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Question Number 103266 Answers: 0 Comments: 1

How can we solve second-order differential equations with non-constant coefficients ? Any idea, please ?

$$\mathrm{How}\:\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{second}-\mathrm{order}\:\mathrm{differential}\:\mathrm{equations}\:\mathrm{with} \\ $$$$\mathrm{non}-\mathrm{constant}\:\mathrm{coefficients}\:?\:\mathrm{Any}\:\mathrm{idea},\:\mathrm{please}\:? \\ $$$$ \\ $$

Question Number 103260 Answers: 1 Comments: 0

Question Number 103255 Answers: 1 Comments: 1

lim_(x→0) (((sin^6 (√2)x)/(tan^7 (x/2))))^(cos^(12) x) =???

$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left(\frac{{sin}^{\mathrm{6}} \sqrt{\mathrm{2}}{x}}{{tan}^{\mathrm{7}} \frac{{x}}{\mathrm{2}}}\right)^{{cos}^{\mathrm{12}} {x}} =??? \\ $$

Question Number 103254 Answers: 0 Comments: 2

2^(3x) −f(x+4)=k∙f(x) ; k=????

$$\:\mathrm{2}^{\mathrm{3}{x}} −{f}\left({x}+\mathrm{4}\right)={k}\centerdot{f}\left({x}\right)\:\:\:\:\:;\:\:\:{k}=???? \\ $$

Question Number 103253 Answers: 3 Comments: 0

(i)^(1/i) =?????

$$\:\sqrt[{{i}}]{{i}}=????? \\ $$

Question Number 103241 Answers: 0 Comments: 0

∫x^(−x) dx

$$\int{x}^{−{x}} {dx} \\ $$

Question Number 103239 Answers: 1 Comments: 0

Question Number 103236 Answers: 0 Comments: 3

e^e^(e.....∞) =?

$${e}^{{e}^{{e}.....\infty} } =? \\ $$

Question Number 103228 Answers: 1 Comments: 1

Question Number 103223 Answers: 0 Comments: 0

In a card game for four players, a pack of fifty-two cards is dealt round so that each player receives thirteen cards. A hand that contains no card greater than nine is called a yarborough. How many deals are necessary for the probability of at least one hand being a yarborough to be greater than (1/2) ? (Ace ranks high)

$$\:\:\:\:\:\:\:\:\:\mathrm{In}\:\mathrm{a}\:\mathrm{card}\:\mathrm{game}\:\mathrm{for}\:\mathrm{four}\:\mathrm{players},\:\mathrm{a}\:\mathrm{pack}\:\mathrm{of}\:\mathrm{fifty}-\mathrm{two}\:\mathrm{cards}\:\mathrm{is}\:\mathrm{dealt}\:\mathrm{round}\:\mathrm{so} \\ $$$$\:\:\:\:\:\:\mathrm{that}\:\mathrm{each}\:\mathrm{player}\:\mathrm{receives}\:\mathrm{thirteen}\:\mathrm{cards}.\:\mathrm{A}\:\mathrm{hand}\:\mathrm{that}\:\mathrm{contains}\:\mathrm{no}\:\mathrm{card}\:\mathrm{greater}\:\mathrm{than}\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{nine}\:\mathrm{is}\:\mathrm{called}\:\mathrm{a}\:\mathrm{yarborough}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{deals}\:\mathrm{are}\:\mathrm{necessary}\:\mathrm{for}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\:\:\:\:\:\:\mathrm{of}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{hand}\:\mathrm{being}\:\mathrm{a}\:\mathrm{yarborough}\:\mathrm{to}\:\mathrm{be}\:\mathrm{greater}\:\mathrm{than}\:\frac{\mathrm{1}}{\mathrm{2}}\:?\:\left(\mathrm{Ace}\:\mathrm{ranks}\:\mathrm{high}\right)\:\:\:\:\:\: \\ $$

Question Number 103220 Answers: 1 Comments: 0

∫_0 ^∞ (x^3 /(e^x +1))dx

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{{e}^{{x}} +\mathrm{1}}{dx} \\ $$

Question Number 103219 Answers: 1 Comments: 3

1+1+1+1+1+1+1+....=S_n S_n =1+1+1+1+1+1+1+.... 2S_n = 2 + 2 + 2+....... .......... subtracting −S_n =1−1+1−1+1−1+1−1+1−1+.... −S_n =(1/2) S_n =−(1/2) I have found this while experiment . I know the sum diverges but is it pretty cool? Kindly rectify me if there is any fault on this non rigorous process I have found some Ramanujan proof S_n =1+2+3+4+5+6+7+... 4S_n = 4+ 8 + 12+... −3S_n =1−2+3−4+5−6+7−8+...... −3S_n =(1/4) S_n =−(1/(12)) Ramanujan had done this on his notebook

Question Number 103209 Answers: 3 Comments: 0