)\mathrm{On}\:\mathrm{a}\:\mathrm{single}\:\mathrm{Argand}\:\mathrm{diagram},\:\mathrm{sketch} \\ $$$$\mathrm{the}\:\mathrm{loci}\:\rightarrow\begin{cases}{\mid\mathrm{z}−\mathrm{z}_{\mathrm{1}} \mid=\mathrm{2}}\\{\mathrm{arg}\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)=\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$

Question Number 112430    Answers: 0   Comments: 0

Π_(n=1) ^∞ (1/(1−x^n ))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\mathrm{1}}{\mathrm{1}−{x}^{{n}} } \\ $$

Question Number 112245    Answers: 0   Comments: 0

(1/(1+x^2 ))+(1/(1+x^4 ))+(1/(1+x^8 ))+..... ( x>1)

$$\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{8}} }+.....\:\:\:\:\:\:\:\:\:\:\left(\:{x}>\mathrm{1}\right) \\ $$

Question Number 112162    Answers: 1   Comments: 3

Question Number 111975    Answers: 1   Comments: 0

Question Number 111957    Answers: 0   Comments: 31

“MATHEMATICS” CONTAINS ALL THE LETTERS OF “ETHICS”. IS THERE ANY LESSON FOR US IN ABOVE SAYING? FOR “math-lovers”? FOR “math-giants”? FOR “overflow-mathematicians”? ........ ...... _( BTW this saying belongs to me)

$$\:\:\:\:\:\:\:\:\:\:``\mathrm{MATHEMATICS}'' \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{CONTAINS} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{ALL}\:\mathrm{THE}\:\mathrm{LETTERS}\:\mathrm{OF} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:``\mathrm{ETHICS}''. \\ $$$$ \\ $$$$\mathrm{IS}\:\:\mathrm{THERE}\:\mathrm{ANY}\:\mathrm{LESSON}\:\mathrm{FOR}\:\mathrm{US} \\ $$$$\mathrm{IN}\:\mathrm{ABOVE}\:\mathrm{SAYING}? \\ $$$${FOR}\:``{math}-{lovers}''? \\ $$$${FOR}\:``{math}-{giants}''? \\ $$$${FOR}\:``{overflow}-{mathematicians}''? \\ $$$$........ \\ $$$$......\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:_{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{BTW}\:{this}\:{saying}\:{belongs}\:{to}\:{me}} \\ $$

Question Number 111774    Answers: 1   Comments: 0

Question Number 111668    Answers: 0   Comments: 6

Question Number 111643    Answers: 1   Comments: 0

((7/3))!(with out calculator)

$$\left(\frac{\mathrm{7}}{\mathrm{3}}\right)!\left({with}\:{out}\:{calculator}\right) \\ $$

Question Number 111536    Answers: 2   Comments: 2

Let 2,3,5,6,7,10,11,... be increasing sequence of positive integers that are neither the square nor cube of an integer. Find the 2016th term of this sequence.

$$\mathrm{Let}\:\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{10},\mathrm{11},...\:\mathrm{be}\:\mathrm{increasing} \\ $$$$\mathrm{sequence}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{that}\:\mathrm{are} \\ $$$$\mathrm{neither}\:\mathrm{the}\:\mathrm{square}\:\mathrm{nor}\:\mathrm{cube}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{integer}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{2016th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{this} \\ $$$$\mathrm{sequence}. \\ $$

Question Number 111466    Answers: 1   Comments: 2

Σ_(n=1) ^∞ (n^n /(n!))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{{n}} }{{n}!} \\ $$

Question Number 111450    Answers: 0   Comments: 1

Question Number 111428    Answers: 0   Comments: 1

Σ_(n=1) ^∞ (n^3 /(n!))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{{n}!} \\ $$

Question Number 111132    Answers: 1   Comments: 0

prove by mathematical induction ⇒ 7^n −(3n+4)×4^(n−1) divided by 9

$$\mathrm{prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction} \\ $$$$\Rightarrow\:\mathrm{7}^{\mathrm{n}} −\left(\mathrm{3n}+\mathrm{4}\right)×\mathrm{4}^{\mathrm{n}−\mathrm{1}} \:\mathrm{divided}\:\mathrm{by}\:\mathrm{9} \\ $$

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