20040809, 01:53  #1 
Dec 2003
Hopefully Near M48
2×3×293 Posts 
Applying the Binomial Theorem More Than Once
I already know how to expand an expression like sqrt(b+c) into an infinite series using the binomial theorem.
But what if I have to apply the process again? This time, I am trying to expand an expression that already has an infinite number of terms. The simplest example of this would be sqrt(a+sqrt(b+c)) Thanks 
20040810, 17:28  #2  
"Bob Silverman"
Nov 2003
North of Boston
2^{3}·937 Posts 
Quote:
sqrt(a + d). What else might you want? Please specify. 

20040811, 13:26  #3  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Applying the Binomial Theorem more than once
Quote:
As I understand it that if we put b+c = d then what is meant is that the value of the expression sqrt(a+sqrt.d) is required. This is a surd (irrational no.) and does not need the Binomial Theorem for its solution. If a straight forward value of sqrt (a + Sqrt (b+c) ) is required assuming that a,b,c,d, are natural nos. then the theory, method, and solution can be provided by me Mally 

20040812, 01:07  #4 
Dec 2003
Hopefully Near M48
11011011110_{2} Posts 
If I use Bob Silverman's suggestion, I will end up with an infinite series where each term is itself an infinite series. Is that supposed to happen?

20040812, 10:25  #5  
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
11580_{10} Posts 
Quote:
Now multiply out the series, dropping those terms which have an exponent larger than those in which you are interested. Paul 

20040812, 16:58  #6  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Applying the binomial theorem more than once
Quote:
Excepting the first termYes After the 18t century mathematicians were forced to break away from the ancient Greek practice of picturing formulae as in their geometry, which explains the comparative stagnation for 2000 years till modern maths arrived on the scene. Please don’t fall into the same error. I give below a worked example and the method used. The propositions I mention can be proved. If required please consult a good text book on elementary Algebra on surds (irrationals) Eg: Find sqrt.(10 + 2 sqrt. 21)= (A) say, Let (A) be = sqrt. x + Sqrt. yProposition (1) Then ( sqrt 10  2sqrt. 21 = sqrt. x – sqrt y ) Proposition(2) By multiplication Sqrt ( 100 – 84 ) = x  y Therefore 4 = x – y (B) By squaring (A) we get 10 + 2 sqrt. 21 = x + y + 2 sqrt ( x* y ) By equating rational partsProposition ( 3 ) We get x + y = 10 From (B) x  y = 4 Therefore x = 7 ; y = 3 Hence (A) =sqrt 7 + sqrt 3 Any difficulty please let me know. Mally 

20040819, 17:29  #7 
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
Applying the Binomial Theorem more than once
Quote:Originally Posted by jinydu If I use Bob Silverman's suggestion, I will end up with an infinite series where each term is itself an infinite series. Is that supposed to happen? unquote If you still insist on the Binomial Theorem derivation try solving this problem Simplify: sqrt (1+ sqrt[1a^2]) + Sqrt (1 sqrt [1a^2]) Hint: both terms are related thus: sqrt(x) +sq rt (y) and sqrt(x)Sqrt(y) Ans: sqrt (2[1+a]) Try it by the method I have given Mally 
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