Isnin, 26 Julai 2010

Here is an extra reading for those who wonders where does 'e' comes from.

I found this article and it is very interesting. I got it from here.

Where does e come from and what does it do?

Suppose you put £1 in a bank. The bank pays 4% interest a year, and this is credited to your account at the end of a year. A little thought shows that at the end of five years an amount of money equal to £$(1+0.04)^5$ will sit in the bank (this bank charges no fees).

However, if the interest (still at an annual rate of 4%) was "compounded" every quarter, then the amount at the end of five years would be £$(1+0.04/4)^{4 \times 5}$.

If the bank gave an interest rate of 100% annually, then after one year the bank balance would then be £$(1+1)^1$, and if the interest were compounded quarterly it would be £$(1+1/4)^{4 \times 1} = \pounds 2.43$. If you were even luckier and found a bank that compounded monthly, the 100% annual rate of interest would then gives you £$(1+1/12)^{12 \times 1} = \pounds 2.61$ after one year. Likewise, daily compounding would give you £$(1+1/365)^{365 \times 1} = \pounds 2.72$.

It’s obvious that compounding more frequently results in more money in the bank. So it is natural to ask whether compounding at every instant in time (that is, continuously) leads to an infinite amount in the bank.

To answer this question we need to evaluate

\[  \lim _{n \rightarrow \infty } (1 + 1/n)^ n. \]

This quantity turns out again to be $e$ - the same base value with the property that the gradient of the graph is unity at $x = 0$.

Now $\lim _{n \rightarrow \infty } (1 + 1/n)^ n$ can be expanded very nicely using the trusty old Binomial Theorem. We find that

\[  e = 1+ \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{r!} + \dots . \]

This series is convergent, and evaluating the sum far enough to give no change in the fourth decimal place (this occurs after the seventh term is added) gives an approximation for $e$ of 2.718.

It was that great mathematician Leonhard Euler who discovered the number e and calculated its value to 23 decimal places. It is often called Euler's number and, like pi, is a transcendental number (this means it is not the root of any algebraic equation with integer coefficients). Its properties have led to it as a "natural" choice as a logarithmic base, and indeed e is also known as the natural base or Naperian base (after John Napier).

There is the remarkable property that if the function $e^ x$ (known as the exponential function and also denoted as "$\exp (x) $") is differentiated with respect to $x$, then the result is the same function $e^ x$. The proof of this can be seen in many textbooks on elementary calculus. It’s also true that for $y = Ae^ x$,

\[ \frac{dy}{dx} = \frac{d(Ae^ x)}{dx} = A\frac{d(e^ x)}{dx} = Ae^ x = y. \]

In fact, it is true that given the equation:

\[ \frac{dy}{dx} = y, \]

then a solution for $y$ is $y = Ae^ x$ where $A$ is a constant.

It’s also possible to relate the general exponential function, $y = ba^ x$, to the exponential function. This is because some number $c$ can always be chosen so that $e^ c = a$. Then

\[ y = b\left(e^ c\right)^ x = b e^{cx}. \]

Differentiating with respect to $x$ gives

\[ \frac{dy}{dx} = cb e^{cx} = cy, \]

which tells us that the general exponential relation, $y = ba^ x$, that so often crops up in applications, is in fact a solution of the differential equation

\[ \frac{dy}{dx} = cy. \]

(Actually it happens to be the only type of solution to this important equation.)

It is this equation that emerges naturally when attempting to model various processes.

Sabtu, 24 Julai 2010

Back again to share my knowledge.

Assalammualaikum and hi everyone! Well i'm not sure there is actually someone reading my blog :P I've abandoned this blog for quite a long time!

Anyhoots, will try my best to share you some knowledge in mathematics and additional mathematics. If you would like to to contact me, just email me addmathsd@gmail.com.Surely will reply you as soon as possible, because i do check my email everyday.

Yeay yeayyy! I miss blogging!

Sabtu, 10 Januari 2009

Gradient


Imagine a broom laying on a wall. Refer to diagram. Observe what is the difference before and after, and describe.

You will notice that the height of the broom from the floor to the upper end of the broom decreases and the distance of the lower end of the broom from the wall increases when tilted further downwards.

We say there is a change in the slope of the broom. In mathematics, we can measure slope and help us to measure the steepness of any slope. This is how we calculate slope or in mathematics we call it 'GRADIENT'


Before, the gradient is 5/2 = 2.5 and after that the gradient is 3/6= 0.5. We say 2.5 is steeper than 0.5. Hence by looking at numbers, we can imagine which slope is steeper and which is not.

We can conclude that gradient is the 'rate' of changes in height against changes in the horizontal distance of two points.

Now, let us look at the following diagram.

The first diagram, the broom is vertically upright, hence the gradient is 6/0 =? What do you think 6 divided by 0 is? Try to figure it out using a calculator. Next the broom is laying horizontal on the floor and the gradient is 0/6=? What do you think 0 divided by 6 is? Try to figure out using a calculator. You should get the first answer is math error and the second answer is 0.

We say, anything which is vertically upright, the gradient is undefined. Meaning, take a cliff as an example. Cliff has no slope but it is very steep right? Yeah it is very steep. So when anything is horizontal , there is no slope, or the gradient is 0. A flat road has no slope, therefore there is no gradient.
To make our lesson more fun, try to figure out the standard gradient of a ramp in buildings!
Thats all i want to share for now have fun and we will continue in the next lesson.

Ahad, 9 November 2008

Khamis, 30 Oktober 2008

Mengapa tah inda ku terpikir selama ani huhu


Bismillahhirahmann nirrrahim...

Assalammualaikum..

Ehe, topik ku hari ani, aku kan share pasal integers. Ahahaaa sungguh ku cigu ani ah, baru tah ku teklik cara mudah untuk mengingati macam mana kan mencampur menolak ani.

Inda wah, kan kalau tolak campur ani simple saja kan, tapi kalau integers ah, selalu penuntut paning bila kan campur, bila kan tolak? Sampai kadang2 dua tahun udah ngajar anak ani, masih ia lum get it. Fewhhh* sangal mengexplen.....inda bahh...ikhlass ikhlass ngajar ani eheheh..

Ani bah, cua kamu liat ah...

Kalau minus (-) bertamu dengan minus (-), symbol sign berubah jadi plus(+)

2- (-3)
= 2 + 3
=5


Alasan: -(-) maksudnya - x - = +, '( )' ertinya di 'kali'.


Kalau plus (+) bertamu dengan minus (-), symbol sign berubah jadi minus(-)

2+ (-3)
= 2 - 3
=-1

Alasan: +(-) maksudnya + x - = -, '( )' ertinya di 'kali'.

Untuk mudah mengingati, kalau 'sign' nya sama, ia akan jadi '+', kalau sign nya inda sama ia jadi '-'.


6+(+5)=6 +5 = 11
6- (-5) =6 + 5 = 11
6- (+5)= 6 - 5 = 1
6+ (-5)= 6 - 5 = 1


Ok atu baru pasal sign berjumpa dengan sign yang dalam bracket ah, cara mengingati yang tadi atu bulih jua di apply untuk contoh seterusnya. It really helps you to remember when to add and when to subtract.

Kalau signnya sama, tani campur nombor atu, kemudian jawapannya, tani ambil signnya yang asal.

Contoh 1: -2- 3

( sign nya sama so tani campur)
side working: 2 +3 = 5
( tani ambil jawapan tadi kemudian -2-3
signnya masih sama minus '-' ) = -5

Contoh 2:
+2+ 3
( sign nya sama so tani campur)
side working: 2 +3 = 5
( tani ambil jawapan tadi kemudian +2+3
signnya masih sama plus '+' ) = +5

Kalau signnya inda sama, tani cari difference nombor atu, kemudian jawapannya, tani ambil sign nombor yang basar.

Contoh 1: -2+ 3
( sign nya inda sama so tani cari differencenya)
side working: 3-2 = 1
( tani ambil jawapan tadi kemudian signnya tani pilih '+' -2+3
sebab nombor 3 lebih besar dari 2, so tani pilih sign 3, '+') = +1

Contoh 2:
+6- 8

( sign nya inda sama so tani cari differencenya)
side working: 8-6 = 2
( tani ambil jawapan tadi kemudian signnya tani pilih '-' +6- 8
sebab nombor 8 lebih besar dari 6, so tani pilih sign 8, '-') = -2

Untuk mudah mengingati, kalau 'sign' nya sama, tani campur, kalau sign nya inda sama tani cari differencenya, then tani pilih sign nombor yang basar untuk jawapan tani.

+7+8=+15
-7-8=-15
-7+8=+1
+7-8=-1