## Product rule

And here’s a quick recap of a demonstration of the product rule (after Leibniz): $\frac{d}{dx}(u(x)v(x)) = (u(x) + \frac{du}{dx})(v(x) + \frac{dv}{dx}) - u(x)x(x)$ $= u(x)v(x) - u(x)v(x) + u(x)\frac{dv}{dx} + v(x)\frac{du}{dx}$

Which is, of course, the product rule.

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## Integration by parts

Saw this referred to in A Most Incomprehensible Thing: Notes Towards a Very Gentle Introduction to the Mathematics of Relativity and it made me shudder – as I always seemed to struggle with it at ‘A’ level maths – so here, for my own benefit, is a quick proof/explanation of the method.

The general rule is: $\int v(x) \frac{du}{dx} dx = u(x)v(x) - \int u(x) \frac{dv}{dx} dx$

And why is this so?

From the product rule we know: $\frac{d}{dx} (u(x)v(x)) = \frac{du}{dx} v(x) + \frac{dv}{dx}u(x)$

So $\frac{du}{dx} v(x) = \frac{d}{dx}(u(x)v(x)) - \frac{dv}{dx}u(x)$

And, integrate both sides and we have: $\int v(x) \frac{du}{dx} dx = u(x)v(x) - \int u(x) \frac{dv}{dx} dx$

Of course, we still have to apply this sensibly to make our problem easier to integrate!

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## Euler’s formula proof

Reading An Introduction to Laplace Transforms and Fourier Series I reach the point where it is stated, rather axiomatically, that: $e^{ix} = \cos x + i \sin x$.

This is a beautiful formula and has always suggested to me some sort of mystical inner mathematical harmony (yes, I am a materialist, but I cannot help it).

But these days I also want to see the proof, so here is one:

We know that complex numbers can be described in polar co-ordinates: $z = |z| (\cos \theta + i\sin \theta)$

So too $e^{ix} = r(\cos \psi + i \sin \psi)$ where $r$ and $\psi$ depend on $x$.

Now (and applying the product rule) $\frac{d}{dx}e^{ix} = ie^{ix} = \frac{dr}{dx}(\cos \psi + i\sin \psi) + \frac{d \psi}{dx}r(-\sin \psi + i \cos \psi)$

So we equate the real and imaginary sides of both sides of this equality we have: $ie^{ix} = (\cos \psi \frac{dr}{dx} - r \sin \psi \frac{d \psi}{dx}) + i(\sin \psi \frac{dr}{dx} + r \cos \psi \frac{d\psi}{dx})$

Then, recalling $e^{ix} = r(\cos \psi + i \sin \psi)$, we have $ir (\cos \psi + i \sin \psi) = ir \cos \psi - r sin \psi = ( \cos \psi \frac{dr}{dx} - r \sin \psi \frac{d\psi}{dx}) + i( \sin \psi \frac{dr}{dx} + r \cos \psi \frac{d\psi}{dx})$

By inspection we can see that $\frac{dr}{dx} = 0$ and $\frac{d\psi}{dx} = 1$, giving us: $ie^{ix} = - r \sin \psi + ir \cos \psi$ and multiplying both sides by $i$ we have: $-e^{ix} = -r \cos \psi - i r \sin \psi$

Reversing the signs: $e^{ix} = r \cos \psi + i r \sin \psi$

But what of $r$ and $\psi$? Well, we have $\frac{dr}{dx} = 0$ and $\frac{d \psi}{dx} = 1$.

So $r$ is constant with respect to $x$ while $\psi$ varies as $x$.

If we set $x = 0$ then $e^{i \times 0} = 1$ – a wholly real number, so $\sin \psi = 0$ and $\psi = 0$. Thus $r ( \cos 0 ) = e^0 = r = 1$ and we can replace $\psi$ with $x$ throughout.

Hence: $e^{ix} = \cos x + i \sin x$.

## Derivative of any number raised to the power of x

I know this is a piece of elementary calculus but I just worked it out from (more or less) first principles (as I knew what the answer was but did not know why).

Let $y = r^x$ what is $\frac{dy}{dx}$? $r = e. k$ where $k$ is some constant. Hence $y=(ke)^x$ or $k^xe^x$ $\frac{r}{e} = k$ and so $\frac{e^{ln(r)}}{e} = k$ and so $e^{ln(r) - 1} = k$.

And $e^{x(ln(r) - 1)}e^x = r^x$, so $e^{xln(r)} = r^x$

Applying the chain rule $u = xln(r)$ and $y = e^u$ so $\frac{dy}{du}\frac{du}{dx} = e^u \times ln(r)$

Which is $e^{xln(r)}ln(r) = r^xln(r)$

Now I have written it all out I know there are redundant steps in there, but this is how I (re)discovered it…

## Good maths or bad maths?

Not sure, so maybe someone who knows can tell me.

Following on from the last blog, can we show $e = \lim_{n \to 0}(1 +n)^{\frac{1}{n}}$?

Assume a constant, $e$ exists such that $\frac{d}{dx}e^x=e^x$, could $\lim_{n \to 0}(1 + n)^{\frac{1}{n}}$ give us this $e$? $\frac{d}{dx}e^x = \lim_{\delta \to 0} \frac{e^{x+\delta} - e^x}{\delta} =^{\lim_{\delta \to 0}}_{\lim_{n \to 0}}\frac{(1 +n)^{\frac{x+\delta}{n}} - (1 +n)^{\frac{x}{n}}}{\delta}$ $= (1+n)^{\frac{x}{n}}\frac{(1+n)^{\frac{\delta}{n}}-1}{\delta}$

Edit: Professor Rubin (see comments) tells me that, as I feared, what follows is not supportable:

Now, and this is the bit I have most doubts over, as $\delta \to 0$ and $n \to 0$ then $\frac{\delta}{n} \to 1$ and $1 + n - 1 \to 0$ and assuming $\frac {0}{0} = 1$ then we are left with $(1 + n)^{\frac{x}{n}}$.

Hence $\lim_{n \to 0} \frac{d}{dy} (1 + n)^{\frac{x}{n}} = (1 + n)^{\frac{x}{n}}$ and therefore $\lim_{n \to 0} (1+ n)^{\frac{1}{n}} = e$, the constant we are seeking.

But there are a lot of assumptions in there: anybody able to tell me how valid they are?

## Differential calculus reminder

This is just an online note to myself about differential calculus. A level maths again…

Calculating $\frac{d}{dx}2^x$ $2^x = e^u$ where $u=\ln(2^x)$

Using the chain rule: $\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$ $\frac{dy}{du} = e^u$ (as $\frac{d}{dx}e^x = e^x$) $\ln(2^x) = x\ln(2)$ hence $\frac{d}{dx}\ln(2^x) = \ln(2)$.

So $\frac{d}{dx}2^x = e^{\ln(2^x)}\ln(2) = 2^x\ln(2)$

And $e$, Euler’s number, really is magical.