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 Math problem - any takers??
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gnkarthik
Junior Member

USA
3 Posts

Posted - 10/05/2007 :  09:42:05  Show Profile
I have an equation I am trying to feed into excel.

E = a*p^c +b*p^d

I want the equation to be transformed such that I can express p in terms of E. Any expertise in the forum??
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tkhunny
Advanced Member

USA
1001 Posts

Posted - 10/05/2007 :  11:02:05  Show Profile
There is expertise, but you are not going to like the answer.

There is no such general solution. For specific values of c and d, there can be solutions, but not for arbitrary c and d.

So, there are a couple of ways to go...

1) What is the nature of the process. What are c and d likely to be? If they are sufficiently consistent and useful, there may be an answer to your question. For example, if c is always twice d (c = 2d), that would be useful.

2) In MS Excel you can use the "Goal Seek" function for one-time solutions or you can write a Visual Basic function to handle larger numbers of cases.

Where does that leave us?
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gnkarthik
Junior Member

USA
3 Posts

Posted - 10/05/2007 :  12:22:51  Show Profile

Thanks for the quick turn around.

I know c and d are constants. So are a and b. I know the values for all a,b,c and d. Since this is a speicific problem, I can take ratios c/d and a/b.
p and E are variables.

Where do I go from there?
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gnkarthik
Junior Member

USA
3 Posts

Posted - 10/05/2007 :  14:34:01  Show Profile
OK.

My question would nowbe... What is a good resource to learn about multinomial equations??
If someone wants to help me......here is my real problem

a =1124/211000
b=0.37495
c=-0.10853
d=-0.63541

E varies from 0.0164637 to 0.00103254. Please look at the table below.
I need to design for a particular value of p for multiple situations. Goalseek in MS excel is an additional step which I preferably want to avoid (because it is a manual operation unless I write a macro).

p E
100 0.0164637
112.202 0.0155073
125.893 0.0146157
141.254 0.0137846
158.489 0.0130095
177.828 0.0122867
199.526 0.0116125
223.872 0.0109835
251.188 0.0103964
281.838 0.00984845
316.227 0.00933684
354.813 0.00885904
398.107 0.00841271
446.683 0.00799564
501.187 0.00760581
562.34 0.0072413
630.956 0.00690038
707.944 0.0065814
794.327 0.00628283
891.249 0.00600327
999.998 0.0057414
1122.02 0.005496
1258.92 0.00526592
1412.53 0.00505012
1584.89 0.00484761
1778.27 0.00465747
1995.26 0.00447886
2238.72 0.00431099
2511.88 0.00415312
2818.37 0.00400458
3162.27 0.00386472
3548.12 0.00373295
3981.06 0.00360873
4466.82 0.00349154
5011.85 0.00338092
5623.39 0.00327641
6309.55 0.00317762
7079.43 0.00308415
7943.25 0.00299565
8912.47 0.00291179
9999.96 0.00283226
11220.1 0.00275678
12589.2 0.00268508
14125.3 0.00261691
15848.9 0.00255205
17782.8 0.00249028
19952.6 0.00243139
22387.2 0.00237521
25118.8 0.00232156
28183.8 0.00227028
31622.8 0.00222122
35481.4 0.00217425
39810.8 0.00212922
44668.4 0.00208602
50118.8 0.00204455
56234.3 0.00200468
63095.9 0.00196634
70794.8 0.00192941
79433.1 0.00189383
89125.4 0.00185951
100000 0.00182638
112202 0.00179437
125893 0.00176342
141255 0.00173346
158490 0.00170444
177829 0.00167631
199528 0.00164902
223874 0.00162253
251191 0.00159679
281841 0.00157176
316231 0.00154741
354817 0.00152371
398111 0.00150061
446688 0.0014781
501193 0.00145614
562348 0.00143471
630965 0.00141378
707954 0.00139333
794338 0.00137333
891262 0.00135378
1.00E+06 0.00133464
1.12E+06 0.00131591
1.26E+06 0.00129756
1.41E+06 0.00127958
1.58E+06 0.00126196
1.78E+06 0.00124468
2.00E+06 0.00122772
2.24E+06 0.00121109
2.51E+06 0.00119475
2.82E+06 0.00117871
3.16E+06 0.00116296
3.55E+06 0.00114748
3.98E+06 0.00113226
4.47E+06 0.0011173
5.01E+06 0.0011026
5.62E+06 0.00108813
6.31E+06 0.0010739
7.08E+06 0.00105989
7.94E+06 0.00104611
8.91E+06 0.00103254
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tkhunny
Advanced Member

USA
1001 Posts

Posted - 10/05/2007 :  15:54:30  Show Profile
I set up a simple example like this:

a 0.0053270
b 0.3749500
c (0.1085300)
d (0.6354100)
E 0.0164637
p 188.2338165

a is in a1
b is in a2
etc.

The value for a is in b1
The value for b is in b2
etc.

The formula for p in b6 is =FindP(b1,b2,b3,b4,b5)

FindP is defined in Visual Basic as:

Function FindP(aVal, bVal, cVal, dVal, EVal)
Static HoldVal, BackVal, TolVal, FofP, FPrimeofP As Double
TolVal = 0.000001
BackVal = EVal / (aVal + bVal) ' Surrogate Initial p-value
For CountVal = 1 To 15
FofP=EVal-(aVal*(BackVal^cVal))-(bVal*(BackVal^dVal))
FPrimeofP=0-(aVal*(cVal*(BackVal^(cVal-1))))-(bVal*(dVal*(BackVal^(dVal-1))))
HoldVal=BackVal-FofP/FPrimeofP
If Abs(HoldVal - BackVal) < TolVal Then
Exit For
Else
BackVal = HoldVal
End If
Next
FindP = HoldVal
End Function

It's just Newton's method and should work most of the time. I found 15 iterations to be sufficient im many cases.

Be warned. I don't get anywhere near your values. I would have to question the values you provided.

I get E = 0.164637 and p = 188.233816

Be additionally warned, I don't know what you are doing, here. I know only that you asked a decent question, although you could have provided better information. If you are doing something dangerous, don't blame me. I can't be held responsible for your coding errors or for your behavior.

Note: This is one of the reasons tutors should offer only hints, rather than complete solutions. Liability. Just something to think about. It's a crazy world we live in.
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Jason
Advanced Member

USA
1235 Posts

Posted - 10/11/2007 :  12:11:09  Show Profile
I get what tkhunny gets

When E = .0164637 then p = 188.233816465834272926
When E = .00103254 then p = 4495384.844865500925


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