real and imaginary part of e^(sin(x+iy))

August 19, 2021Posted inAdvanced Math, Complex Analysis, Computer science, Mathematics, Mechanical Engineering, Multivariable Calculus
Question : Find the real and imaginary part of e^(sin(x+iy)) Solution: Let \[z = e^(sin(x+iy))\] We shall write it as \[z = e^{Sin\left(x+iy\right)}=e^{Sin\left(x\right)Cos\left(iy\right)+\cos{\left(x\right)}\sin{\left(iy\right)}}\] Since \[Cos\left(iy\right)=\cosh{\left(y\right)}\ ,\ \ \sin{\left(iy\right)}=isinh\left(y\right)\] we shall…

Solved Sum of the series 1/1.2-1/2.3+1/3.4-1/4.5+…

August 19, 2021Posted inAbstract Algebra, Advanced Math, Multivariable Calculus, Number Theory, Real Analysis, Trigonometry
Question : Sum of the series 1/1.2-1/2.3+1/3.4-1/4.5+… Solution : Let the Sum of the series is \[S=\ \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\ldots..\] \[=\left(\frac{1}{1}-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots.\] \[=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots.\right)+\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots..\right)\] \[=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots.\right)+\left(-1+1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots..\right)\] We shall use formula of the series \[\log{\left(1+x\right)}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots..\] where…

solved evaluate the following limit

July 13, 2021Posted inMultivariable CalculusTags:
Question : Evaluate the following limit \[\lim_{x\rightarrow\frac{\pi}{2}\ }{\frac{\sin{\left(2x\right)}}{x^2+\frac{\pi}{2}x-\frac{\pi^2}{2}}}\] Solution : To evaluate the limit \[\lim_{x\rightarrow\frac{\pi}{2}\ }{\frac{\sin{\left(2x\right)}}{x^2+\frac{\pi}{2}x-\frac{\pi^2}{2}}}\] Since \[sin(2\ast\frac{\pi}{2})=sin(\pi)=0 \] and \[\left(\frac{\pi}{2}\right)^2+\frac{\pi}{2}\ast\frac{\pi}{2}-\frac{\pi^2}{2}=\frac{\pi^2}{4}+\frac{\pi^2}{4}-\frac{\pi^2}{2}=\frac{\pi^2}{2}-\frac{\pi^2}{2}=0\] That is \[\lim_{x\rightarrow\frac{\pi}{2}\ }{\frac{\sin{\left(2x\right)}}{x^2+\frac{\pi}{2}x-\frac{\pi^2}{2}}} \] Is in \[\frac{0}{0}\ form\]…