version 1.1, 1999/09/28 21:26:20
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version 1.3, 2000/08/30 15:02:30
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New functions |
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array_min(array_name) |
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array_max(array_name) |
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will calculate the min or max value from the |
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array given by array_name. All elements in |
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array array_name should be of numerical type, |
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if there is one element assigned a string value, |
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then array_min() and array_max() will give an |
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error, indicating that it could not calculate |
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the min or max value for strings. |
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/LET arr[123]=123 |
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/LET arr[456]=456 |
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/LET arr[1]=1 |
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/LET arr[34]=34 |
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/LET arr[56]=56 |
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/LET arr[78]=78 |
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/LET arr[12]=12 |
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/LET arr[4]=4 |
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/LET arr["124"]=124 |
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/LET arr[1.2]=1.01234567 |
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/LET max = array_max(arr) |
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/LET min = array_min(arr) |
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The results of max will be 456 and min will be 1. |
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array_moments(result_array, input_array) |
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/LET elements = array_moments(result_array, data_array) |
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The input array is data_array and the calculated results will be placed in a newly |
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created array named result_array. This resulting array contains exactly five elements, |
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result_array[0] = number of elements in input array |
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result_array[1] = mean value of elements in input array |
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result_array[2] = variance of elements in input array |
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result_array[3] = skewness of elements in input array |
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result_array[4] = Kurtosis value of elements in input array |
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Suppose all values in array data_array is denoted by $X$, |
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the $i$-th element in the array is denoted by $x_i$, and |
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the number of elements in the array is denoted by $n$. |
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The formula of mean value is given by $\Sigma_{i=0}^{n-1} x_i / n$. |
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Let $\mu$ represents the mean value of the array. |
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The variance is defined as $\frac{\Sigma_{i=0}^{n-1}(x_i - \mu)^2}{n-1}$. |
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The standard deviation of the array can be calculated by taking the square root |
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of variance. Let $\sigma$ denotes the standard deviation of the array. |
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Skewness is calculated from $\frac{\Sigma_{x=0}^{n-1}((x_i - \mu)/\sigma)^3}{n}$ |
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The Kurtosis value is from the formula |
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\frac{\Sigma_{i=0}^{n-1}(x_i - \mu)/\sigma)^4}{n} - 3$ |
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The constant $3$ is used to make the normal distribution appear to have a zero Kurtosis. |
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Formula answer |
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As of CAPA 5.1, a new type of answer can be used by the instructor. |
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A formula as an answer to a problem. That is, the instructor defines a string of |
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formula and ask the students to enter the formula, as long as the entered formula |
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is equivalent to the answer formula, the CAPA system will check and |
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issue correctness or incorrectness based on their equivalence. |
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The underlying mechanism behind this type of answer is that besides the formula string, |
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two additional pieces of informations have to be provided |
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by the instructor, (1) the list of variables used in the answer string and |
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(2) the values of these variables to be used in evaluating formula equivalence. |
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Those two pieces of information are given within a pair of angle brackets appearing |
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as the right hand side of the keywork "eval =". |
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The list of variables is entered as a string or a variable containing a |
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string value. Within that string, each variable is separated by a comma. |
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The symbol '@' then follows. |
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Two forms of variable values can be used. A string with comma separated numerical values |
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or two comma separated numerical values divided by a ':' symbol and followed by |
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a '#' symbol and an integer indicating the number of values to be |
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interpolated within the two values given previously. Both form can be replaced by |
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a variable containing the proper string value. |
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Tolerence can be given to allow the instructor fine tune the results |
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of acceptable values when checking the equivalence of two formulae. |
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/LET f="x^2+y*y^(2)" |
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/LET vlist = "x,y" |
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/LET pts = "1,4:4,5#5" |
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/ANS(f,str=FML ,eval = <"x,y" @ "-1.0,-1.0":"1,1"#4, pts, "0.0,0.0"> ,tol=1e-9) |
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-- |
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Fixed rad != 1/s in capaUnit.c |
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add init_array() function to the user |
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