Applying the Binomial Theorem More Than Once

[jinydu]

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

[R.D. Silverman]

Originally Posted by jinydu

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

I am not sure that I understand. just let d = sqrt(b+c), and expand
sqrt(a + d). What else might you want? Please specify.

[mfgoode]

Originally Posted by Bob Silverman

I am not sure that I understand. just let d = sqrt(b+c), and expand
sqrt(a + d). What else might you want? Please specify.

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

[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?

[xilman]

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?

Yes, it is supposed to happen.

Now multiply out the series, dropping those terms which have an exponent larger than those in which you are interested.


Paul

[mfgoode]

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?

Excepting the first term-Yes
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. y---------------Proposition (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 parts----------------------------------------Proposition ( 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

[mfgoode]

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[1-a^2]) + Sqrt (1- sqrt [1-a^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