Confusing velocity distribution result

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zn5252
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hello,
Please see attached excerpt of a solution to a problem. I do not understand why the result obtained is to second order ? (this is a diffusion equation they say)
Please let me know if you would need more info.
Thanks,
 
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  • Printed book page 260, Problem Book in Relativity and Gravitation, showing equation (1): P(v,n) = <P(v+Δv,n-1)> over a sphere = (1/6)[P(v+Δv e_x,n-1)+P(v+Δv e_y,n-1)+P(v+Δv e_z,n-1)+P(v-Δv e_x,n-1)+P(v-Δv e_y,n-1)+P(v-Δv e_z,n-1)] ≈ P(v,n-1) + (Δv)² ∇²P/6.
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n-1 means at the step which is prior to step n where n is the number of boosts . I have attached the first paragraph for more details.
 
Attachments
  • Printed text of a solution to problem 7.22 on a random-walk velocity distribution: P(v, n) is the probability of velocity v after n boosts, with a differential equation derived from a step n-1 at distance Δv.
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It's because the first order differences will cancel out: f(x+dx) + f(x-dx) ≈ f(x) + f'(x)dx + f''(x)dx2/2 + f(x) - f'(x)dx + f''(x)dx2/2 = 2f(x) + f''(x)dx2