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**barandaman****Guest**

tinypic.com/r/2u6gyef/5

Hi,

Could someone please help me with this problem?

I have done some of the deriving up to this point, so if we cannot move from the LHS to the RHS on the attached image, I have done something wrong up to this point and then I will take another photo to show the previous steps, but I really think they were right as it was much more simple.

So we have to get from the LHS to the RHS, dividing by a gives the first two terms, but then I really do not understand how the LAST term on the RHS becomes b/a. Could anyone shed some insight? Thank you so much! Image attached.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,674

Hi barandaman;

What have you done so far?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**barandaman****Guest**

I will write it out neatly and take a photo because I have not learnt latex 5 minutes!

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,674

Hi;

See if you can write the problem out larger too.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,258

hi barandaman

here's the LHS using Latex.

I think you need to do two things.

(i) tell us what the question asks

(ii) show how you got this far.

Please use Latex:

You start with square brackets math and end with square brackets /math

The code for the above is:

` [math]a \pi^e = a \pi^{\Psi} + \frac{b(y^{ \Psi} - {ybar})}{a+b^2}[/math]`

There is a code for the line over the y but I forget at the moment. But ybar is clear I think.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,258

hi bobbym,

Didn't see your post as I was taking a long time over mine.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**barandaman****Guest**

tinypic.com/r/rgxlwy/5

It is the problem we looked at yesterday, to solve it we set PI=PI^e, so all that changes is the LHS term is PI^e.

Then I expand the RHS as seen, and get the result as shown initially in the first image. I have no idea how to convert that into the solution, which is:

Pi^e = Pi^* + (b/a)(y^* - ybar)

I hope this is clearer!

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,674

Hi bob bundy;

That is okay.

Use

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,258

hi barandaman

That last line is not quite correct.

Your pi terms have lost their denominator.

Put that in and you can cancel it throughout. The expression is then much simpler.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**barandaman****Guest**

bob bundy wrote:

hi barandaman

That last line is not quite correct.

Your pi terms have lost their denominator.

Put that in and you can cancel it throughout. The expression is then much simpler.

Bob

I really am having trouble with the latex...

The pi terms are different, one is pi^e one is pi^*

**barandaman****Guest**

GOT IT! tghank you so much now i know what you mean, ahhhh thank you