Naturual Logarithms

bobbym · 2010-10-03 11:19:40

Hi;

Your problem is much diferent than what you have given me.
You can not just drop those sigmas.
They mean sum, They are very important!
All the above work is not correct because the form you gave me was incomplete.
Let's start fresh with the exact problem.

Is this what your problem really looks like? Please check carefully.

Neetu · 2010-10-04 04:17:09

ok... sory for any inconvenience i created. i thought we can leave sigma for derivatives.

my problem is like that wht u hv written. only difference is of sigmas. rest of the things are same

BUT, in 2nd term there is costant 'a' as well it is like a*(∑1 y 1) and summation is only for y, 1 is not included in summation.

please try to solve it soon. as i hv deadline tomorrow.

Last edited by Neetu (2010-10-04 04:17:52)

Neetu · 2010-10-04 04:27:28

and there is third term as well. An exact problem is as follows:

F(x, y) = ∑1 ∑2 y log2(1+ x*z/N*B* y) + c*(∑2 x -r) + a *(∑1 y 1)

∑1 represents summation ranges from p=1 to P

∑2 represents summation ranges from q=1 to Q

x, y and z are functions of p and q

'c' is a constant like 'a' and 'r' is another variable.

previosly i hvnt included c*(∑2 x -r) because i thought it wil be zero, when we do derivative of F(x, y) w.r.t y

i have given you exact problem now.

Last edited by Neetu (2010-10-04 04:34:33)

bobbym · 2010-10-04 15:00:45

Hi Neetu;

Why are we taking the derivative of this?

Neetu · 2010-10-04 20:46:23

according to my project. it is required

if we do derivative of F(x, y) w.r.t y and equate it to 0 then it yields as follows:

ln (x*z / y) - x*z / y = __?___ ..............(2)

and i want to derive 2nd equation from it. please solve it.

Last edited by Neetu (2010-10-04 20:46:51)

bobbym · 2010-10-04 22:03:42

Hi Neetu;

Is this finally what we are dealing with?

On top of everything else there is a deadline too!

I hate to sound like a mathematician but differentiation under the summation sign has rules.
Since I have no idea what the interval of convergence is for the first one it may not be justified.
But I will just go ahead blindly and stupidly. Boy does this bring back memories!
Also I am highly skeptical about the double summation having a closed form.
Unless this comes from some physical process where you are sure there is an answer in that form,
I must say it is 1 in a 10000. I must point out I hope you are not building anything with this analysis.

Okay, here goes: I differentiated under the summation operator with respect to y as you wanted.

bobbym wrote:

Also I am highly skeptical about the double summation having a closed form.

Okay I was wrong about the closed form. Here it is!

Frankly, I am amazed. I got this far! Let 's set it to 0.

That is where I am at right now! Now comes the hard part!

Neetu · 2010-10-04 22:33:15

Yes, The Problem Is Same What U Have Wriiten.

Thank U So Much For Trying It

Please Try To Reach Required Form. I Am Only Left With This Part Of The Project.

Cheers!

bobbym · 2010-10-04 22:34:47

Hi Neetu;

I am working on it right now. I have to stop for a minute to attend to some moderator duties.

Neetu · 2010-10-04 22:38:45

ok... NO PROBLEM. I LL BACK TO YOU IN 2 HOURS. I HOPE TILL THEN WE ACHIEVE DESIRED RESULT. I KNO ITZ HARDER. ESPECIALLY WHEN U DONT KNO ALL D DETAILS.

BUT STILL PLEASE TRY TO DO.... CHEERS!

bobbym · 2010-10-04 22:43:20

Hi;

Can I still assume x and z are between 0 and 1 if I need to? Any additional info on y?

Neetu · 2010-10-04 22:47:11

Yes, u can still assume x and z are beween range of 0 to 1.

y can be 0 or 1. this is only info i have abt y

bobbym · 2010-10-04 23:34:07

This is what I can do:

Take a look at forms A) and B) everything on the right is a constant.
They are simpler than the form you requested. y will have to be 1 as 0 is out of the question.
This is as close as I can get.

Neetu · 2010-10-05 01:46:18

Bobbym thank u so much for tryin it. I ll thnk over it

i want to ask

is this some identity ln(x) = x +constant ?

bobbym · 2010-10-05 01:59:37

Hi Neetu;

There is an inequality:

Neetu · 2010-10-05 02:31:29

hi, Apart from inequality is this similar to the form whch i hv written whch is

ln(x) = x +constant

and wht is Hx? is this constant?

DaveRobinsonUK · 2011-06-26 22:47:02

Hello all.

It has been so long since i was last on here. I fell out with maths as I was stuck on the original problem of this post. However since picking it all up after a break I have now got it

Just shows that a break works wonders.

And thank you Bobby, the raising both sides to e^x works a treat.

Got to take the little one to school now.

bobbym · 2011-06-26 23:12:40

Hi Dave;

Have not seen you in a while. How is everything? Glad your math is still moving forward.

DaveRobinsonUK · 2011-06-27 22:56:27

I know, I was bad and didn't do any maths for months. Never been good at time management
and seem to throw myself completely at things, in this case trying to get a a job.

I was always thinking about maths and felt bad that I wasn't doing any, so about a month ago I
picked up the books again and decided to just do 2 hrs a day. So now I can do both look for a job
and do maths.

Winning Combination, and I am much happier now

bobbym · 2011-06-27 23:18:27

Hi Dave;

A day without math is like a day without sunshine.

Maiya · 2011-11-06 20:33:22

i don't happen to get this thing....
how does logX to base e become 2.303loge to base 10................

Last edited by Maiya (2011-11-06 20:34:31)

bobbym · 2011-11-06 20:45:07

Hi Maiya;

logX to base e that is just log(x) or ln(x). So you have:

On the left you have a variable on the right you have a constant. What do you want to do here? Solve for x?

Maiya · 2011-11-07 22:51:19

there was a mistake in the question
how is log(x)to base e=2.303logx to base 10

Bob · 2011-11-08 01:15:00

hi Maiya,

Bob

Maiya · 2011-11-08 02:12:25

whats the value of that exponential constant e and how does that come
if u can please help me out with that...................

bobbym · 2011-11-08 03:08:23

Hi Maiya;

whats the value of that exponential constant e

e or Eulers number is generated by the following series.

Its value is approximately 2.71828182845904523536028747135266249775724709369995957496697

Math Is Fun Forum

#76 2010-10-03 11:19:40

Re: Naturual Logarithms

#77 2010-10-04 04:17:09

Re: Naturual Logarithms

#78 2010-10-04 04:27:28

Re: Naturual Logarithms

#79 2010-10-04 15:00:45

Re: Naturual Logarithms

#80 2010-10-04 20:46:23

Re: Naturual Logarithms

#81 2010-10-04 22:03:42

Re: Naturual Logarithms

#82 2010-10-04 22:33:15

Re: Naturual Logarithms

#83 2010-10-04 22:34:47

Re: Naturual Logarithms

#84 2010-10-04 22:38:45

Re: Naturual Logarithms

#85 2010-10-04 22:43:20

Re: Naturual Logarithms

#86 2010-10-04 22:47:11

Re: Naturual Logarithms

#87 2010-10-04 23:34:07

Re: Naturual Logarithms

#88 2010-10-05 01:46:18

Re: Naturual Logarithms

#89 2010-10-05 01:59:37

Re: Naturual Logarithms

#90 2010-10-05 02:31:29

Re: Naturual Logarithms

#91 2011-06-26 22:47:02

Re: Naturual Logarithms

#92 2011-06-26 23:12:40

Re: Naturual Logarithms

#93 2011-06-27 22:56:27

Re: Naturual Logarithms

#94 2011-06-27 23:18:27

Re: Naturual Logarithms

#95 2011-11-06 20:33:22

Re: Naturual Logarithms

#96 2011-11-06 20:45:07

Re: Naturual Logarithms

#97 2011-11-07 22:51:19

Re: Naturual Logarithms

#98 2011-11-08 01:15:00

Re: Naturual Logarithms

#99 2011-11-08 02:12:25

Re: Naturual Logarithms

#100 2011-11-08 03:08:23

Re: Naturual Logarithms

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