Math Problem Statement

d<<<<D apply binomial expansion

Solution

To apply the binomial expansion to the given expression 2kq((d2+D2)3/2(D2+4d2)3/2)2kq \left( (d^2 + D^2)^{-3/2} - (D^2 + 4d^2)^{-3/2} \right) where dDd \ll D, we'll use the binomial series expansion for terms of the form (1+x)n(1 + x)^n for x1|x| \ll 1.

The binomial expansion for (1+x)n(1 + x)^n where nn can be any real number and x1|x| \ll 1 is given by: (1+x)n1+nx+n(n1)2!x2+n(n1)(n2)3!x3+(1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots

Let's rewrite each term in a suitable form for the binomial expansion. First, note that d2D2d^2 \ll D^2 and 4d2D24d^2 \ll D^2.

For (d2+D2)3/2(d^2 + D^2)^{-3/2}: (d2+D2)3/2=(D2(1+d2D2))3/2=D3(1+d2D2)3/2(d^2 + D^2)^{-3/2} = (D^2(1 + \frac{d^2}{D^2}))^{-3/2} = D^{-3} \cdot (1 + \frac{d^2}{D^2})^{-3/2} Using the binomial expansion (1+x)n1+nx(1 + x)^n \approx 1 + nx for x1|x| \ll 1: (1+d2D2)3/2132d2D2(1 + \frac{d^2}{D^2})^{-3/2} \approx 1 - \frac{3}{2}\frac{d^2}{D^2} Thus: (d2+D2)3/2D3(132d2D2)=D332d2D5(d^2 + D^2)^{-3/2} \approx D^{-3} \left(1 - \frac{3}{2}\frac{d^2}{D^2}\right) = D^{-3} - \frac{3}{2} \frac{d^2}{D^5}

For (D2+4d2)3/2(D^2 + 4d^2)^{-3/2}: (D2+4d2)3/2=(D2(1+4d2D2))3/2=D3(1+4d2D2)3/2(D^2 + 4d^2)^{-3/2} = (D^2(1 + \frac{4d^2}{D^2}))^{-3/2} = D^{-3} \cdot (1 + \frac{4d^2}{D^2})^{-3/2} Using the binomial expansion (1+x)n1+nx(1 + x)^n \approx 1 + nx for x1|x| \ll 1: (1+4d2D2)3/21324d2D2=16d2D2(1 + \frac{4d^2}{D^2})^{-3/2} \approx 1 - \frac{3}{2}\frac{4d^2}{D^2} = 1 - 6\frac{d^2}{D^2} Thus: (D2+4d2)3/2D3(16d2D2)=D36d2D5(D^2 + 4d^2)^{-3/2} \approx D^{-3} \left(1 - 6\frac{d^2}{D^2}\right) = D^{-3} - 6 \frac{d^2}{D^5}

Now substitute these approximations back into the original expression: 2kq((d2+D2)3/2(D2+4d2)3/2)2kq((D332d2D5)(D36d2D5))2kq \left( (d^2 + D^2)^{-3/2} - (D^2 + 4d^2)^{-3/2} \right) \approx 2kq \left( \left(D^{-3} - \frac{3}{2} \frac{d^2}{D^5}\right) - \left(D^{-3} - 6 \frac{d^2}{D^5}\right) \right) =2kq(D332d2D5D3+6d2D5)= 2kq \left( D^{-3} - \frac{3}{2} \frac{d^2}{D^5} - D^{-3} + 6 \frac{d^2}{D^5} \right) =2kq(6d2D532d2D5)= 2kq \left( 6 \frac{d^2}{D^5} - \frac{3}{2} \frac{d^2}{D^5} \right) =2kq(122d2D532d2D5)= 2kq \left( \frac{12}{2} \frac{d^2}{D^5} - \frac{3}{2} \frac{d^2}{D^5} \right) =2kq(92d2D5)= 2kq \left( \frac{9}{2} \frac{d^2}{D^5} \right) =9kqd2D5= \frac{9kqd^2}{D^5}

So, the binomial expansion of the given expression is: 2kq((d2+D2)3/2(D2+4d2)3/2)9kqd2D52kq \left( (d^2 + D^2)^{-3/2} - (D^2 + 4d^2)^{-3/2} \right) \approx \frac{9kqd^2}{D^5}

Would you like further details or have any questions?

Here are 8 related questions for further exploration:

  1. What is the general form of the binomial expansion?
  2. How does the binomial expansion change for negative exponents?
  3. What is the range of validity for the binomial expansion used in this context?
  4. How does the error term affect the accuracy of the binomial expansion?
  5. Can the binomial expansion be used for any values of dd and DD?
  6. What are other applications of the binomial theorem in physics and engineering?
  7. How would the expansion change if higher-order terms were included?
  8. How can this approximation be visualized graphically?

Tip: When applying the binomial expansion, always ensure that the terms being expanded are small enough to justify truncating the series after a few terms for an accurate approximation.

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Math Problem Analysis

Mathematical Concepts

Binomial Expansion
Approximations

Formulas

Binomial series expansion

Theorems

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Suitable Grade Level

Advanced Mathematics