# To make maths fun

#### thayes93

Well basically, I've always been thought maths from a technical (operations) point of view. For example, we were taught the quadratic equation but I have no idea how or why it works. However, I am amazed and fascinated by maths at times when I wander off and mess about with numbers and theories, or at things like the omnipresence of e or the trigonometric functions.

I suppose what I'd like to know is whether there are any books you can recommend that I should read that explore maths from an interesting perspective and explain the theory behind things and whatnot, I'm sure you know what I mean!

edit: in case it helps, I am 16 and in school, I have a strong interest in physics and have represented Ireland at Science Olympiads, just if that makes it any easier to gauge my interests and ability.

#### Char. Limit

Gold Member
Hmm... well, I'm 17, and I struggle to understand most of his books, but you could try taking a look at "An Imaginary Tale: The story of $$\sqrt{-1}$$" or "Dr. Euler's Fabulous Formula: Cures Many Mathematical Ills".

But like I said, I can hardly understand them. They don't deal with physics much, I think. Mainly pure math.

#### fourier jr

Mathematics and the Search for Knowledge - Morris Kline

#### Werg22

Mathematics and the Search for Knowledge - Morris Kline
I found Kline's borderline racism in his account of the history of mathematics quite distasteful.

Anyways, to the OP: maybe you should try something like Algebra by Gelfand? I think it might help answer some of your questions.

#### fourier jr

I found Kline's borderline racism in his account of the history of mathematics quite distasteful.
...what borderline racism? I've never heard of that before

#### Ulagatin

Hi,

I'm 17 and will be going into year 12 next year in Australia. To understand the quadratic formula, it really is quite simple: take the general form for $$a\neq0$$ of $$ax^2 + bx + c = 0$$ and divide by the factor a. We then get the following: $$x^2 + \frac{b}{a}x + \frac{c}{a} = 0.$$ This then continues with the "completing the square" method, as follows: $$(x+\frac{b}{2a})^2 = x^2 + \frac{b}{a}x + (\frac{b}{2a})^2.$$

To return to our original function, after dividing by a, we must equate both sides as such: $$x^2 + \frac{b}{a}x + \frac{c}{a} = (x+\frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2.$$ We are well aware that both equations are equal to zero, so then we can use simple mathematics to find that this holds:
$$\text{As } (x+\frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2 = 0\text{, then }(x+\frac{b}{2a})^2 = (\frac{b}{2a})^2 - \frac{c}{a}.$$

We now get to the stage where we begin to recognise the steps as bringing about the quadratic formula: $$x + \frac{b}{2a} = \pm\sqrt{(\frac{b}{2a})^2 - \frac{c}{a}}.$$

Simple steps of manipulation follow: $$x = -\frac{b}{2a} \pm\sqrt{(\frac{b}{2a})^2 - \frac{c}{a}}.$$

And again: $$x = -\frac{b}{2a} \pm\sqrt{\frac{b^2}{4a^2} - \frac{c}{a}}.$$

Rearrange this as follows (slightly more complex): $$x = -\frac{b}{2a} \pm\sqrt{b^2-4ac}.$$

One more step in the process and you're done!

Here's the derived Quadratic Formula: $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} .$$

Now, the $$\sqrt{b^2 - 4ac}$$ element of the equation is called the discriminant, and when it is less than zero, there exist no real solutions to the quadratic - note though that the operative word in this sentence is 'real' with respect to the 'real numbers'. When it is equal to zero, there is one real solution (see the formula - this root is obvious), and when it is greater than zero, there are clearly two real solutions.

As a side note, quadratics become much more interesting when you deal with complex numbers (in the form of x + iy where $$i = \sqrt{-1}$$ and $$x, y \in \Re$$). This occurs for a quadratic when the discriminant is less than zero i.e you take the square root of a negative number, an operation clearly not defined in the reals. If you would like to discuss complex numbers, let me know, you're more than welcome to (and if you're interested, take a look at the polar and exponential forms of a complex number, and most definitely, Euler's identity)!

Just so that my post can demonstrate some sheer mathematical beauty, here is Euler's identity: $${e^{i\pi} + 1 = 0.}$$

Cheers ,
Davin

"To make maths fun"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving