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Help with a formula
01-20-2015, 11:02 PM
Post: #1
Help with a formula
Trying to solve the following for "t" - but hitting bumps.

G(t)

...and WolframAlpha ignores me...

solve for t

Any hints or pointers?
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01-20-2015, 11:10 PM
Post: #2
RE: Help with a formula
(01-20-2015 11:02 PM)Geir Isene Wrote:  Any hints or pointers?

I don't see any algebraic solution (sum of exponentials, logs with added constants... I don't think you can solve this one).
My suggestion is to give up.
(use a numeric method to find roots for 't' if you need to solve for a specific value).

Claudio
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01-21-2015, 02:27 PM
Post: #3
RE: Help with a formula
(01-20-2015 11:02 PM)Geir Isene Wrote:  Trying to solve the following for "t" - but hitting bumps.

G(t)

...and WolframAlpha ignores me...

solve for t

Any hints or pointers?
Very interesting! Try to represent ln(e^###) in some other form. Even AFX-2.0 tells "syntax error", not HP-50G.

HP-50g, HP-48gii, TI-83 plus, FX-2.0, CFX-9850GB plus
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01-21-2015, 08:09 PM
Post: #4
RE: Help with a formula
Hi,

Could you please explain further your problem ?
Solving t of G(t): what does it refer to ?

Dominique
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01-21-2015, 08:13 PM (This post was last modified: 01-21-2015 08:15 PM by Geir Isene.)
Post: #5
RE: Help with a formula
(01-21-2015 08:09 PM)Dominique Wrote:  Hi,

Could you please explain further your problem ?
Solving t of G(t): what does it refer to?

Here: http://isene.me/2015/01/10/hidden-risk-o...urcing-it/ (bottom equation)

I specifically want to know what "t" is when G(t)=0
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01-22-2015, 04:37 AM
Post: #6
RE: Help with a formula
(01-21-2015 08:13 PM)Geir Isene Wrote:  I specifically want to know what "t" is when G(t)=0

\(0 = \frac{T}{9}(\ln(e^\frac{8t}{T}+e^4)-4)(1+D) - Dt\)

\(Dt = \frac{T}{9}\ln(e^{\frac{8t}{T}-4}+1)(1+D)\)

\(t = \frac{T(1+D)}{9D}\ln(1+e^{\frac{8t}{T}-4})\)

This fixed point equation can be solved iteratively. With T=24 and D=0.5 I got t=0.1527. That's probably not what you want. However there's another fixed point at t~18.71 but unfortunately it's not attractive.
In these situations we can use \(f^{-1}(t)=t\) instead:

\(\frac{9D}{T(1+D)}t=\ln(1+e^{\frac{8t}{T}-4})\)

\(e^{\frac{9D}{T(1+D)}t}-1=e^{\frac{8t}{T}-4}\)

\(\ln(e^{\frac{9D}{T(1+D)}t}-1)=\frac{8t}{T}-4\)

\(\frac{T}{8}(\ln(e^{\frac{9D}{T(1+D)}t}-1)+4)=t\)

After a few iterations we get t=18.713393.

You can use this program for the HP-42S:
Code:
00 { 20 Byte Prgm }
01 9
02 *
03 RCL* 01
04 RCL/ 00
05 E^X-1
06 LN
07 4
08 +
09 RCL* 00
10 8
11 /
12 END

You just have to store \(T\) in register 00 and \(\frac{D}{1+D}\) in register 01, enter a guess and hit the [R/S] button a couple of times.

Cheers
Thomas
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01-22-2015, 08:15 AM
Post: #7
RE: Help with a formula
(01-22-2015 04:37 AM)Thomas Klemm Wrote:  
(01-21-2015 08:13 PM)Geir Isene Wrote:  I specifically want to know what "t" is when G(t)=0

\(0 = \frac{T}{9}(\ln(e^\frac{8t}{T}+e^4)-4)(1+D) - Dt\)

\(Dt = \frac{T}{9}\ln(e^{\frac{8t}{T}-4}+1)(1+D)\)

\(t = \frac{T(1+D)}{9D}\ln(1+e^{\frac{8t}{T}-4})\)

This fixed point equation can be solved iteratively. With T=24 and D=0.5 I got t=0.1527. That's probably not what you want. However there's another fixed point at t~18.71 but unfortunately it's not attractive.
In these situations we can use \(f^{-1}(t)=t\) instead:

\(\frac{9D}{T(1+D)}t=\ln(1+e^{\frac{8t}{T}-4})\)

\(e^{\frac{9D}{T(1+D)}t}-1=e^{\frac{8t}{T}-4}\)

\(\ln(e^{\frac{9D}{T(1+D)}t}-1)=\frac{8t}{T}-4\)

\(\frac{T}{8}(\ln(e^{\frac{9D}{T(1+D)}t}-1)+4)=t\)

After a few iterations we get t=18.713393.

You can use this program for the HP-42S:
Code:
00 { 20 Byte Prgm }
01 9
02 *
03 RCL* 01
04 RCL/ 00
05 E^X-1
06 LN
07 4
08 +
09 RCL* 00
10 8
11 /
12 END

You just have to store \(T\) in register 00 and \(\frac{D}{1+D}\) in register 01, enter a guess and hit the [R/S] button a couple of times.

Cheers
Thomas

Neat. Thanks Thomas.
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01-24-2015, 05:00 PM
Post: #8
RE: Help with a formula
Adopted to the HP-41 and with a better user interface. Start the program, enter the values for D and T and the program spits out "t" to the number of decimal places you have set with FIX:

Code:

001*LBL "↑="
002 RCL 01
003 "D="
004 ARCL X
005 PROMPT 
006 STO 01
007 RCL 02
008 "T="
009 ARCL X
010 PROMPT 
011 STO 02
012 10
013*LBL 00
014 STO 03
015 9
016 *
017 RCL 01
018 *
019 RCL 01
020 1
021 +
022 /
023 RCL 02
024 /
025 E^X-1
026 LN
027 4
028 +
029 RCL 02
030 *
031 8
032 /
033 RCL 03
034 X<>Y
035 RND
036 X!=Y?
037 GTO 00
038 "↑="
039 ARCL X
040 AVIEW
041 END
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02-22-2015, 08:31 PM
Post: #9
RE: Help with a formula
An idea for a rough estimation - may not works for all D and T: With little algebra you can get: exp(m×t+b)+1=exp(a×t), where m=8/T, b=-4 and a=9×D÷((1+D)×T)
For a rough estimation of t you can "forget" the +1 on the left side and the approximately solution for t is: t=b÷(a-m) = (-4)÷(1÷8-1÷3) = 19.2
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02-22-2015, 08:57 PM
Post: #10
RE: Help with a formula
Thanks :-)
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