what is dimension of the vector space C(R) of the complex numbers over real
Question : What is The dimension of the vector space C(R) of the complex numbers over real numbers ? Solution : We know that a complex number z is written…
limit tan^-1(x)/x as x approaches 0
Question : Find the limit tan^-1(x)/x as x approaches 0 Solution : We shall let the limit is \[L={\rm limit}_{x\rightarrow0}{\frac{\tan^{-1}{\left(x\right)}}{x}} Since \tan^{-1}{\left(0\right)}=0\ Hence \frac{\tan^{-1}{\left(0\right)}}{0}=\frac{0}{0}\] We can apply L’Hospital rule, that…
limit of sin^-1(x)/x as x approaches to 0
Question : limit of sin^-1(x)/x as x approaches to 0 Solution : We shall let the limit is \[L={\rm limit}_{x\rightarrow0}{\frac{\sin^{-1}{\left(x\right)}}{x}}\] Since \[\sin^{-1}{\left(0\right)}=0\ Hence \frac{\sin^{-1}{\left(0\right)}}{0}=\frac{0}{0}\] We can apply L’Hospital rule, that…
limit of sin^-1(xy-2)/tan^-1(3xy-6) as (x y) approaches to (2 1)
Question : Find the limit of sin^-1(xy-2)/tan^-1(3xy-6) as (x y) approaches to (2 1) \[{\rm limit}_{\left(x,y\right)\rightarrow(2,1)}{\frac{\sin^{-1}{\left(xy-2\right)}}{\tan^{-1}{\left(3xy-6\right)}}} \] Solution : We shall let the limit is \[ L={\rm limit}_{\left(x,y\right)\rightarrow(2,1)}{\frac{\sin^{-1}{\left(xy-2\right)}}{\tan^{-1}{\left(3xy-6\right)}}}\] Multiplying and…
how to Find sinh(x+iy) formula
Question : Find sinh(x+iy) formula Solution : Let the complex number as \[z=sinh\left(x+iy\right)\] We know that \[\sin{\left(iz\right)}=isinh\left(z\right)\] So we shall multiply and divide by i we get \[z=\frac{isinh\left(x+iy\right)}{i}=\frac{\sin{\left(i\left(x+iy\right)\right)}}{i}=\frac{\sin{\left(ix-y\right)}}{i}\] Now we…
real and imaginary part of e^(sin(x+iy))
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+…
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…
The area of the region bounded between the line x =4 and the parabola y^2=16x is equal to
QUESTION : The area of the region bounded between the line x =4 and the The area of the region bounded between the line x =4 and the parabola y^2=16x…
solved evaluate the following limit
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 \[\frac{0}{0}\ form\]…
how to find Taylor series expansion of log(1-x^2) about x=0 ?
Question : Find Taylor series expansion of \[log(1-x^2) \] about x=0 Solution : We find Taylor series expansion of Log(1-x^2) about x =0 as follows Let \[g\left(x\right)=Log{\left(1-x^2\right)}\] \[g\left(0\right)\ =0\] \[g^\prime\left(x\right)=-\frac{2x}{1-x^2}\…