Help with a formula
01-20-2015, 11:02 PM
Post: #1
 Geir Isene Senior Member Posts: 503 Joined: Dec 2013
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?
01-20-2015, 11:10 PM
Post: #2
 Claudio L. Senior Member Posts: 1,175 Joined: Dec 2013
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
01-21-2015, 02:27 PM
Post: #3
 Hlib Member Posts: 120 Joined: Jan 2015
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
01-21-2015, 08:09 PM
Post: #4
 Dominique Junior Member Posts: 4 Joined: Jan 2015
RE: Help with a formula
Hi,

Solving t of G(t): what does it refer to ?

Dominique
01-21-2015, 08:13 PM (This post was last modified: 01-21-2015 08:15 PM by Geir Isene.)
Post: #5
 Geir Isene Senior Member Posts: 503 Joined: Dec 2013
RE: Help with a formula
(01-21-2015 08:09 PM)Dominique Wrote:  Hi,

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
01-22-2015, 04:37 AM
Post: #6
 Thomas Klemm Senior Member Posts: 953 Joined: Dec 2013
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
01-22-2015, 08:15 AM
Post: #7
 Geir Isene Senior Member Posts: 503 Joined: Dec 2013
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.
01-24-2015, 05:00 PM
Post: #8
 Geir Isene Senior Member Posts: 503 Joined: Dec 2013
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
02-22-2015, 08:31 PM
Post: #9
 Csaba Tizedes Member Posts: 229 Joined: May 2014
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
02-22-2015, 08:57 PM
Post: #10
 Geir Isene Senior Member Posts: 503 Joined: Dec 2013
RE: Help with a formula
Thanks :-)
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