Mathematics Compilations

Imaginary numbers are not 'imaginary'. Real numbers are not 'real'. In fact, they're both in our heads... We are quite familiar with the logarithmic function. We define it as the inverse of the exponential function. That is, we say: $$\log{e^x} = e^{\log{x}} = x$$ (In fact, this is how the inverse of a function is even defined.) Now, we're all comfortable with the natural log. We use it for various results, but we always keep in mind that the number we're evaluating is POSITIVE. Why though? Let's consider some negative number $a$. What does it mean to take it's natural logarithm? It means to find a number such that $e$ raised to that number gives $a$. And surely since $e$ raised to any power always gives a positive value, it makes no sense to take the log of a negative number, right? Right? Complex Numbers: Function Domains Remember how we  extended  the existence of real numbers to the complex number system? What did we do? ...

Using the Taylor series is a straight forward method, but let us try to approach the problem in a different manner. Integration by Parts NOTE: This method requires working knowledge of  integration by parts . When I tried to solve this problem, I first started to expand it using the integration by parts method. The pattern I observed was the thing which helped me put myself on the right track to solve this problem. Here is the pattern: $$\int e^{x^2}dx = xe^{x^2} - 2\int x^2e^{x^2}dx$$ $$3\int x^2e^{x^2}dx = x^3e^{x^2} - 2\int x^4e^{x^2}dx$$ $$5\int x^4e^{x^2}dx = x^5e^{x^2} - 2\int x^6e^{x^2}dx$$ To make sure this pattern sustains, I decided to find the integral of $x^{2n}e^{x^2}$. $$Y = \int x^{2n}e^{x^2}dx$$ $$Y = x^{2n+1}e^{x^2} - \int x[2nx^{2n-1}e^{x^2}+2x^{2n+1}e^{x^2}]dx$$ $$(2n+1)Y = x^{2n+1}e^{x^2}-2\int x^{2n+2}e^{x^2}dx$$ $$\therefore Y = \frac{x^{2n+1}e^{x^2}}{2n+1}-\frac{2}{2n+1}\int x^{2n+2}e^{x^2}dx$$ Note that the $\int x^{2n+2}e^{x^2}dx$ is sim...

We know from the product rule that $d(xy) = xdy+ydx$. Integrating it, $$\int d(xy) = \int xdy + \int ydx$$ $$\therefore \int xdy = xy- \int ydx$$ This has applications in many different problems, and is known as integration by parts.

Consider the given equation... $$y^2-7=x^3=(x^{\frac{3}{2}})^2$$ $$y^2-(x^{\frac{3}{2}})^2=7$$ If $x^{\frac{3}{2}}$ is a decimal, then $y$ should also be some decimal to make the equation an integer, i.e., $7$. So $y$ will be a decimal in this case. But consider when $x^{\frac{3}{2}}$ is an integer. The only integers whose difference of squares is $7$, is $4$ and $3$ (why?). $$4^2 - 3^2 =7$$ But compare it to the equation $y^2-(x^{\frac{3}{2}})^2=7$. Then $$x^{\frac{3}{2}} = 3$$ $$x = 3^{\frac{2}{3}}$$ But here $x$ is a decimal! Therefore, at least one of the variables $x$ and $y$ is a decimal in the given equation, thus proving the fact that it has no integral solutions.

Assume there  is  a complex number whose square root is purely imaginary. $$\sqrt{a+ib} = ic$$ Squaring both sides, $$a+ib = (ic)^2 = -c^2$$ The RHS is a real number, whereas the LHS is a complex number. The derived equation is false. Therefore, our original assumption that  there can be a complex number whose square root is purely imaginary  is false.