First Order Error Analysis (Taylor Series)

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SUMMARY

The discussion focuses on performing a first-order error analysis using Taylor series to estimate the velocity of a falling parachutist, represented by the equation v(t) = (gm/c)(1-e^(-(c/m)t)). The parameters used are g = 9.8, m = 9.8, and c = 12.5 ± 1.5. Participants clarify the method of approximation by differentials, which allows for estimating values of functions based on their derivatives at known points. The key takeaway is the application of the formula Δy ≈ f'(x)Δx to approximate the error in the velocity calculation.

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  • Familiarity with differential calculus and derivatives
  • Basic knowledge of physics, particularly the equations of motion
  • Proficiency in mathematical notation and function approximation
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Homework Statement



The equation for the velocity of a falling parachutist can be computed by,

v(t) = \frac{gm}{c}(1-e^{-(\frac{c}{m})t})

Use a first-order error analysis to estimate the error of v at t = 6, if g = 9.8 and m = 9.8, c = 12.5 plus or minus 1.5.

Homework Equations





The Attempt at a Solution



I've never done "first order error analysis" using taylor series so I've looked at the solution to try and do the problem backwards but I can't make sense of what's going on. (See figure attached)

attachment.php?attachmentid=31519&stc=1&d=1295722310.jpg


Can someone clarify to me what they're doing here? I need the ideas behind what's being done.

Thanks again!
 

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He is using approximation by differentials. In one variable, say you have a function f(x) and you know it and its derivative at x0. You can approximate its value at some nearby point x1 by

f(x1) ≈ f(x0) + f'(x0)(x1-x0)

You may see it written as

f(x1) - f(x0) ≈ f'(x0)(x1-x0)

or

Δy ≈ f'(x)Δx

For example, the second line in your image is exactly this last equation with y and x replaced by v and c.
 
Last edited:
LCKurtz said:
He is using approximation by differentials. In one variable, say you have a function f(x) and you know it and its derivative at x0. You can approximate its value at some nearby point x1 by

f(x1) ≈ f(x0) + f'(x0)(x1-x0)

You may see it written as

f(x1) - f(x0) ≈ f'(x0)(x1-x0)

or

Δy = f'(x)Δx

For example, the second line in your image is exactly this last equation with y and x replaced by v and c.

EDIT: I think I got it now.
 

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