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!

 

Enhanced by Zemanta

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.