Proving a=d: Conditional Identity

AI Thread Summary
The discussion focuses on proving the identity a = d given the relationships a = 1/(1-b), b = 1/(1-c), and c = 1/(1-d). Participants explore algebraic manipulations to derive the connection between these variables. A hint is provided to simplify the expression for b in terms of d, which aids in the proof. The conversation also touches on the use of LaTeX for writing fractions in mathematical expressions. Ultimately, the participant successfully solves the problem after receiving guidance.
rama
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1. The pratement, all variables and given/known data
If a =1÷(1-b) ,b=1÷(1-c),c=1÷(1-d) prove that a=d


Homework Equations





The Attempt at a Solution


a=1÷(1-b)
a-1÷(1-b)=0
{a(1-b)-1)}÷1-b=0
a-ab-1=0
a-ab=1
similarly
b-bc=1
c-cd=1
could any of you please give a hint, this was on test for 4 marks
 
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rama said:
1. The pratement, all variables and given/known data
If a =1÷(1-b) ,b=1÷(1-c),c=1÷(1-d) prove that a=d


Homework Equations





The Attempt at a Solution


a=1÷(1-b)
a-1÷(1-b)=0
{a(1-b)-1)}÷1-b=0
a-ab-1=0
a-ab=1
similarly
b-bc=1
c-cd=1
could any of you please give a hint, this was on test for 4 marks

You are given
<br /> b = \frac{1}{1 - c} = \frac{1}{1 - \frac{1}{1 - d}}<br />
Simplify the right hand side and then use
<br /> a = \frac{1}{1 - b}.<br />
 
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pasmith thanks for your time I finally got it
actually I hesitated to post such easy problems (for this PF level)
good to see you taking time on helping me
by the way how do you write fractions?
 
Use Latex
Click on the sigma sign on the Advanced editor for references.
Use two #s around the code.
For example \frac{1}{2} and putting two #s around will give ##\frac{1}{2}##
If you want to see what others used,right click on the rendered Latex code,>Show Math as> Tex commands.
Pasmith used this Code:
b = \frac{1}{1 - c} = \frac{1}{1 - \frac{1}{1 - d}}
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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