(Special relativity) Binomial Approximation

[ak345]

Use the binomial approximation to derive the following:
A) γ=1+.5(β^2)
B)1/γ=1-.5(β^2)
C)1-(1/γ)=.5(^2)

I know the approximation is 1+(.5β^2)+(3/8)β^4+...
A) is self explanatory but not sure how to derive B) and C)

 

[PAllen]

For C, from B, arithmetic.

 

[PAllen]

For B from A, what is expansion of 1/(1+x) ?

 

[ChrisVer]

[itex] \gamma = ( 1 - \beta^{2}) ^ {-\frac{1}{2}} = \frac{1}{ \sqrt{1+x}}, ~~ x= - \beta^{2}[/itex]
Why not use the binomial for the:
[itex] \gamma^{-1}[/itex]?
http://en.wikipedia.org/wiki/Binomial_theorem

 

[HalfLostAndSearching]

I would like to clarify that the binomial approximation method is commonly used in mathematics to estimate the value of a function when its exact value is difficult to calculate. It is based on the binomial theorem which states that for any real number x and any positive integer n, (1+x)^n = 1 + nx + (n(n-1)/2!)x^2 + (n(n-1)(n-2)/3!)x^3 + ....

Now, let us apply this method to derive the given equations:

A) Starting with the given equation γ = 1 + 0.5β^2, we can rewrite it as γ = (1 + β^2/2)^1. Using the binomial theorem, we can approximate this as γ ≈ 1 + (1/2)(β^2/2) = 1 + 0.5β^2. This is the same as the given equation, hence the approximation is valid.

B) Similarly, for the equation 1/γ = 1 - 0.5β^2, we can rewrite it as 1/γ = (1 - β^2/2)^1. Using the binomial theorem, we can approximate this as 1/γ ≈ 1 - (1/2)(β^2/2) = 1 - 0.5β^2. This is the same as the given equation, hence the approximation is valid.

C) Finally, for the equation 1 - (1/γ) = 0.5β^2, we can rewrite it as 1 - (1/γ) = (1 - 1/γ)^1. Using the binomial theorem, we can approximate this as 1 - (1/γ) ≈ 1 - (1/2)(1/γ)^2 = 1 - 0.5(1/γ)^2. This is the same as the given equation, hence the approximation is valid.

In conclusion, the binomial approximation method can be used to estimate the values of complex equations, such as those involving special relativity, when the exact calculation is not feasible. However, it is important to note that this method only provides an approximation and may not give the exact value.