i have an equation for solving but can not copy&paste from word. any help
- Thread starter ibrahim
- Start date
Thank you Sir, it is as follows:
summation[n=0, inf](Pm)^n(1-Pm)nTm+ws
+summation[m=0, inf](Ps)^m(1-Ps)mTs
where:
Pm and Ps are very small.
Tm=5
Ts=1
ws=10
(Ans:between 12 and 14)
summation[n=0, inf](Pm)^n(1-Pm)nTm+ws
+summation[m=0, inf](Ps)^m(1-Ps)mTs
where:
Pm and Ps are very small.
Tm=5
Ts=1
ws=10
(Ans:between 12 and 14)
Bob Brown MSEE
Full Member
- Joined
- Oct 25, 2012
- Messages
- 598
\(\displaystyle \sum _{n=0}^{\infty } P_m{}^n \left( 1 - P_m \right)n T_m+ W_s\)
+ \(\displaystyle \sum _{m=0}^{\infty } P_s{}^m \left( 1 - P_s \right)m T_s\)
where:
\(\displaystyle P_m
\text{ and }
P_s
\text{ are very small.}\)
\(\displaystyle T_m
\text{ = 5}\)
\(\displaystyle T_s
\text{ = 1}\)
\(\displaystyle W_s
\text{ = 10}\)
Is this it?
+ \(\displaystyle \sum _{m=0}^{\infty } P_s{}^m \left( 1 - P_s \right)m T_s\)
where:
\(\displaystyle P_m
\text{ and }
P_s
\text{ are very small.}\)
\(\displaystyle T_m
\text{ = 5}\)
\(\displaystyle T_s
\text{ = 1}\)
\(\displaystyle W_s
\text{ = 10}\)
Is this it?
Yes exactly Mr. Bob Brown
Thank you Mr. Brown.
Yes, this is the equation. If some one can help me to solve it with some details.
Thank you Mr. Brown.
What equation? That is just an expression.
The left hand side is X
using the given parameters, can you or anybody else; give me the answer for X? thank you.
Last edited:
Since you have not quantified small, I shall assume it means at most of absolute value less than 1. For |x|<1,
\(\displaystyle \sum_{n=1}^{\infty}nx^{n-1} = \dfrac{d}{dx} \sum_{n=0}^{\infty}x^n = \dfrac{d}{dx}\left(\dfrac{1}{1-x}\right)\)
With some manipulations, that will allow you to simplify your sums.
edit: Your editied post:
What is X? You mare making no sense whatsoever.
[FONT=MathJax_Math][/FONT]X=[FONT=MathJax_Size1]∑[/FONT][FONT=MathJax_Main]∞[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]W[/FONT][FONT=MathJax_Math]s[/FONT]
+ [FONT=MathJax_Size1]∑[/FONT][FONT=MathJax_Main]∞[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Math]s[/FONT]
where:
[FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main] and [/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Main] are very small.(10^-6)[/FONT]
[FONT=MathJax_Math]T[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main] = 5[/FONT]
[FONT=MathJax_Math]T[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Main] = 1[/FONT]
[FONT=MathJax_Math]W[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Main] = 10[/FONT]
The answer i found for X is approx. 13
but i could not figure out how it was calculated from the above. thank you.
Last edited:
If Pm and Pn are that small, your answer will barely be bigger than 10.
A partial sum calculation of the first sum will show you that nothing before the 6th decimal place will change.
Unless you have written the problem wrong and it should be \(\displaystyle P_s^n(1-P_s)^n\).
Last edited:
Thank you
You almost make me reach what i am looking for, but please can you show me:
1-How did you get the answer is near 10? because my main struggle is how to apply the rules to get the answer.
2- If we increase Pm and Ps to say 10^-4 or 10^-3, will this have any significance? e.g. will the answer will increase from 10 to say 13 or 14?
Thank you for your help and time. really appreciated.
Note: ( the problem was not written wrong).
You almost make me reach what i am looking for, but please can you show me:
1-How did you get the answer is near 10? because my main struggle is how to apply the rules to get the answer.
2- If we increase Pm and Ps to say 10^-4 or 10^-3, will this have any significance? e.g. will the answer will increase from 10 to say 13 or 14?
Thank you for your help and time. really appreciated.
Note: ( the problem was not written wrong).
Its easiest just to plug it into a calculator. But look at the terms: Say \(\displaystyle P_m = 10^{-k}\) for arbitrary integral \(\displaystyle k>0\).
Then the first few terms of the first sum are \(\displaystyle 5(1*10^{-k}(1-10^{-k}) + 2*10^{-2k}(1-10^{-2k}) +3*10^{-3k}(1-10^{-3k}) + ...)\) As n increases, you affect the sum exponentially less.
I know when you apply the rule a/1-r for this part [FONT=MathJax_Size1]∑[/FONT][FONT=MathJax_Main]∞[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Math]n[/FONT]
, (a=1,r=Pm), and as Pm is very small, we get 1.
[FONT=MathJax_Main] But what confuses me is this part: ([/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]P[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Math]T[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]W[/FONT][FONT=MathJax_Math]s[/FONT]
as n is not the power here. so i can not apply the rule above. what is a and what is r?
thank you.
Last edited:
It is no longer a/(1-r) because this is not a geometric series. As I stated before, however, it is the derivative of one. Forget about the Tm, it can be factored out of the sum.
\(\displaystyle \sum r^n(1-r)n = (1-r)r\sum nr^{n-1} = (1-r)r\cdot\dfrac{d}{dr}\sum r^n = (1-r)r\cdot\dfrac{d}{dr}\left(\dfrac{1}{1-r}\right) = \dfrac{(1-r)(r)}{(1-r)^2} = \dfrac{r}{1-r}\).
Thank you all
Thank you Sir.
All are clear to me now.Thank you for your time, help and all the information you gave me.
All the best for you.
Also, thank you Mark FL and Bob Brown MSEE.
Your help is appreciated so much.
Thank you Sir.
All are clear to me now.Thank you for your time, help and all the information you gave me.
All the best for you.
Also, thank you Mark FL and Bob Brown MSEE.
Your help is appreciated so much.