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Question Number 141685 Answers: 1 Comments: 0
$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:......{nice}\:...\:...\:...\:{calculus}..... \\ $$$$\:\:\mathrm{I}{f}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{tan}\left({x}\right)}{{x}}\:=\:\mathrm{1}\:,\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:{lim}\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{tan}\left({x}\right)}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Question Number 141681 Answers: 1 Comments: 0
$$\mathrm{On}\:\mathrm{the}\:\mathrm{Argand}\:\mathrm{Diagram},\:\mathrm{the}\:\mathrm{variable}\:\mathrm{point} \\ $$$$\mathrm{Z}\:\mathrm{represents}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\:{z}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{a}\:\mathrm{point} \\ $$$$\mathrm{Z}\:\mathrm{which}\:\mathrm{moves}\:\mathrm{such}\:\mathrm{that}\:\mid\frac{{z}−\mathrm{1}}{{z}+\mathrm{2}}\mid=\mathrm{2} \\ $$
Question Number 143167 Answers: 2 Comments: 0
$$\int\boldsymbol{{arctan}}\left(\sqrt{\sqrt{\boldsymbol{{x}}}+\mathrm{1}}\right)\boldsymbol{{dx}}=??? \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{propose}}'\:\boldsymbol{{par}}\:\boldsymbol{{Rodrigue}} \\ $$
Question Number 141672 Answers: 1 Comments: 0
$$\:\:{Solve}\:{the}\:{equation}\: \\ $$$$\:\:{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{3}=\mathrm{0} \\ $$
Question Number 141668 Answers: 1 Comments: 0
$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:.......{Advanced}\:...\bigstar\:...\bigstar\:...\:{Calculus}....... \\ $$$$\:\:\:\:\:\:\:\:\:{if}\:\:\:\:\Omega\:=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} }\:{then}\:{prove} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{that}\:::\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:{e}^{\Omega−\mathrm{1}} \:\: \\ $$$$\:\:\:\:{proof}\::: \\ $$$$\:\:\:\:{method}\:\left(\mathrm{1}\right): \\ $$$$\:\:\:\:\:\psi\:\left(\mathrm{1}+{x}\:\right)=\:−\gamma+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right){x}^{{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\left(\:{Maclaurin}\:{series}\:{for}\:\psi\left({x}+\mathrm{1}\right)\:\right) \\ $$$$\:\:\:\:{x}:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\psi\:\left(\frac{\mathrm{3}}{\mathrm{2}}\:\right)=−\gamma\:+\:\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} }\:\:\left(\ast\:\right) \\ $$$$\:\:\:\:{we}\:{know}\:{that}\:::\:\psi\left(\mathrm{1}+{x}\right)=\frac{\mathrm{1}}{{x}}+\psi\left({x}\right) \\ $$$$\:\:\:\:\:\:\left(\:\ast\:\right)\:\:\Rrightarrow\:\psi\:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{2}+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\gamma+\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} } \\ $$$$\:\:\:\:\:\:\:\left(\ast\right)\:\:\:\:\:\Rrightarrow\:\:\:\:\:\:\:\:\:\mathrm{2}−\gamma−{ln}\left(\mathrm{4}\right)=−\gamma+\mathrm{2}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ln}\left(\frac{{e}}{\mathrm{2}}\right)=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \zeta\left({n}\right)}{\mathrm{2}^{{n}} }\:=\Omega \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:{e}^{\Omega\:−\mathrm{1}} \:\:\:....\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.{july}.\mathrm{1970}... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Question Number 141665 Answers: 0 Comments: 1
Question Number 141663 Answers: 0 Comments: 1
$$\:\underset{\:\:−−−−−−−−−−−−−−−−−−−} {\:} \\ $$$$\:\:{x}^{\mathrm{2}} \left({x}−\mathrm{12}\right)\left({x}−\mathrm{15}\right)={k}\left({x}−\mathrm{16}\right) \\ $$$$\:\:\:\:{find}\:{x}\:{in}\:{terms}\:{of}\:{k}\:\left(>\mathrm{0}\right). \\ $$$$\:\:\overset{−−−−−−−−−−−−−−−−−−−} {\:} \\ $$
Question Number 141658 Answers: 1 Comments: 0
Question Number 141654 Answers: 0 Comments: 2
$${Solve}\:{for}\:{real}\:{numbers} \\ $$$$\mathrm{5}\centerdot\left(\sqrt[{\mathrm{5}}]{\mathrm{1}−{z}}\:+\:\sqrt[{\mathrm{5}}]{\mathrm{1}+{z}}\:=\:\mathrm{2}+\mathrm{4}\centerdot\left(\sqrt[{\mathrm{4}}]{\mathrm{1}−{z}}\:+\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{z}}\right)\right. \\ $$
Question Number 141653 Answers: 0 Comments: 0
$${what}\:{is}\:{condition}\:{to}\:{have} \\ $$$${log}\left(\:{I}\:+{A}\right)=\Sigma\:{a}_{{n}} {A}^{{n}} \\ $$$${and}\:{determine}\:{the}\:{sequence}\:\left({a}_{{n}} \right) \\ $$$${A}\:\in\:{M}_{{n}} \left({C}\right) \\ $$
Question Number 141652 Answers: 0 Comments: 0
$${A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${find}\:{e}^{{A}\:} \:{and}\:{e}^{{tA}} \\ $$$${find}\:{ch}\left({A}\right)\:{and}\:{sh}\left({A}\right) \\ $$
Question Number 141651 Answers: 0 Comments: 2
Question Number 141649 Answers: 1 Comments: 0
$$\:\:\:\:\:\:\:\:\:.......{advanced}\:\:{calculus}...... \\ $$$$\:\:\:\:{prove}\:\:{that}−:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\phi:=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{cos}\left(\mathrm{2}\pi{x}^{\mathrm{2}} \right)}{{cosh}^{\mathrm{2}} \left(\pi{x}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\checkmark \\ $$
Question Number 141937 Answers: 0 Comments: 2
$${Write}\:{and}\:{graph}\:{the}\:{equation}\:{of}\:{the}\:{graph}\:{of}\:{y}={sin}\left(\pi{x}\right) \\ $$$${It}\:{is}\:{stretched}\:{up}\:{by}\:{a}\:{factor}\:{of}\:\mathrm{5}\:{and}\:{shifted}\:\frac{\mathrm{1}}{\mathrm{2}}\:{unit}\:{to}\:{the}\:{right} \\ $$$${Help}\:{me}\:{please} \\ $$$$ \\ $$
Question Number 141645 Answers: 0 Comments: 0
Question Number 141643 Answers: 1 Comments: 0
Question Number 141642 Answers: 0 Comments: 0
Question Number 141640 Answers: 1 Comments: 0
Question Number 141635 Answers: 0 Comments: 0
$${solve}\:{the}\:{differential}\:{equation}\:\left({PDE}\right), \\ $$$${z}\left(\frac{\partial{z}}{\partial{x}}−\frac{\partial{z}}{\partial{y}}\right)={z}^{\mathrm{2}} +\left({x}+{y}\right)^{\mathrm{2}} . \\ $$
Question Number 141633 Answers: 0 Comments: 0
$$\mathrm{1}<\mathrm{a}<\mathrm{b}\:,\mathrm{prove}\:\mathrm{that}\:: \\ $$$${b}^{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \mathrm{C}_{{n}} ^{{k}} \:{a}^{\frac{{ln}\left(\underset{{p}=\mathrm{0}} {\overset{{n}−{k}} {\sum}}\mathrm{C}_{{n}−{k}} ^{{p}} {a}^{{n}−{p}} {b}^{{p}} \right)}{{ln}\left({a}\right)}} \\ $$
Question Number 141632 Answers: 0 Comments: 0
$$\mathrm{Let}\:{f}\left({x}\right)=\frac{{sin}\left({x}\right)}{{x}}\:,\:\mathrm{prove}\:\mathrm{that}\:: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\:{f}\left({n}\pi+\alpha\right)+{f}\left({n}\pi−\alpha\right)\:\right]=\:\mathrm{1}+{f}\left(\alpha\right) \\ $$
Question Number 141669 Answers: 1 Comments: 0
Question Number 141628 Answers: 0 Comments: 3
Question Number 141627 Answers: 2 Comments: 0
Question Number 141623 Answers: 1 Comments: 0
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} }{\mathrm{1}+{x}}{dx}\right)^{\mathrm{2}} ={ln}\:\mathrm{2} \\ $$
Question Number 141614 Answers: 1 Comments: 2
$$\Sigma\frac{\mathrm{1}}{{k}+\mathrm{1}}{C}_{{n}} ^{{k}} \:.\: \\ $$
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