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Question Number 107113    Answers: 0   Comments: 0

Question Number 107108    Answers: 1   Comments: 3

y′′+y=(1/(cosx))

$$\mathrm{y}''+\mathrm{y}=\frac{\mathrm{1}}{\mathrm{cosx}} \\ $$

Question Number 107107    Answers: 1   Comments: 2

Question Number 107105    Answers: 1   Comments: 0

Question Number 107102    Answers: 0   Comments: 0

Question Number 107101    Answers: 0   Comments: 0

Fun time 1+2x+3x^2 +4x^3 +....=(1/((1−x)^2 )) 1+4+12+32+...=(1/((1−2)^2 )) 4+12+32+....=0 (No 1 fun) 5+11+17+23+...=0 Σ_(n=1) ^∞ 6n−1=6Σ_(n=1) ^∞ n−Σ^∞ 1=6.(−(1/(12)))−(−(1/2))=0 Σ^∞ n=−(1/(12)) (Ramanujan sum) Σ^∞ 1=1+1+1+1+1+...=−(1/2) Σ^∞ n^2 .Σ^∞ (1/n^2 )≥(Σ^∞ 1)^2 (Cauchy schwarz ineqality) Σ^∞ n^2 .(π^2 /6)≥(1/4) Σ^∞ n^2 ≥(3/(2π^2 ))

$$\mathrm{Fun}\:\mathrm{time} \\ $$$$ \\ $$$$\mathrm{1}+\mathrm{2x}+\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}^{\mathrm{3}} +....=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} } \\ $$$$\mathrm{1}+\mathrm{4}+\mathrm{12}+\mathrm{32}+...=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}+\mathrm{12}+\mathrm{32}+....=\mathrm{0}\:\:\left(\mathrm{No}\:\mathrm{1}\:\mathrm{fun}\right) \\ $$$$ \\ $$$$\mathrm{5}+\mathrm{11}+\mathrm{17}+\mathrm{23}+...=\mathrm{0}\:\:\: \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{6n}−\mathrm{1}=\mathrm{6}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{n}−\overset{\infty} {\sum}\mathrm{1}=\mathrm{6}.\left(−\frac{\mathrm{1}}{\mathrm{12}}\right)−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\overset{\infty} {\sum}\mathrm{n}=−\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\left(\mathrm{Ramanujan}\:\mathrm{sum}\right) \\ $$$$\overset{\infty} {\sum}\mathrm{1}=\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+\mathrm{1}+...=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\overset{\infty} {\sum}\mathrm{n}^{\mathrm{2}} .\overset{\infty} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\geqslant\left(\overset{\infty} {\sum}\mathrm{1}\right)^{\mathrm{2}} \:\:\:\left(\mathrm{Cauchy}\:\mathrm{schwarz}\:\mathrm{ineqality}\right) \\ $$$$\overset{\infty} {\sum}\mathrm{n}^{\mathrm{2}} .\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\geqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\overset{\infty} {\sum}\mathrm{n}^{\mathrm{2}} \geqslant\frac{\mathrm{3}}{\mathrm{2}\pi^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 107092    Answers: 1   Comments: 2

Question Number 107088    Answers: 3   Comments: 1

Question Number 107080    Answers: 1   Comments: 0

Question Number 107073    Answers: 0   Comments: 0

Question Number 107071    Answers: 0   Comments: 0

Question Number 107069    Answers: 2   Comments: 0

Question Number 107065    Answers: 1   Comments: 0

Solve in R^3 { ((x^2 +2xy+y^2 −4z^2 =0)),((3x−2y+z=3)) :} x^2 +y^2 +3z^2 −4xy+yz+x+2y−3z+7=0

$${Solve}\:{in}\:\mathbb{R}^{\mathrm{3}} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} −\mathrm{4}{z}^{\mathrm{2}} =\mathrm{0}}\\{\mathrm{3}{x}−\mathrm{2}{y}+{z}=\mathrm{3}}\end{cases} \\ $$$$\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{3}{z}^{\mathrm{2}} −\mathrm{4}{xy}+{yz}+{x}+\mathrm{2}{y}−\mathrm{3}{z}+\mathrm{7}=\mathrm{0} \\ $$

Question Number 107064    Answers: 1   Comments: 0

Solve in R x+(√(−x^2 +x+6))=(√(2x(√(x^2 +x+6))−x))

$${Solve}\:{in}\:\mathbb{R} \\ $$$${x}+\sqrt{−{x}^{\mathrm{2}} +{x}+\mathrm{6}}=\sqrt{\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{6}}−{x}} \\ $$

Question Number 107055    Answers: 0   Comments: 0

Question Number 107036    Answers: 2   Comments: 1

∫ (((√(1+x^2 )) dx)/x^4 ) ?

$$\int\:\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} }\:? \\ $$

Question Number 107033    Answers: 0   Comments: 0

Question Number 107030    Answers: 4   Comments: 1

lim_(x→0) (2+(3/x))^(1/(4x−2))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{2}+\frac{\mathrm{3}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{2}}} \\ $$

Question Number 107028    Answers: 3   Comments: 1

@bemath@ lim_(x→∞) (2+3x)^(1/(4x−2))

$$\:\:\:@{bemath}@ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{2}+\mathrm{3}{x}\right)^{\frac{\mathrm{1}}{\mathrm{4}{x}−\mathrm{2}}} \\ $$

Question Number 107018    Answers: 1   Comments: 2

Given f(x)=((10)/(2−sin 2x)). Find maximum value f(x).

$$\mathcal{G}{iven}\:{f}\left({x}\right)=\frac{\mathrm{10}}{\mathrm{2}−\mathrm{sin}\:\mathrm{2}{x}}.\:{Find}\:{maximum}\:{value}\: \\ $$$${f}\left({x}\right). \\ $$

Question Number 107034    Answers: 0   Comments: 6

Question Number 107002    Answers: 3   Comments: 0

∫_0 ^π ((cosx)/(√(2−sin^2 x)))dx

$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cosx}}{\sqrt{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}}\mathrm{dx} \\ $$

Question Number 106997    Answers: 0   Comments: 3

Question Number 106990    Answers: 1   Comments: 4

Two places A and B both on a parallel of latitude α°N differs in longitudes by θ°. Show that the shortest distance between them is: (([2 sin^(− 1) (cos α sin (θ/2))])/(360)) × 2πR Topic: Longitude and Latitude

$$\mathrm{Two}\:\mathrm{places}\:\:\mathrm{A}\:\:\mathrm{and}\:\:\mathrm{B}\:\:\mathrm{both}\:\mathrm{on}\:\mathrm{a}\:\mathrm{parallel}\:\mathrm{of}\:\mathrm{latitude}\:\:\alpha°\mathrm{N} \\ $$$$\mathrm{differs}\:\mathrm{in}\:\mathrm{longitudes}\:\mathrm{by}\:\:\theta°.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance} \\ $$$$\mathrm{between}\:\mathrm{them}\:\mathrm{is}:\:\:\:\:\:\frac{\left[\mathrm{2}\:\mathrm{sin}^{−\:\mathrm{1}} \left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right)\right]}{\mathrm{360}}\:\:×\:\:\mathrm{2}\pi\mathrm{R} \\ $$$$ \\ $$$$\mathrm{Topic}:\:\:\mathrm{Longitude}\:\mathrm{and}\:\mathrm{Latitude} \\ $$

Question Number 106983    Answers: 2   Comments: 0

Question Number 106977    Answers: 3   Comments: 0

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