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Science Forum Index » Math - Symbolic Forum » Is this a LambertW Question? Need some Python Code.
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| charlesrkiss |
 Posted: Sat Apr 26, 2008 8:24 pm |
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I need to find m; X, and Y is known:
X = (m*Y)^(1/m+1)
Is this a LambertW function question?
Bcause I need some Python code; I like the Python code in Wikipedia
for LambertW, is this a similar problem?
BTW, in my application, Y is actually a function of two parameters,
distance "a" and time, "t": f(a,t) = a^3/2t, where a < 1 is known. So,
I need an X for every Y input. Does that make any sense? :)
The problem is that X is subject to other constraints. When m = 2,
one can say that:
1/Y = 2/X^3, the derivative of a negative inverse square.
but sadly, understanding m, as a function of time, is being difficult.
So I'm having trouble figuring out how to solve for m, or even
conceptualizing how to approach the problem, for that matter... |
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| charlesrkiss |
| Posted: Sun Apr 27, 2008 5:54 am |
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On Apr 27, 7:22 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
Quote: "charlesrkiss" <charles.k...@gmail.com> wrote in message
news:3b73d61d-ad1f-44e2-8667-3e63c550c2c1@27g2000hsf.googlegroups.com
I need to find m; X, and Y is known:
X = (m*Y)^(1/m+1)
Is that (1/m + 1) or [1/(m+1)] ?
Is this a LambertW function question?
Could be.
SORRY!! It is (m*Y)^[1/(m+1)] !!! I'm so sorry! |
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| Greg Neill |
| Posted: Sun Apr 27, 2008 6:22 am |
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"charlesrkiss" <charles.kiss@gmail.com> wrote in message
news:3b73d61d-ad1f-44e2-8667-3e63c550c2c1@27g2000hsf.googlegroups.com
Quote: I need to find m; X, and Y is known:
X = (m*Y)^(1/m+1)
Is that (1/m + 1) or [1/(m+1)] ?
Quote:
Is this a LambertW function question?
Could be. |
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| Chip Eastham |
| Posted: Sun Apr 27, 2008 6:41 am |
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On Apr 27, 11:54 am, charlesrkiss <charles.k...@gmail.com> wrote:
Quote: On Apr 27, 7:22 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"charlesrkiss" <charles.k...@gmail.com> wrote in message
news:3b73d61d-ad1f-44e2-8667-3e63c550c2c1@27g2000hsf.googlegroups.com
I need to find m; X, and Y is known:
X = (m*Y)^(1/m+1)
Is that (1/m + 1) or [1/(m+1)] ?
Is this a LambertW function question?
Could be.
SORRY!! It is (m*Y)^[1/(m+1)] !!! I'm so sorry!
I suspect that you are interested in a numeric solution,
rather than a symbolic one.
Note that if Y > 0 is given, then varying m >= 0 in the
expression:
X = (m*Y)^(1/(m+1))
can only produce a limited range of values for X. At
m=0 we get X = 0. The limit as m --> +oo is X = 1.
For clarity I'm going to treat X as X(m), assuming some
fixed value of Y. Notice that when m = 1/Y we get X=1.
The range of X(m) can be pictured this way. On [0,1/Y]
the function is increasing and provides all values in
[0,1]. On the interval (1/Y,+oo) the function has a
global maximum where:
(1/m) - ln(m) = ln(Y)
A way to see this is by taking natural logarithm of X(m):
ln(X(m)) = (ln(m) + ln(Y))/(m+1)
Since the natural logarithm is monotone increasing,
this preserves all the information about where X(m)
is increasing or decreasing. Now:
d ln(X(m))/dm = ((1/m) - (ln(m) + ln(Y))/(m+1)^2
The critical point where d ln(X(m))/dm = 0 is thus
where (1/m) - ln(m) = ln(Y) as claimed above.
Thus for feasible values of X > 1, there will be two
values of m that produce it (except for the globally
maximum value of X(m), which has unique m).
regards, chip |
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| Greg Neill |
| Posted: Sun Apr 27, 2008 11:44 am |
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"charlesrkiss" <charles.kiss@gmail.com> wrote in message
news:6576332a-8248-4ddb-9b99-b9d38a33426f@b1g2000hsg.googlegroups.com
Quote: On Apr 27, 7:22 am, "Greg Neill" <gneill...@OVEsympatico.ca> wrote:
"charlesrkiss" <charles.k...@gmail.com> wrote in message
news:3b73d61d-ad1f-44e2-8667-3e63c550c2c1@27g2000hsf.googlegroups.com
I need to find m; X, and Y is known:
X = (m*Y)^(1/m+1)
Is that (1/m + 1) or [1/(m+1)] ?
Is this a LambertW function question?
Could be.
SORRY!! It is (m*Y)^[1/(m+1)] !!! I'm so sorry!
Then you're in luck.
m = (-1/ln(x))*W(-ln(X)*X/Y) |
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