Question Number 221447 by Nicholas666 last updated on 05/Jun/25 $$ \\ $$$$\:\:\:\:\mathrm{if}\:\:\:\:\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\:\left({x}\:+\:{r}_{{i}} \right)\:\equiv\:\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}\:{a}_{{j}} {x}^{{n}−{i}} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{show}\:\mathrm{that}\:; \\ $$$$\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{tan}^{−\mathrm{1}}…
Question Number 221416 by wewji12 last updated on 04/Jun/25 $$\mathrm{ex3}. \\ $$$$\mathrm{prove} \\ $$$${f}^{\left({n}\right)} \left(\alpha\right)=\frac{{n}!}{\mathrm{2}\pi\boldsymbol{{i}}}\:\oint_{\:\partial{S}} \:\frac{{f}\left({z}\right)}{\left({z}−\alpha\right)^{{n}+\mathrm{1}} }\:\mathrm{d}{z} \\ $$$$\mathrm{ex4}. \\ $$$$\mathrm{Let}\:{z}_{\mathrm{0}} \:\mathrm{be}\:\mathrm{any}\:\mathrm{point}\:\mathrm{interior}\:\mathrm{to}\:\mathrm{a}\:\mathrm{positively} \\ $$$$\mathrm{oriented}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}\:\mathcal{C} \\…
Question Number 221417 by leromain last updated on 04/Jun/25 $${Let}\:\:{S}\:{be}\:{delimited}\:{by}\:{the}\:{equations}\: \\ $$$${x}=\mathrm{0};\:{y}=\mathrm{0}\:;\:{z}=\mathrm{0}\:{and}\:{x}+{y}+{z}=\mathrm{0} \\ $$$${Find}\:{the}\:{flux}\:{of}\:{vector}\:{field}\: \\ $$$${V}\left({x},{y},{z}\right)=\left({x},{y},{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:{through}\:{S} \\ $$ Terms of Service Privacy Policy…
Question Number 221413 by BHOOPENDRA last updated on 04/Jun/25 Answered by mr W last updated on 05/Jun/25 $${x}=\mathrm{4}\:\mathrm{cos}\:\theta \\ $$$${y}=\mathrm{2}\:\mathrm{sin}\:\theta \\ $$$${x}+{y}=\mathrm{2}\left(\mathrm{2}\:\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right)=\mathrm{2}\sqrt{\mathrm{5}}\:\mathrm{sin}\:\left(\theta+\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\right) \\ $$$$\left({x}+{y}\right)_{{min}}…
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Question Number 221415 by wewji12 last updated on 04/Jun/25 $$\int\:\:\mathrm{d}{z}\:\left[−\frac{\mathrm{1}}{\pi}\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \centerdot\underset{{k}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\:\frac{\Gamma\left(\nu−{k}\right)}{{k}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\frac{\mathrm{2}}{\pi}\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{z}\right){J}_{\nu} \left({z}\right)−\frac{\mathrm{1}}{\pi}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}} \left(\psi^{\left(\mathrm{0}\right)} \left({k}+\nu+\mathrm{1}\right)+\psi^{\left(\mathrm{0}\right)} \left({k}+\mathrm{1}\right)\right)}{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \right] \\ $$ Answered…
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Question Number 221406 by ajfour last updated on 03/Jun/25 Commented by ajfour last updated on 03/Jun/25 $${Find}\:\mathrm{tan}\:\phi\:\:\:{if}\:\mathrm{tan}\:\theta=\mathrm{1}/\mathrm{4}. \\ $$ Answered by mr W last updated…
Question Number 221407 by fantastic last updated on 03/Jun/25 $${A}\:{and}\:{B}\:{are}\:{two}\:{angles}\:{such}\:{that}\:\mathrm{0}^{\mathrm{0}} <{B}<{A}<\mathrm{90}^{\mathrm{0}} \:{then}\:{prove}\:{geometrycaly}\:{that}\: \\ $$$$\mathrm{cos}\:\left({A}+{B}\right)=\mathrm{cos}\:{A}\mathrm{cos}\:{B}−\mathrm{sin}\:{A}\mathrm{sin}\:{B} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com