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When you this symbol [tex]\rightarrow[/tex] in the middle of an equation, what does it mean?

Thanks

- Thread starter aricho
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- #1

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When you this symbol [tex]\rightarrow[/tex] in the middle of an equation, what does it mean?

Thanks

- #2

honestrosewater

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[tex]\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}[/tex].

Are you seeing it used in this way? Ex. P -> Q?

- #3

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

- #4

honestrosewater

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I think it loosely means "approaches". So

[tex]\lim_{x\rightarrow \infty} f(x) = L[/tex]

would be read "the limit of f(x) as x approaches (positive) infinity is L", I think.

For an explanation, see

http://en.wikipedia.org/wiki/Limit_(mathematics)

http://mathworld.wolfram.com/Limit.html

[tex]\lim_{x\rightarrow \infty} f(x) = L[/tex]

would be read "the limit of f(x) as x approaches (positive) infinity is L", I think.

For an explanation, see

http://en.wikipedia.org/wiki/Limit_(mathematics)

http://mathworld.wolfram.com/Limit.html

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cool, cheers...

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lurflurf

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Say you had something likearicho said:i saw it under "lim" and it was pointing to infinity.....what does it mean there?

[tex]e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n[/tex]

that mean if you chose any positive real number x

[tex]|e-(1+\frac{1}{n})^n|<x[/tex]

for all n>N(x)

That is to say loosely [tex](1+\frac{1}{n})^n[/tex] can be made as close as desired by chosing n sufficiently large.

You can imagine a function above and one below [tex](1+\frac{1}{n})^n[/tex] 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.

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

- #8

lurflurf

Homework Helper

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e is a number that appears alot in math. For example as the base of the natural log.aricho said:wow....i'm totally lost....sorry, could u explain that?.....thanks....

It can be shown that

[tex](1+\frac{1}{n})^n<(1+\frac{1}{n})^{n+1}[/tex]

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

[tex]e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n[/tex]

and when we do so what we mean is

[tex](1+\frac{1}{n})^n[/tex]

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

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got it thanks for that...

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