Mathable

Good morning fellow mathematicians! Another week, another post. Just like the title says, we are going to "derive" the Time-Dependent Schrödinger Equation today in a very intuitive way, using well-known concepts that you might encounter in classical mechanics. At first I would like you to remember the Maxwell Equations in free space in their differential form: $$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B} \qquad\qquad \vec{\nabla} \times \vec{B}=\frac{1}{c^{2}}\partial_t \vec{E}\\ \vec{\nabla}\circ\vec{E}=0 \qquad\qquad \vec{\nabla}\circ\vec{B}=0$$ Now, what would happen if we would apply a curl to $\vec{\nabla} \times \vec{E}$ ? On the one hand we could make use of the other Maxwell Equations and by interchanging the differentials to get: $$ \begin{align*} \vec{\nabla}\times(\vec{\nabla} \times \vec{E})&=\vec{\nabla}\times(-\partial_t \vec{B})\\ &=-\partial_t (\vec{\nabla}\times\vec{B})\\ &=-\partial_t (\frac{1}{c^{2}}\partial_t \vec{E})\\ ...

An Integral with a Surprising Result!

Good morning fellow mathematicians! Once again it is time for some sweet, sweet mathemagic. In today's post we are going to deal with an integral, which is going to deliver quite the surprising result. Without further ado, let us take a look at what we are going to deal with today: $$\int_{0}^{1}\frac{\ln{(x)}}{x-1}dx$$ This in itself might not seem to like the most exciting integral in mankind's glorious history, but let us continue and see what we get. At first I would like to introduce a substitution, namely $t=x-1$. But don't forget your differentials! This being said, we are going to arrive at $$\int_{-1}^{0}\frac{\ln{(t+1)}}{t}dt\text{ ,}$$ where $dt=dx$, with our boundaries changing respectively. Next, we would like to derive the Taylor Series of $\ln{(1+x)}$ using some simple observations. Consider the derivative of $\ln{(t+1)}$ with respect to $t$. Naturally we are going to get $\frac{1}{1+t}$. This result though, looks suspiciously like the geometric se...

Integration using a System of Equations

Good morning fellow mathematicians! Today we are going to discuss an integration method, which I find to be pretty helpful and applicable in many situations. After deriving the formula we are interested in, I'm going to provide you with an example, where this very method might come in handy. Let us consider the following integral $$\mathcal{I}=\int\limits_{a}^{b}\frac{h(x)}{f(x)+g(x)}dx\text{ ,}$$ where $f,g,h$ are real or complex valued functions which are differentiable in the interval $[a,b]$. We would like to construct ourselves another integral $\mathcal{J}$, such that $\mathcal{I}+\mathcal{J}=(b-a)$. Let us thus define some real or complex valued function $i$, such that $h(x)+i(x)=f(x)+g(x)$ and $$\mathcal{J}=\int\limits_{a}^{b}\frac{i(x)}{f(x)+g(x)}dx\text{ .}$$ It is now quite obvious, that $\mathcal{I}+\mathcal{J}$ is going to give us $(b-a)$, since: $$\begin{align*} \mathcal{I}+\mathcal{J}&=\int\limits_{a}^{b}\frac{h(x)+i(x)}{f(x)+g(x)}dx\\ &=\int\...

One Symmetric Integral!

Good morning fellow mathematicians! I'm finally back and we are going to dive right in. Today we are going to work with quite the amazing integral, but before we can actually get started, I would like to prove a well-known fact regarding functions. Let us consider some arbitrary function, for example $f:\mathbb{R} \rightarrow \mathbb{R}$. We are now going to go through some simple algebraic manipulations and I hope that you are going to agree with all the steps I'm now doing: \begin{align*} f(x)&=\frac{f(x)}{2}+\frac{f(x)}{2}\\ &=\frac{f(x)}{2}+\frac{f(x)}{2}+\frac{f(-x)}{2}-\frac{f(-x)}{2}\\ &=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2} \end{align*} Obviously, adding $f(-x)$ only holds, if we are not leaving the domain of our function. For reasons that will become apparent in a second, we are going to call the first fraction $e(x)$ and the second one $o(x)$, leaving us with the decomposition $f(x)=e(x)+o(x)$. I am now going to claim, that $e(x)$ is an even f...

Good morning fellow mathematicians :v

Good morning fellow visitor! What you can witness here is my new website. Long story short: My last website has been created using Wordpress. The biggest downside to this service was, that there was no way to implement a proper Latex script without spending quite some money. Thank god there exists blogger! With this provider I can simply implement the language using a simple HTML header, which allows me to create posts way easier and more frequently than before (it was seriously annoying to use Wordpress, it just wasn't any fun). With that out of the way, what is to be expected from this website? I have quite a big folder at home, where I collected everything maths and physics related which came to my mind those past 2 years. My goal is it to translate everything into the latex language, aka. fill this website to the brimm with various math concepts and ideas! I hope this short description of my website's content piqued your interest a bit. Make sure to follow this blog...

A PDF Editor with many useful features!

Before we go ahead, here's a little disclaimer: The creators of PDFelement Pro were so kind to sponsor this post! As a recurring sponsor of the Flammable Maths YouTube channel I am proud to also feature them on my blog. All the impressions I give in this review are based upon my own experiences with the program and are in now way influenced by iSkysoft. Also, I do not earn any commissions from any of the links being featured below! iSkysoft's official Logo $\underline{\Large{\text{PDFelement Pro review}}}$ $\large{\text{An overview}}$ PDFelement Pro is a PDF editing software that is available for both Mac and Windows. The application in itself has been created by iSkysoft and will provide their users with a variety of amazing features that will make handling PDF files a great experience! $\large{\text{The main features}}$ As mentioned above, PDFelement Pro comes with a whole lot of amazing features that will make your life easi...

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