In general, the length of a differentiable function can be computed using the integral:

$$\begin{aligned} \displaystyle \int_a^b \sqrt{1+f'(x)^2} \, dx \\ \end{aligned}$$

To find the length of a semi-circle, we can start with:

$$\begin{aligned} x^2 + y^2 &= r^2 \\ \end{aligned}$$

which leads to:

$$\begin{aligned} & \displaystyle \int_{-r}^r \sqrt{1+\frac{x^2}{r^2-x^2}} \, dx \\ &= \displaystyle r \int_{-\pi/2}^{\pi/2} \cos\theta \sqrt{1+\frac{r^2\sin^2\theta}{r^2-r^2\sin^2\theta}} \, d\theta \\ &= \displaystyle r \int_{-\pi/2}^{\pi/2} \cos\theta \sec \theta \, d\theta \\ &= \pi r \end{aligned}$$

$\Box$

# posted by rot13(Unafba Pune) @ 12:52 PM 0 comments

Starting from:

$$\begin{aligned} y^2 + x^2 &= r^2 \\ y &= \sqrt{r^2 - x^2} \\ \end{aligned}$$

The area of the circle is therefore:

$$\begin{aligned} \int_{-r}^{r} 2 \sqrt{r^2 - x^2} \, dx \\ \end{aligned}$$

Let $\displaystyle \, x = r\sin \theta$, so $dx = r \cdot \cos{\theta} \, d\theta$:

$$\begin{aligned} \int_{-r}^{r} 2 \sqrt{r^2 - x^2} \, dx &= \int_{-\pi/2}^{\pi/2} 2 r \cdot \cos{\theta} \sqrt{r^2 - r^2 \sin^2 \theta} \, d\theta \\ &= 2 r^2 \int_{-\pi/2}^{\pi/2} \cos^2{\theta} \, d\theta \\ &= 2 r^2 \int_{-\pi/2}^{\pi/2} \frac{1 + \cos 2{\theta}}{2} \, d\theta \\ &= r^2 \int_{-\pi/2}^{\pi/2} (1 + \cos 2{\theta}) \, d\theta \\ \end{aligned}$$

Let $\displaystyle \, u = 2 \theta$, so $du = 2 \, d\theta$:

$$\begin{aligned} r^2 \int_{-\pi/2}^{\pi/2} (1 + \cos 2{\theta}) \, d\theta &= \frac{r^2}{2} \int_{-\pi}^{\pi} (1 + \cos{u}) \, du \\ &= \frac{r^2}{2} (u + \sin u) \, \biggr|_{-\pi}^{\pi} \\ &= \pi \, r^2 \\ \end{aligned}$$

$\Box$

# posted by rot13(Unafba Pune) @ 4:47 PM 0 comments

Matrix

Vector

Product

# posted by rot13(Unafba Pune) @ 4:25 PM 0 comments

By the fundamental theorem of calculus,

$$\begin{aligned} f(n) = f(0) + \int_0^n f'(n-x) \, dx \end{aligned}$$

If we integrate by parts once, with $u = f′(n−x)$ and $dv = dx$, we can deduce:

$$\begin{aligned} f(n) = f(0) + n \cdot f'(0) + \int_0^n x \cdot f''(n-x) \, dx \end{aligned}$$

What formula can we prove if we repeat this trick ?

Give it a try and it's not too hard to show that:

$$\begin{aligned} \displaystyle f(n) = f(0) + n \cdot f'(0) + \frac{n^2f''(0)}{2!} + \frac{n^3f'''(0)}{3!} + \cdots + \frac{n^kf^{(k)}(0)}{k!} + \cdots \end{aligned}$$

Or,

$$\begin{aligned} \displaystyle f(n) = \sum_{i=0}^\infty \frac{n^i f^{(i)}(0)}{i!} \end{aligned}$$

This turns out to be known as the Maclaurin Series, or a specific case of Taylor Series when it's centered at zero. Why would this be actually useful ? Well, this means we can approximate the value of any function as long as we can differentiate the function to any arbitrary level of depth, and if the function and all it's derivatives are defined at zero!

For example, as Professor Jim Fowler pointed out, "just the behavior of cosine near $x=0$ is enough to recover the value of cosine at any other point --- it's nuts".

Special thanks to Jim for his generous insight and humorous teaching of Calculus One at Coursera.

Note:

$$\begin{aligned} \dfrac{d}{dx} f(n-x) &= - \, f'(n-x) \\ f(n-x) &= - \int f'(n-x) \, dx \\ \biggl[ \, {f(n-x)}\biggr]_n^0 &= - \int_n^0 f'(n-x) \, dx \\ f(n) - f(0) &= \int_0^n f'(n-x) \, dx \\ \therefore \; f(n)&= f(0) + \int_0^n f'(n-x) \, dx \end{aligned}$$

# posted by rot13(Unafba Pune) @ 2:24 PM 0 comments