Easy question

1. Jul 7, 2005

aricho

sorry, this might seem like a stupid question for all you, but i'm only in year 10.

When you this symbol $$\rightarrow$$ in the middle of an equation, what does it mean?

Thanks

2. Jul 7, 2005

honestrosewater

What type of equation? In a logical context, $$\rightarrow$$ is the implication (or conditional) symbol, representing the truth function
$$\begin{array}{|c|c|c|}\hline P&Q& P\rightarrow Q\\ \hline T&T&T \\ \hline T&F&F \\ \hline F&T&T \\ \hline F&F&T \\ \hline \end{array}$$.
Are you seeing it used in this way? Ex. P -> Q?

3. Jul 7, 2005

aricho

i saw it under "lim" and it was pointing to infinity.....what does it mean there?

4. Jul 7, 2005

honestrosewater

Last edited: Jul 7, 2005
5. Jul 7, 2005

aricho

cool, cheers...

6. Jul 7, 2005

lurflurf

$$e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n$$
that mean if you chose any positive real number x
$$|e-(1+\frac{1}{n})^n|<x$$
for all n>N(x)
That is to say loosely $$(1+\frac{1}{n})^n$$ can be made as close as desired by chosing n sufficiently large.
You can imagine a function above and one below $$(1+\frac{1}{n})^n$$ that represent respectively the largest and smallest values the function takes for all values greater than itself. The upper and lower functions become ever closer, meaning the function becomes confined to ever smaller intervals containing e as n becomes ever larger.

7. Jul 7, 2005

aricho

wow....i'm totally lost....sorry, could u explain that?.....thanks....

8. Jul 7, 2005

lurflurf

e is a number that appears alot in math. For example as the base of the natural log.
It can be shown that
$$(1+\frac{1}{n})^n<(1+\frac{1}{n})^{n+1}$$
also
it can be shown that there difference is less than e/n
and we know e is between the numbers
thus if we want to know e with in 1% we can take n=100 for example.
As we take n larger and larger we can trap it in smaller and smaller ranges.
We can then know it with more and more accuracy.
All of calculus is based on this type of situation. We know how to calculate something with more and more accurate approximations, but we can not calculate it exactly.
Here are some numbers to hopefully make it more clear
for each n we can generate a range for e with the above formula. As n becomes larger the range becomes smaller.
n=1 2<e<4
n=5 4.48832<e<2.985984
n=10 2.593742<e<2.853117
n=100 2.704813<e<2.7318619
n=1000 2.716923<e<2.7196408
n=10^6 2.718280<e<2.7182831
n=10^9 2.718281827<e<2.718281829
The actual value of e to 14 decimal places is 2.71828182846459
Remember that since e is irrational it has an infinite number of digits with no repeating digit patterns.
all this means that we can write
$$e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n$$
and when we do so what we mean is
$$(1+\frac{1}{n})^n$$
can be made as close to e as we want by making n big enough
or conversley that if n is large it will be close to e and as n becomes larger it will become closer.

9. Jul 7, 2005

aricho

got it thanks for that...