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Legendre transform

2008-02-15T10:01:10Z

163.117.132.133:

The '''Legendre transform''' (Adrien-Marie Legendre)
is used to perform a change ''change of variables''
(see, for example, Ref. 1, Chapter 4 section 11 Eq. 11.20 - 11.25):

If one has the function <math>f(x,y);</math> one can write

:<math>df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy</math>

Let <math>p= \partial f/ \partial x</math>, and <math>q= \partial f/ \partial y</math>, thus

:<math>df = p~dx + q~dy</math>

If one subtracts <math>d(qy)</math> from <math>df</math>, one has

:<math>df- d(qy) = p~dx + q~dy -q~dy - y~dq</math>
or
:<math>d(f-qy)=p~dx - y~dq </math>

Defining the function <math>g=f-qy</math>
then

:<math>dg = p~dx - y~dq</math>

The partial derivatives of <math>g</math> are

:<math>\frac{\partial g}{\partial x}= p, ~~~ \frac{\partial g}{\partial q}= -y</math>.

==Example==
==See also==
*[[Thermodynamic relations]]
==References==
#Mary L. Boas "Mathematical methods in the Physical Sciences" John Wiley & Sons, Second Edition.
#[http://www.iupac.org/publications/pac/2001/7308/7308x1349.html Robert A. Alberty "Use of Legendre transforms in chemical thermodynamics", Pure and Applied Chemistry '''73''' pp. 1349-1380 (2001)]
[[category: mathematics]]

Stirling's approximation

2007-03-28T13:50:27Z

163.117.132.133:

James Stirling (1692-1770, Scotland)

:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k .</math>

Because of [http://en.wikipedia.org/wiki/Euler-Maclaurin_formula Euler-MacLaurin formula]

:<math>\sum_{k=1}^N \ln k=\int_1^N \ln x\,dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R ,</math>

where ''B''1 = −1/2, ''B''2 = 1/6, ''B''3 = 0, ''B''4 = −1/30, ''B''5 = 0, ''B''6 = 1/42, ''B''7 = 0, ''B''8 = −1/30, ... are the [http://en.wikipedia.org/wiki/Bernoulli_numbers Bernoulli numbers], and ''R'' is an error term which is normally small for suitable values of ''p''.

Then, for large ''N'',

:<math>\ln N! \sim \int_1^N \ln x\,dx \sim N \ln N -N .</math>
[[Category: Mathematics]]

Stirling's approximation

2007-03-28T13:49:42Z

163.117.132.133:

James Stirling (1692-1770, Scotland)

:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k</math>

Because of [http://en.wikipedia.org/wiki/Euler-Maclaurin_formula Euler-MacLaurin formula]

:<math>\sum_{k=1}^N \ln k=\int_1^N \ln x\,dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R</math>

where ''B''1 = −1/2, ''B''2 = 1/6, ''B''3 = 0, ''B''4 = −1/30, ''B''5 = 0, ''B''6 = 1/42, ''B''7 = 0, ''B''8 = −1/30, ... are the [http://en.wikipedia.org/wiki/Bernoulli_numbers Bernoulli numbers], and ''R'' is an error term which is normally small for suitable values of ''p''.

Then, for large ''N'',

:<math>\ln N! \sim \int_1^N \ln x\,dx \sim N \ln N -N</math>
[[Category: Mathematics]]

Stirling's approximation

2007-03-28T13:49:27Z

163.117.132.133:

James Stirling (1692-1770, Scotland)

:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k</math>

Because of [http://en.wikipedia.org/wiki/Euler-Maclaurin_formula Euler-MacLaurin formula]

:<math>\sum_{k=1}^N \ln k=\int_1^N \ln x\,dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R</math>

where ''B''1 = −1/2, ''B''2 = 1/6, ''B''3 = 0, ''B''4 = −1/30, ''B''5 = 0, ''B''6 = 1/42, ''B''7 = 0, ''B''8 = −1/30, ... are the [http://en.wikipedia.org/wiki/Bernoulli_numbers Bernoulli numbers], and ''R'' is an error term which is normally small for suitable values of ''p''.

Then, for large ''N'',

:<math>\ln N! \sim \int_1^N \ln x dx \sim N \ln N -N</math>
[[Category: Mathematics]]

Stirling's approximation

2007-03-28T13:48:56Z

163.117.132.133:

James Stirling (1692-1770, Scotland)

:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k</math>

Because of [http://en.wikipedia.org/wiki/Euler-Maclaurin_formula Euler-MacLaurin formula]

:<math>\sum_{k=1}^N \ln k=\int_1^N \ln x\,dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R</math>

where ''B''1 = −1/2, ''B''2 = 1/6, ''B''3 = 0, ''B''4 = −1/30, ''B''5 = 0, ''B''6 = 1/42, ''B''7 = 0, ''B''8 = −1/30, ... are the [http://en.wikipedia.org/wiki/Bernoulli_numbers Bernoulli numbers], and ''R'' is an error term which is normally small for suitable values of ''p''.

Then, for large ''N'',

:<math>\ln N! \approx \int_1^N \ln x dx = N \ln N -N</math>
[[Category: Mathematics]]

Stirling's approximation

2007-03-28T13:47:10Z

163.117.132.133:

James Stirling (1692-1770, Scotland)

:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k</math>

Because of [http://en.wikipedia.org/wiki/Euler-Maclaurin_formula Euler-MacLaurin formula]

:<math>\sum_{k=1}^N \ln k=\int_1^N \ln x\,dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R</math>

where ''B''1 = −1/2, ''B''2 = 1/6, ''B''3 = 0, ''B''4 = −1/30, ''B''5 = 0, ''B''6 = 1/42, ''B''7 = 0, ''B''8 = −1/30, ... are the [http://en.wikipedia.org/wiki/Bernoulli_numbers Bernoulli numbers], and ''R'' is an error term which is normally small for suitable values of ''p''.

Then

:<math>\ln N! \approx \int_1^N \ln x dx = N \ln N -N +1</math>

Thus, for large ''N''

:<math>\ln N! \approx N \ln N -N</math>
[[Category: Mathematics]]

Stirling's approximation

2007-03-28T13:43:33Z

163.117.132.133:

James Stirling (1692-1770, Scotland)

:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k</math>

Because of [http://en.wikipedia.org/wiki/Euler-Maclaurin_formula Euler-MacLaurin formula]

:<math>\sum_{k=1}^N \ln k=\int_1^N \ln x\,dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R</math>

where ''B''1 = −1/2, ''B''2 = 1/6, ''B''3 = 0, ''B''4 = −1/30, ''B''5 = 0, ''B''6 = 1/42, ''B''7 = 0, ''B''8 = −1/30, ... are the [http://en.wikipedia.org/wiki/Bernoulli_numbers Bernoulli numbers], and ''R'' is an error term which is normally small for suitable values of ''p''.

:<math>~\approx \int_1^N \ln x dx</math>

:<math>~= \left[ x \ln x - x \right]_1^N</math>

:<math>~= N \ln N -N +1</math>

Thus, for large ''N''

:<math>\ln N! \approx N \ln N -N</math>
[[Category: Mathematics]]

Stirling's approximation

2007-03-28T13:37:05Z

163.117.132.133:

James Stirling (1692-1770, Scotland)

:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k</math>

Because of [Euler-MacLaurin formula]

:<math>\sum_{k=1}^N \ln k=\int_1^N \ln x\,dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R</math>

where ''B''1 = −1/2, ''B''2 = 1/6, ''B''3 = 0, ''B''4 = −1/30, ''B''5 = 0, ''B''6 = 1/42, ''B''7 = 0, ''B''8 = −1/30, ... are the [[Bernoulli numbers]], and ''R'' is an error term which is normally small for suitable values of ''p''.

:<math>~\approx \int_1^N \ln x dx</math>

:<math>~= \left[ x \ln x - x \right]_1^N</math>

:<math>~= N \ln N -N +1</math>

Thus, for large ''N''

:<math>\ln N! \approx N \ln N -N</math>
[[Category: Mathematics]]

Stirling's approximation

2007-03-28T13:36:44Z

163.117.132.133:

James Stirling (1692-1770, Scotland)

:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k</math>

Because of [Euler-MacLaurin formula]

:<math>\sum_{k=1}^N \ln k=\int_1^N \ln x dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R</math>

where ''B''1 = −1/2, ''B''2 = 1/6, ''B''3 = 0, ''B''4 = −1/30, ''B''5 = 0, ''B''6 = 1/42, ''B''7 = 0, ''B''8 = −1/30, ... are the [[Bernoulli numbers]], and ''R'' is an error term which is normally small for suitable values of ''p''.

:<math>~\approx \int_1^N \ln x dx</math>

:<math>~= \left[ x \ln x - x \right]_1^N</math>

:<math>~= N \ln N -N +1</math>

Thus, for large ''N''

:<math>\ln N! \approx N \ln N -N</math>
[[Category: Mathematics]]

Stirling's approximation

2007-03-28T13:28:31Z

163.117.132.133:

James Stirling (1692-1770, Scotland)

:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k</math>

:<math>~\approx \int_1^N \ln x dx</math>

:<math>~= \left[ x \ln x - x \right]_1^N</math>

:<math>~= N \ln N -N +1</math>

Thus, for large ''N''

:<math>\ln N! \approx N \ln N -N</math>
[[Category: Mathematics]]