i need a solid explanation of Natural log

[Hades]

When i inquired to my professors, they were at first a little negligent to answer but eventually came to say that when using natural systems, like we typically do in orgo chem, natural log is much like an average that we use.... yet i still dont see the relationship. for example, if i type in LN 5, i get something like 1.609.... if i use LN 6 ill get another number... LN 7 etc.. but if i graph the line, its not linear by any means which explains why its not proportionate either. What purpose does performing LN 5 do?

Also, what purpose does this have when im using LN to determine graphically Ea of a reaction?

[fafalone]

ln is just a logarithm with a base of e (~2.71) instead of 10, e is just a naturally recurring number that represents the area under the curve of y=1/x.

[NSX]

Hades said in post # :

like we typically do in orgo chem, natural log is much like an average that we use.... yet i still dont see the relationship.

 

Like faf said.

 

b/c we can easily do stuff like :lint:xⁿdx, where n :neq: -1.

 

But for n = -1, we can't do :lint:dx/x. i think i remember reading about a letter between euler and another mathematician (?) talking about this discrepancy. then, they created the function you now know as LN.

 

Hades said in post # :

Also, what purpose does this have when im using LN to determine graphically Ea of a reaction?

 

YOu mean E_a? ACtivation Energy?

[Dave]

NSX said in post # :

then, they created the function you now know as LN.

 

I'm not entirely sure whether logarithmic functions were defined first, or whether they were looking for a function that, when the derivative was taken, gave the original function as itself (as in e^x).

 

But anyway, it turns out that for things like radioactive decay, cooling curves, etc that the logarithmic function and e have a large role to play. For example, take your bog standard radioative decay equation:

 

dB/dt = -:lclambda: t

 

where :lclambda: is the decay constant.

 

(not entirely sure about the B - may be Q instead?)

 

to solve this equation, you need to do seperation of variables, which basically boils it down to this equation:

 

:int: 1/(-:lclambda: t) = :int: dB

 

and from the definition of the logarithmic function, the solution is something of the form:

 

Q = Q₀e^(-:lclambda: t)

 

This equation crops up in a lot of places, but i'm not sure whether it crops up in your particular case.

[NSX]

Actually, I was mistaken.

The symbol e was first used by mathematician Leonhard Euler to represent the base of natural logarithms in a letter to another mathematician, Christian Goldbach, in 1731

p. 341

Logarithms were invented by the Scottish mathematician John Napier. Although he did not introduce the natural[i/] logarithmic functino, it is sometimes called the Napierian logarithm.

p. 341

Napier coined the term logarithm, from the two Greek words logos (or ratio) and arithmos (or number) to describe the theory that he spent 20 years developing and that first appeared in the book Mirifici Logarithmorum canonis descriptio (A description of the Marvelous Rule of Logarithms)

p .315

 

Larson, Hostetler, Edwards. Calculus 7^th ed.

 

InTeReStInG eH?

[ski_power]

And there was a series summation for e, right?

What is it? And in normal life is e significant?

[Dave]

e^(x) = 1 + x + x^(2)/2! + x^(3)/3! + ... + x^(n)/n! + ...

 

As I said before, it's important for things like radioactive decay (nuclear reactors, medical uses, etc), thermal properties of materials, capacitor discharge and various other things.

 

It plays a quite important role in the solutions for many mathematical problems, so inevitably it is involved in real life problems as well.

[AntiMagicMan]

The natural number e was I believe first discovered in studying the limit as x tends to infinity of (1 + 1/n)^n and it can be shown that e^x is equal to the limit as x tends to infinity of (1 + x/n)^n.

 

The natural log is the logarithm to the base e.

 

 

It is possible to express any decay formula in terms of any base but e is used for the property that e^x is equal to it's derivative and integral, this naturally makes differentiation and integration involving exponential functions a lot easier.

[bloodhound]

it can be shown that

 

(1+x/n)^n < e^x< (1+x/n)^(n+1)

[Dave]

Had to do this question for my analysis. The basic proof behind it is showing that [math]\lim_{n\to\infty} \left( 1+\frac{1}{n} \right)^{n} \leq e[/math] and then [math] \lim_{n\to\infty} \left (1+\frac{1}{n} \right)^{n} \geq e[/math], so the limit must be e.

[AntiMagicMan]

Theorem: For [math]\[

c \in R

\][/math] [math]\[

(1 + \frac{c}{n})^n \to e^c

\]

[/math]

 

Proof: Let

[math]

\[

x_n = (1 + \frac{c}{n})^n

\]

[/math]

 

[math]\[

\log x_n = n\log (1 + \frac{c}{n})

\]

[/math]

 

Use [math]\[

\log x = \int\limits_1^x {\frac{1}{t}} dt

\]

[/math]

 

So [math]\[

\log (1 + \frac{c}{n}) = \int\limits_1^{1 + \frac{c}{n}} {\frac{{dx}}{x}}

\]

[/math]

We know that the area under the graph between these two points is log(1 + c/n) and we can pin down an upper and lower bound simply by taking the rectangles with width 1 and 1+c/n with height f(1) and f(1+c/n), then you get the inequality

 

[math]\[

\frac{c}{n}\left( {1 + \frac{c}{n}} \right)^{ - 1} < \log (1 + \frac{c}{n}) < \frac{c}{n}

\]

[/math]

 

which is equivalent to

 

[math]

\[

\frac{{cn}}{{c + n}} < \log x_n < c

\]

[/math]

 

now it is true that

 

[math]

\[

\mathop {\lim }\limits_{n \to \infty } \frac{{cn}}{{c + n}} = c

\]

[/math]

 

So [math]\[

\mathop {\lim }\limits_{n \to \infty } \log x_n = c

\]

[/math] by squeezing

 

and all the remains to be done is apply the exponential function and we have [math]\[

\mathop {\lim }\limits_{n \to \infty } x_n = e^c

\]

[/math] QED!

[Dave]

Nice proof.

[AntiMagicMan]

I should have added that the last line is a result of the COLT theorem, but I guess that is pretty much implied. This was an exercise in our Analysis module.

[bloodhound]

nice one.

 

 

if i was asked to "show" that lim n -> inf of (1+x/n)^n = e^x

 

i would have taken f(n) = (1+x/n)^n

 

and g(n)=ln [f(n)]

 

found the limit of g(n) by using L'Hopitals Rule . =x

 

then limit of f(n) is an exponential of something which is tending to x

 

therefore lim of f(n)=e^x

 

I am not sure this constitutes a formal proof.

 

you probably have to prove L'hopitals rule and other stuff as well.

[bloodhound]

also quite impressed with the notation ur created by Tex. I cant be bothered to lean it

[bloodhound]

I quite used to like QED at the end. but i am starting to take to the Empty of Filled in square. which most of my lecturers use.

[AntiMagicMan]

I must admit, I cheated with the TeX, I used MathType to create it.

 

One of my friends can't decide which is best so he uses a tombstone with QED written inside it

[bloodhound]

cool. i got mathtype. havent explorered it properly. i didnt know it can convert o tex. going to have a look now.

 

by the way. which uni do u go to?

[bloodhound]

$$

\eqalign{

& r = {{\sum\limits_{i = 1}^n {x_i y_i } - {{\sum\limits_{i = 1}^n {x_i } \sum\limits_{i = 1}^n {y_i } } \over n}} \over {\sqrt {\left( {\sum\limits_{i = 1}^n {x_i^2 } - {{\left( {\sum\limits_{i = 1}^n {x_i } } \right)} \over n}^2 } \right)\left( {\sum\limits_{i = 1}^n {y_i^2 - {{\left( {\sum\limits_{i = 1}^n {y_i } } \right)} \over n}^2 } } \right)} }} \cr

& \chi ^2 = \sum {{{\left( {O_i - E_i } \right)^2 } \over {E_i }}} \cr

& r_s = 1 - {{6\sum {d^2 } } \over {n\left( {n^2 - 1} \right)}} \cr

& f(x,y) = \int_x^y {{\textstyle{1 \over {k^2 }}}} dk = \left. { - {1 \over k}} \right|_{k = x}^{k = y} = {1 \over x} - {1 \over y} = {{y - x} \over {xy}} \cr

& D(f) = \{ x,y \in R:x,y \ne 0\} \cr}

$$

just testing my new found TeX skills

[bloodhound]

y isnt it working?

 

[math]

 

\frac{{ - b \pm \sqrt {b^2 - 4ac} }}

{{2a}}

 

[/math]

<math>\sqrt{x^2+7}</math>

[bloodhound]

sorry found out why

[math]

$$

\eqalign{

& r = \frac{{\sum\limits_{i = 1}^n {x_i y_i } - \frac{{\sum\limits_{i = 1}^n {x_i } \sum\limits_{i = 1}^n {y_i } }}

{n}}}

{{\sqrt {\left( {\sum\limits_{i = 1}^n {x_i^2 } - \frac{{\left( {\sum\limits_{i = 1}^n {x_i } } \right)}}

{n}^2 } \right)\left( {\sum\limits_{i = 1}^n {y_i^2 - \frac{{\left( {\sum\limits_{i = 1}^n {y_i } } \right)}}

{n}^2 } } \right)} }} \cr

& \chi ^2 = \sum {\frac{{\left( {O_i - E_i } \right)^2 }}

{{E_i }}} \cr

& r_s = 1 - \frac{{6\sum {d^2 } }}

{{n\left( {n^2 - 1} \right)}} \cr

& f(x,y) = \int_x^y {\tfrac{1}

{{k^2 }}} dk = \left. { - \frac{1}

{k}} \right|_{k = x}^{k = y} = \frac{1}

{x} - \frac{1}

{y} = \frac{{y - x}}

{{xy}} \cr

& D(f) = \{ x,y \in R:x,y \ne 0\} \cr}

$$

[/math]

 

what am i doing wrong. doesnt work!!

[Dave]

You need to replace < with [ and > with ], then split it up a bit, or else it's just going to die because it only allows images of a certain size to be generated for security purposes.

[AntiMagicMan]

You need to go into preferences and set the translator to use LaTeX 2.09.

 

I go to Durham University.

[bloodhound]

testing

[math]

\[

\sqrt {a^2 + b^2 }

\]

[/math]

[math]

\[

\begin{array}{l}

\bigcup\limits_{i = 1}^n {X_i } = R \\

X_i = \{ \frac{k}{i}:k = Z\} \\

\end{array}

\]

[/math]

[Dave]

That's better.