Difference between revisions of "TF ErrorAna PropOfErr"

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;<math>=1 + \frac{x}{2} - \frac{x^2}{4}</math>
 
;<math>=1 + \frac{x}{2} - \frac{x^2}{4}</math>
  
 +
 +
<math>f(x_1, x_2)=f({x_o}_1, {x_o}_2)+(x_1-{x_o}_1) \frac{\partial f}{\partial x_1}</math>
  
 
<math>f(x_1, x_2)=f({x_o}_1, {x_o}_2)+(x_1-{x_o}_1) \frac{\partial f}{\partial x_1}|_{x_1=x_o_1 ,x_2=x_o_2}}+(x_2-{x_o}_2) \frac{\partial f}{\partial x_2|_{x_1=x_o_1 , x_2=x_o_2}}</math>
 
<math>f(x_1, x_2)=f({x_o}_1, {x_o}_2)+(x_1-{x_o}_1) \frac{\partial f}{\partial x_1}|_{x_1=x_o_1 ,x_2=x_o_2}}+(x_2-{x_o}_2) \frac{\partial f}{\partial x_2|_{x_1=x_o_1 , x_2=x_o_2}}</math>
  
 
<math>f(x_1, x_2) – f({x_o}_1, {x_o}_2) = (x_1-{x_o}_1) \frac{\partial f}{\partial x_1}\Big\lvert_{\substack {x_1={x_o}_1 \\ x_2={x_o}_2}}+(x_2-{x_o}_2) \frac{\partial f}{\partial x_2}\Big\lvert_{\substack {x_1={x_o}_1 x_2={x_o}_2}}</math>
 
<math>f(x_1, x_2) – f({x_o}_1, {x_o}_2) = (x_1-{x_o}_1) \frac{\partial f}{\partial x_1}\Big\lvert_{\substack {x_1={x_o}_1 \\ x_2={x_o}_2}}+(x_2-{x_o}_2) \frac{\partial f}{\partial x_2}\Big\lvert_{\substack {x_1={x_o}_1 x_2={x_o}_2}}</math>

Revision as of 20:18, 9 January 2010

A quantity which is calculated using quantities with known uncertainties will have an uncertainty based upon the uncertainty of the quantities used in the calculation.

To determine the uncertainty in a quantity which is a function of other quantities, you can consider the dependence of these quantities in terms of a tayler expansion

Consider a calculation of a Table's Area

[math]A= L \times W[/math]

The mean that the Area (A) is a function of the Length (L) and the Width (W) of the table.

[math]A = f(L,W)[/math]


The Taylor series expansion of a function f(x) about the point a is given as

[math]f(x) = f(a) + f^{\prime}(x)|_{x=a} \frac{x}{1!} + f^{\prime \prime}(x)|_{x=a} \frac{x^2}{2!} + ...[/math]

[math]= \sum_{n=0}^{infty} f^{(n)}(x)|_{x=a} \frac{x^n}{n!}[/math]


For small values of x (x << 1) we can expand the function about 0 such that

[math]\sqrt{1+x} = \sqrt{1-0} | \frac{1}{2}(1+x)^{-1/2}|_{x=0} \frac{x^1}{1!}+ \frac{1}{2}\frac{-1}{2}(1+x)^{-3/2}|_{x=0} \frac{x^2}{2!}[/math]

[math]=1 + \frac{x}{2} - \frac{x^2}{4}[/math]


[math]f(x_1, x_2)=f({x_o}_1, {x_o}_2)+(x_1-{x_o}_1) \frac{\partial f}{\partial x_1}[/math]

[math]f(x_1, x_2)=f({x_o}_1, {x_o}_2)+(x_1-{x_o}_1) \frac{\partial f}{\partial x_1}|_{x_1=x_o_1 ,x_2=x_o_2}}+(x_2-{x_o}_2) \frac{\partial f}{\partial x_2|_{x_1=x_o_1 , x_2=x_o_2}}[/math]

[math]f(x_1, x_2) – f({x_o}_1, {x_o}_2) = (x_1-{x_o}_1) \frac{\partial f}{\partial x_1}\Big\lvert_{\substack {x_1={x_o}_1 \\ x_2={x_o}_2}}+(x_2-{x_o}_2) \frac{\partial f}{\partial x_2}\Big\lvert_{\substack {x_1={x_o}_1 x_2={x_o}_2}}[/math]