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Question Number 222937    Answers: 1   Comments: 0

Prove that the following identity holds : Σ_(λ∈Z[i]) ((1/((1 + 3λ)^3 ))) = ((π^(9/2) (√(1 + (√(3 )))))/(2^(5/4) 3^(27/8) Γ((3/4))^6 )) Where Z[i] = {a + bi : a,b ∈ Z} denotes gaussian integers !

$$ \\ $$$$\:\:\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{identity}\:\mathrm{holds}\::\:\:\:\: \\ $$$$\:\:\:\underset{\lambda\in\mathbb{Z}\left[{i}\right]} {\sum}\:\left(\frac{\mathrm{1}}{\left(\mathrm{1}\:+\:\mathrm{3}\lambda\right)^{\mathrm{3}} }\right)\:=\:\frac{\pi^{\mathrm{9}/\mathrm{2}} \:\sqrt{\mathrm{1}\:+\:\sqrt{\mathrm{3}\:}}}{\mathrm{2}^{\mathrm{5}/\mathrm{4}} \:\:\mathrm{3}^{\mathrm{27}/\mathrm{8}} \:\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{6}} \:}\:\:\:\:\: \\ $$$$\:\:\:\:\mathrm{Where}\:\mathbb{Z}\left[{i}\right]\:=\:\left\{{a}\:+\:{bi}\::\:{a},{b}\:\in\:\mathbb{Z}\right\}\:\mathrm{denotes}\:\mathrm{gaussian}\:\mathrm{integers}\:!\:\:\:\:\:\:\:\:\: \\ $$$$\: \\ $$

Question Number 222929    Answers: 0   Comments: 0

Question Number 222927    Answers: 0   Comments: 0

∫∫∫_V ▽∙FdV=∫∫_∂V F∙dS F=−▽φ ▽∙(−▽φ)=−▽^2 φ ∫∫∫_V (−▽^2 φ)dV=∫∫_∂V (−▽φ)∙dS −∫∫∫_V ▽^2 φdV=−∫∫_∂V ▽φ∙dS ∫∫∫_V ▽^2 φdV=∫∫_∂V ▽φ∙dS ρ(r)=Σ_i q_i δ^((3)) −(r−r_i ) ▽^2 ((1/(∣r−r′∣)))=−4πδ^((3)) (r−r′) φ(r)=∫∫∫_V ((ρ(r′))/(4π∣r−r′∣))dV′ ▽^2 φ=▽^2 ((1/(4π))∫∫∫_V ((ρ(r′))/(∣r−r′∣))dV′) =(1/(4π))∫∫∫_V ρ(r′)▽^2 ((1/(∣r−r′∣)))dV′ =(1/(4π))∫∫∫_V ρ(r′)(−4πδ^((3)) (r−r′))dV′ =−∫∫∫_V ρ(r′)δ^((3)) (r−r′)dV′ =−ρ(r) ▽^2 φ=−ρ

$$\int\int\int_{{V}} \bigtriangledown\centerdot\boldsymbol{\mathrm{F}}{dV}=\int\int_{\partial{V}} \boldsymbol{\mathrm{F}}\centerdot\mathrm{d}\boldsymbol{{S}} \\ $$$$\boldsymbol{\mathrm{F}}=−\bigtriangledown\phi \\ $$$$\bigtriangledown\centerdot\left(−\bigtriangledown\phi\right)=−\bigtriangledown^{\mathrm{2}} \phi \\ $$$$\int\int\int_{{V}} \left(−\bigtriangledown^{\mathrm{2}} \phi\right){dV}=\int\int_{\partial{V}} \left(−\bigtriangledown\phi\right)\centerdot\mathrm{d}\boldsymbol{{S}} \\ $$$$−\int\int\int_{{V}} \bigtriangledown^{\mathrm{2}} \phi\mathrm{d}{V}=−\int\int_{\partial{V}} \bigtriangledown\phi\centerdot{d}\boldsymbol{{S}} \\ $$$$\int\int\int_{{V}} \bigtriangledown^{\mathrm{2}} \phi{dV}=\int\int_{\partial{V}} \bigtriangledown\phi\centerdot{d}\boldsymbol{{S}} \\ $$$$\rho\left(\boldsymbol{{r}}\right)=\underset{{i}} {\sum}{q}_{{i}} \delta^{\left(\mathrm{3}\right)} −\left(\boldsymbol{{r}}−\boldsymbol{{r}}_{{i}} \right) \\ $$$$\bigtriangledown^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mid\boldsymbol{{r}}−\boldsymbol{{r}}'\mid}\right)=−\mathrm{4}\pi\delta^{\left(\mathrm{3}\right)} \left(\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\right) \\ $$$$\phi\left(\boldsymbol{\mathrm{r}}\right)=\int\int\int_{{V}} \frac{\rho\left(\boldsymbol{\mathrm{r}}'\right)}{\mathrm{4}\pi\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid}{dV}' \\ $$$$\bigtriangledown^{\mathrm{2}} \phi=\bigtriangledown^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}\pi}\int\int\int_{{V}} \frac{\rho\left(\boldsymbol{\mathrm{r}}'\right)}{\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid}{dV}'\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi}\int\int\int_{{V}} \rho\left(\boldsymbol{\mathrm{r}}'\right)\bigtriangledown^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid}\right){dV}' \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi}\int\int\int_{{V}} \rho\left(\boldsymbol{\mathrm{r}}'\right)\left(−\mathrm{4}\pi\delta^{\left(\mathrm{3}\right)} \left(\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\right)\right)\mathrm{d}{V}' \\ $$$$=−\int\int\int_{{V}} \rho\left(\boldsymbol{\mathrm{r}}'\right)\delta^{\left(\mathrm{3}\right)} \left(\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\right){dV}' \\ $$$$=−\rho\left(\boldsymbol{\mathrm{r}}\right) \\ $$$$\bigtriangledown^{\mathrm{2}} \phi=−\rho \\ $$

Question Number 222924    Answers: 2   Comments: 0

Question Number 222922    Answers: 1   Comments: 0

Prove:∫_0 ^1 J_0 (ln(1/x))dx=((√2)/2)

$$ \\ $$$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} {J}_{\mathrm{0}} \left(\mathrm{ln}\frac{\mathrm{1}}{{x}}\right){dx}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

Question Number 222921    Answers: 2   Comments: 0

Question Number 222908    Answers: 1   Comments: 0

Question Number 222889    Answers: 1   Comments: 10

Question Number 222885    Answers: 3   Comments: 0

Question Number 222881    Answers: 2   Comments: 0

∫_0 ^∞ t^a e^(−t) erf(kt)dt,a>0,k>0

$$ \\ $$$$\:\int_{\mathrm{0}} ^{\infty} {t}^{{a}} {e}^{−{t}} \mathrm{erf}\left({kt}\right){dt},{a}>\mathrm{0},{k}>\mathrm{0} \\ $$

Question Number 222879    Answers: 1   Comments: 0

Question Number 222876    Answers: 1   Comments: 0

∫_0 ^1 ∫_(2y) ^2 e^x^2 dxdy=?

$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{2}{y}} ^{\mathrm{2}} {e}^{{x}^{\mathrm{2}} } {dxdy}=? \\ $$

Question Number 222874    Answers: 0   Comments: 0

evaluate the following integral ∫((x^5 ln(x))/((x^2 +1)^3 )) dx

$$\mathrm{evaluate}\:\mathrm{the}\:\mathrm{following}\:\mathrm{integral} \\ $$$$\int\frac{{x}^{\mathrm{5}} \mathrm{ln}\left({x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }\:{dx} \\ $$

Question Number 222867    Answers: 1   Comments: 0

∫_(−π) ^π x sin x cosnxdx

$$\int_{−\pi} ^{\pi} {x}\:\mathrm{sin}\:{x}\:\mathrm{cos}{nxdx} \\ $$

Question Number 222858    Answers: 1   Comments: 1

Prove: ∫_0 ^(1/2) ((arcsin^2 x)/x)dx=((πi)/6)((π^2 /(36))−Li_2 (((1+i(√3))/2)))−(1/3)ζ(3)

$$\mathrm{Prove}: \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{arcsin}^{\mathrm{2}} {x}}{{x}}{dx}=\frac{\pi{i}}{\mathrm{6}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{36}}−\mathrm{Li}_{\mathrm{2}} \left(\frac{\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)−\frac{\mathrm{1}}{\mathrm{3}}\zeta\left(\mathrm{3}\right) \\ $$

Question Number 222856    Answers: 1   Comments: 0

find the possible root of x^3 −2x^2 −5x+6=0 using the fixed point iteration method?

$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{possible}}\:\boldsymbol{{root}}\:\boldsymbol{{of}}\:\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{{x}}+\mathrm{6}=\mathrm{0} \\ $$$$\boldsymbol{{using}}\:\boldsymbol{{the}}\:\boldsymbol{{fixed}}\:\boldsymbol{{point}}\:\boldsymbol{{iteration}}\:\boldsymbol{{method}}? \\ $$

Question Number 222855    Answers: 1   Comments: 0

Prove:∫_(−π) ^π x ln (1+sin x +cos x)dx=2πG

$$\mathrm{Prove}:\int_{−\pi} ^{\pi} {x}\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right){dx}=\mathrm{2}\pi{G} \\ $$

Question Number 222850    Answers: 1   Comments: 0

∫_0 ^∞ ((x^(−x) e^(−x) )/(Γ(1−x)))dx

$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{−{x}} {e}^{−{x}} }{\Gamma\left(\mathrm{1}−{x}\right)}{dx} \\ $$

Question Number 222848    Answers: 1   Comments: 0

∫_0 ^1 ((arcatn^2 x)/x)dx=(π/2)G−(7/8)ζ(3)

$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arcatn}^{\mathrm{2}} {x}}{{x}}{dx}=\frac{\pi}{\mathrm{2}}{G}−\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right) \\ $$

Question Number 222847    Answers: 1   Comments: 0

let f(x)=1.013x^5 −5.262x^3 −0.01732x^2 +0.8389x −1.912. Evaluate f(2.279) by first calculating (2.279)^2 ,(2.279)^3 ,(2.279)^4 and(2.279)^5 using four−digit round arithmetic. hence,compute the absolute and relative errors.

$$\boldsymbol{{let}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\mathrm{1}.\mathrm{013}\boldsymbol{{x}}^{\mathrm{5}} −\mathrm{5}.\mathrm{262}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{0}.\mathrm{01732}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{0}.\mathrm{8389}\boldsymbol{{x}} \\ $$$$−\mathrm{1}.\mathrm{912}.\:\boldsymbol{{Evaluate}}\:\boldsymbol{{f}}\left(\mathrm{2}.\mathrm{279}\right)\:\boldsymbol{{by}}\:\boldsymbol{{first}}\:\boldsymbol{{calculating}} \\ $$$$\left(\mathrm{2}.\mathrm{279}\right)^{\mathrm{2}} ,\left(\mathrm{2}.\mathrm{279}\right)^{\mathrm{3}} ,\left(\mathrm{2}.\mathrm{279}\right)^{\mathrm{4}} \boldsymbol{{and}}\left(\mathrm{2}.\mathrm{279}\right)^{\mathrm{5}} \:\boldsymbol{{using}} \\ $$$$\boldsymbol{{four}}−\boldsymbol{{digit}}\:\boldsymbol{{round}}\:\boldsymbol{{arithmetic}}.\:\boldsymbol{{hence}},\boldsymbol{{compute}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{absolute}}\:\boldsymbol{{and}}\:\boldsymbol{{relative}}\:\boldsymbol{{errors}}. \\ $$

Question Number 222838    Answers: 2   Comments: 0

Prove:∫_0 ^(1/2) ((ln(2x))/( (√(1+x^2 ))))dx=−(π^2 /(20))

$$\:\:\mathrm{Prove}:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}=−\frac{\pi^{\mathrm{2}} }{\mathrm{20}} \\ $$

Question Number 222845    Answers: 0   Comments: 0

Evaluate; Σ_(k=0) ^n (−1)^k (((2n − k)),(( k)) )

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Evaluate}; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\left(−\mathrm{1}\right)^{{k}} \:\begin{pmatrix}{\mathrm{2}{n}\:−\:{k}}\\{\:\:\:\:\:\:\:\:{k}}\end{pmatrix} \\ $$$$ \\ $$

Question Number 222835    Answers: 0   Comments: 0

Question Number 222830    Answers: 1   Comments: 0

Question Number 222829    Answers: 1   Comments: 0

If f(x) = ((3x + [x])/(2x)) Find lim_(x→−5^+ ) f(x) − lim_(x→−5^− ) f(x) = ?

$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{3x}\:+\:\left[\mathrm{x}\right]}{\mathrm{2x}} \\ $$$$\mathrm{Find}\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow−\mathrm{5}^{+} } {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\:−\:\underset{\boldsymbol{\mathrm{x}}\rightarrow−\mathrm{5}^{−} } {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:? \\ $$

Question Number 222828    Answers: 1   Comments: 0

vector field F^→ ;R^3 →R^3 , F_h ∈C^ω and Let′s define as A^→ =▽^→ ×F^→ can we find vector field F^→ .....??? Curl and Divergence inverse operator dose exist?? (▽_ ^→ ×)^(−1) A^→ , (▽_ ^→ ∗)^(−1) A^→ ex. ( ((d )/dx))^(−1) =∫

$$\mathrm{vector}\:\mathrm{field}\:\:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}};\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R}^{\mathrm{3}} \:,\:{F}_{{h}} \in\mathcal{C}^{\omega} \\ $$$$\mathrm{and}\:\mathrm{Let}'\mathrm{s}\:\mathrm{define}\:\mathrm{as}\:\overset{\rightarrow} {\boldsymbol{\mathrm{A}}}=\overset{\rightarrow} {\bigtriangledown}×\overset{\rightarrow} {\boldsymbol{\mathrm{F}}} \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{vector}\:\mathrm{field}\:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}.....??? \\ $$$$\mathrm{Curl}\:\mathrm{and}\:\mathrm{Divergence}\:\:\mathrm{inverse}\:\mathrm{operator}\:\mathrm{dose}\:\mathrm{exist}?? \\ $$$$\left(\overset{\rightarrow} {\bigtriangledown}_{\:} ×\right)^{−\mathrm{1}} \overset{\rightarrow} {\boldsymbol{\mathrm{A}}}\:,\:\left(\overset{\rightarrow} {\bigtriangledown}_{\:} \ast\right)^{−\mathrm{1}} \overset{\rightarrow} {\boldsymbol{\mathrm{A}}} \\ $$$$\mathrm{ex}.\:\left(\:\frac{\mathrm{d}\:\:\:}{\mathrm{d}{x}}\right)^{−\mathrm{1}} =\int\: \\ $$

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