Simplifying this equation....harder

Karimazer1 · 2012-01-07 11:42:57

Hello again...

I want to simplify this equation to get 'p1' in terms of 'p0'...

The stage I'm at is

...

So I moved the

across....but now I have two 'p1's' on the RHS....I should be able to get a simple answer from this expressing c1 in terms of c0....

Thank you if anyone can provide some insight..! Karim

bobbym · 2012-01-07 13:19:58

Hi Karimazer1;

Is this what you want to solve?

It is not always possible to solve for a variable. Here you can not solve for p1 in terms of elementary functions. It can be solved using the Lambert function but that is not an elementary function.

Karimazer1 · 2012-01-08 02:09:11

That's exactly what I want to solve, but I want an expression for 'c1' =....

All the other examples do it in a 'quite' easy way...but this question has the exponential instead of just c0 etc. Is there no way for me to take logs to get rid of the 'exp'? I don't want to solve it in that I want to find an answer, I just need an expression which is of the format 'c1' = ........
but because of the exp(-c1) it makes it difficult for me..

If there's no way for this to be done, please let me know It may mean I've done the previous work wrong

Karimazer1 · 2012-01-08 02:10:08

Karimazer1 wrote:

That's exactly what I want to solve, but I want an expression for 'c1' =....
All the other examples do it in a 'quite' easy way...but this question has the exponential instead of just c0 etc. Is there no way for me to take logs to get rid of the 'exp'? I don't want to solve it in that I want to find an answer, I just need an expression which is of the format 'c1' = ........
but because of the exp(-c1) it makes it difficult for me..
If there's no way for this to be done, please let me know It may mean I've done the previous work wrong

And by c0/c1 that I refer to here I mean p0/p1...sorry for typos i moved on to a new question and am trying the old one again now

bobbym · 2012-01-08 02:17:21

Hi;

I am not following you on this question. Do you want to solve in terms of p0/p1?

Karimazer1 · 2012-01-08 02:30:44

bobbym wrote:

Hi;
I am not following you on this question. Do you want to solve in terms of p0/p1?

I do apologize Bob; given the expression you posted in latex which is correct, I think what I need is an expression in terms of p0/p1,

So that's why, given the expression you posted, I could multiply (1+t) to the LHS...and then I'm close (I want p1 in terms of p0)...but then I have two (p1's) so I don't know what to do...

Do you follow me? I am sorry for being so incoherent

Thanks

bobbym · 2012-01-08 02:54:37

Hi;

I think you will have the same problem as before. In order to get p0/p1 all on the same side you would have to be able to get p0 by itself on the left side and then divide by p1. Hopefully now the division by p1 eliminates all the p1's on the right.

But to get p0 by itself on one side is not possible in terms of elementary functions. So I do not think this is possible either.

Karimazer1 · 2012-01-08 03:26:34

Thank you for your honesty...

I must have done it wrong...that part explicitly asks to state p1 in terms of p0... I tried taking logs of everything but that didnt work... I understand what you've written, because I tried to do it by dividing through...but nothing got eliminated.

I must've done something wrong before getting to this stage...although before getting here, it's just differentiating -exp(-p0) - l1p0 which gives 'p0exp(-p0) - l1' .... the same with the second part...and they're basically equal because l1=l2 (l2 which is in the second) and then it equates to the two formulas above....but I'm definitely missing something out Thank you for your efforts Bob

bobbym · 2012-01-08 03:33:19

Hi;

Are you sure about your differentiation of -exp(-p0) - l1p0? Are you differentiating with respect to p0? Is l1 a constant?

Remember I said in terms of an elementary function it could be done. But if you use the product log or Lambert function then

Karimazer1 · 2012-01-08 03:56:45

bobbym wrote:

Hi;
Are you sure about your differentiation of -exp(-p0) - l1p0? Are you differentiating with respect to p0? Is l1 a constant?
Remember I said in terms of an elementary function it could be done. But if you use the product log or Lambert function then

Omg, is that differentiation wrong what I did? I am differentiating with respect to p0 and l1 is the Lagrange multiplier....so I assume it is constant!

That could be where I've done it wrong....I think your 100% right, thinking back, there's no 'x' or anything, it's just a constant, so it is itself! I'll try to do it all from this now....

Thank you so much! I'll try it now with the second equation and the equare the two and see if I can get the expression!!!

bobbym · 2012-01-08 04:02:22

Hi Karimazer1;

Okay, hope it works out better this time!

Karimazer1 · 2012-01-08 04:04:02

So I've just worked through it, this means that I have

exp(-p0)=(exp(-p1)/(1+p))

So the same expression as above...just without those nasty p0's and p1's at the start...

Do you think this can be simplified to get an expression for p1 in terms of p0?

(The method I just tried is taking logs, so I have

so now
so now

So that actually works...I think....I am very excited...but also frustrated that if this is correct, I did a question wrong again because of basic fundamentals (how can I not differentiate exponential functions!)

bobbym · 2012-01-08 04:12:24

Hi;

Assuming that this is correct

exp(-p0)=(exp(-p1)/(1+p))

then the correct solution is

Your method of taking logs is not correct. Please check your first step.

Karimazer1 · 2012-01-08 05:19:19

You really are the best!!!

I think this must be right.... I did the initial differentiating of exp wrong! You are great...

I am just working on part 2, which I have the answer for in the back of the book! And so hopefully this leads perfectly onto it...

Thank you!!!

Karimazer1 · 2012-01-08 05:21:19

Sorry again, on a side question..because of my lack of unbderstanding of Exp/log etc,

if I had, for example, exp(-a)/exp(-b)

and I took logs, would it then be -a/-b or -a--b = -a+b = b-a...

You correcting my bad logs made me think I am going to rewrite all the rules of differentiation and integration so I don't do it wrong again

bobbym · 2012-01-08 05:27:39

The answer is b - a.

Karimazer1 · 2012-01-08 05:55:53

bobbym wrote:

The answer is b - a.

Ok, I'll try and find why so I can stop wasting your time with stupid questions, argh I wish I could just have a list of formulae in front of me, on every topic ever

bobbym · 2012-01-08 06:05:14

Here is the one you need.

Your questions are not stupid. Everyone makes mistakes. I only knew of two material beings that do not mistakes. One is a guy that never does anything and Jarrod, the EBE at A*r**e;;;a 5****1.

Talking about lists, I started with the Abramowitz - Stegun book. Picked it up at a library sale for 50 cents!

Karimazer1 · 2012-01-08 06:22:06

50 cents?!

I just ordered my version from Amazon...(thank you:D)...looking inside the cover it has so many good topics! I paid a little more than 50 cents...darn I need to find some bargains local

bobbym · 2012-01-08 06:28:32

It has everything in it including a recipe for chocolate cake.

Karimazer1 · 2012-01-08 08:16:41

Exactly what I need to destress from all this maths one day I will get a PhD ( don't laugh too much hahahaha!)

bobbym · 2012-01-08 13:05:09

Exactly what I need to destress from all this maths one day I will get a PhD

Are you trying for that? If you do then you will.

Math Is Fun Forum

#1 2012-01-07 11:42:57

Simplifying this equation....harder

#2 2012-01-07 13:19:58

Re: Simplifying this equation....harder

#3 2012-01-08 02:09:11

Re: Simplifying this equation....harder

#4 2012-01-08 02:10:08

Re: Simplifying this equation....harder

#5 2012-01-08 02:17:21

Re: Simplifying this equation....harder

#6 2012-01-08 02:30:44

Re: Simplifying this equation....harder

#7 2012-01-08 02:54:37

Re: Simplifying this equation....harder

#8 2012-01-08 03:26:34

Re: Simplifying this equation....harder

#9 2012-01-08 03:33:19

Re: Simplifying this equation....harder

#10 2012-01-08 03:56:45

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#11 2012-01-08 04:02:22

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#12 2012-01-08 04:04:02

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#13 2012-01-08 04:12:24

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#14 2012-01-08 05:19:19

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#15 2012-01-08 05:21:19

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#16 2012-01-08 05:27:39

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#17 2012-01-08 05:55:53

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#18 2012-01-08 06:05:14

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#19 2012-01-08 06:22:06

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#20 2012-01-08 06:28:32

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#21 2012-01-08 08:16:41

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#22 2012-01-08 13:05:09

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