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Question Number 218813 Answers: 3 Comments: 0

Question Number 218812 Answers: 5 Comments: 0

Question Number 218799 Answers: 0 Comments: 1

For those who are interested in cryptography. The below text has been encrypted using Vigenere cipher, such that numbers, punctuation marks and the letter E^(..) have remained the same. A keyword of length 9 has been used, which starts with the letter K. Decrypt the text.

$$\mathrm{For}\:\mathrm{those}\:\mathrm{who}\:\mathrm{are}\:\mathrm{interested}\:\mathrm{in}\:\mathrm{cryptography}. \\ $$$$\mathrm{The}\:\mathrm{below}\:\mathrm{text}\:\mathrm{has}\:\mathrm{been}\:\mathrm{encrypted}\:\mathrm{using} \\ $$$$\mathrm{Vigenere}\:\mathrm{cipher},\:\mathrm{such}\:\mathrm{that}\:\mathrm{numbers},\:\mathrm{punctuation} \\ $$$$\mathrm{marks}\:\mathrm{and}\:\mathrm{the}\:\mathrm{letter}\:\overset{..} {\mathrm{E}}\:\mathrm{have}\:\mathrm{remained}\:\mathrm{the}\:\mathrm{same}. \\ $$$$\mathrm{A}\:\mathrm{keyword}\:\mathrm{of}\:\mathrm{length}\:\mathrm{9}\:\mathrm{has}\:\mathrm{been}\:\mathrm{used},\:\mathrm{which} \\ $$$$\mathrm{starts}\:\mathrm{with}\:\mathrm{the}\:\mathrm{letter}\:\mathrm{K}.\:\mathrm{Decrypt}\:\mathrm{the}\:\mathrm{text}. \\ $$

Question Number 218997 Answers: 1 Comments: 0

Question Number 218792 Answers: 0 Comments: 0

Question Number 218785 Answers: 0 Comments: 0

Question Number 218781 Answers: 1 Comments: 0

prove: ∫_0 ^(π/4) arccos ((√2)/( (√(3−tan^2 x)))) dx=(π^2 /(24))

$$\mathrm{prove}: \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{arccos}\:\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \:{x}}}\:{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\ $$

Question Number 218780 Answers: 2 Comments: 1

Question Number 218779 Answers: 3 Comments: 0

Question Number 218778 Answers: 4 Comments: 0

Question Number 218771 Answers: 1 Comments: 0

∫_0 ^( ∞) J_ν (αt)J_ν (βt)dt=?? (α,β≠0)

$$\int_{\mathrm{0}} ^{\:\infty} \:{J}_{\nu} \left(\alpha{t}\right){J}_{\nu} \left(\beta{t}\right)\mathrm{d}{t}=?? \\ $$$$\left(\alpha,\beta\neq\mathrm{0}\right) \\ $$

Question Number 218775 Answers: 2 Comments: 0

Question Number 218767 Answers: 1 Comments: 0

Σ_(ℓ∈(−∞,∞)) J_ℓ (z)=??

$$\underset{\ell\in\left(−\infty,\infty\right)} {\sum}\:{J}_{\ell} \left({z}\right)=?? \\ $$

Question Number 218766 Answers: 0 Comments: 0

Σ_(ℓ=1) ^∞ (1/ℓ)J_ν (ℓt)=??

$$\underset{\ell=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\ell}{J}_{\nu} \left(\ell{t}\right)=?? \\ $$

Question Number 218765 Answers: 1 Comments: 0

solve ∫_0 ^( ∞) J_ν (kt)e^(−wt) dt=g_(ν,k) (w)

$$\mathrm{solve} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:{J}_{\nu} \left({kt}\right){e}^{−{wt}} \mathrm{d}{t}=\mathrm{g}_{\nu,{k}} \left({w}\right) \\ $$

Question Number 218749 Answers: 1 Comments: 0

Question Number 218748 Answers: 1 Comments: 0

∫_0 ^∞ ((x^2 cos x)/(cosh 2x−cos x))−((2x^2 )/(e^(4x) −2e^(2x) cos x+1))dx,lemma:Σ_(k=1) ^(+∞) ((cos kx)/p^k )=((p cos x−1)/(p^2 −2p cos x+1)),p>1

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} \mathrm{cos}\:{x}}{\mathrm{cosh}\:\mathrm{2}{x}−\mathrm{cos}\:{x}}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{e}^{\mathrm{4}{x}} −\mathrm{2}{e}^{\mathrm{2}{x}} \mathrm{cos}\:{x}+\mathrm{1}}{dx},\mathrm{lemma}:\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{cos}\:{kx}}{{p}^{{k}} }=\frac{\mathrm{p}\:\mathrm{cos}\:{x}−\mathrm{1}}{{p}^{\mathrm{2}} −\mathrm{2}{p}\:\mathrm{cos}\:{x}+\mathrm{1}},{p}>\mathrm{1} \\ $$

Question Number 218739 Answers: 2 Comments: 0

Question Number 218738 Answers: 2 Comments: 0

Question Number 218737 Answers: 2 Comments: 0

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Question Number 218733 Answers: 4 Comments: 1

Question Number 218719 Answers: 0 Comments: 0

Question Number 218713 Answers: 1 Comments: 3

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