# quantile

Quantiles of a data set

## Syntax

``Y = quantile(X,p)``
``Y = quantile(X,N)``
``Y = quantile(___,'all')``
``Y = quantile(___,dim)``
``Y = quantile(___,vecdim)``
``Y = quantile(___,'Method',method)``

## Description

example

````Y = quantile(X,p)` returns quantiles of the elements in data vector or array `X` for the cumulative probability or probabilities `p` in the interval [0,1]. If `X` is a vector, then `Y` is a scalar or a vector having the same length as `p`.If `X` is a matrix, then `Y` is a row vector or a matrix where the number of rows of `Y` is equal to the length of `p`. For multidimensional arrays, `quantile` operates along the first nonsingleton dimension of `X`. ```

example

````Y = quantile(X,N)` returns quantiles for `N` evenly spaced cumulative probabilities (1/(`N` + 1), 2/(`N` + 1), ..., `N`/(`N` + 1)) for integer `N`>1. If `X` is a vector, then `Y` is a scalar or a vector with length `N`. If `X` is a matrix, then `Y` is a matrix where the number of rows of `Y` is equal to `N`.For multidimensional arrays, `quantile` operates along the first nonsingleton dimension of `X`. ```

example

````Y = quantile(___,'all')` returns quantiles of all the elements of `X` for either of the first two syntaxes.```

example

````Y = quantile(___,dim)` returns quantiles along the operating dimension `dim` for either of the first two syntaxes.```

example

````Y = quantile(___,vecdim)` returns quantiles over the dimensions specified in the vector `vecdim` for either of the first two syntaxes. For example, if `X` is a matrix, then `quantile(X,0.5,[1 2])` returns the 0.5 quantile of all the elements of `X` because every element of a matrix is contained in the array slice defined by dimensions 1 and 2.```

example

````Y = quantile(___,'Method',method)` returns either exact or approximate quantiles based on the value of `method`, using any of the input argument combinations in the previous syntaxes.```

## Examples

collapse all

Calculate the quantiles of a data set for specified probabilities.

Generate a data set of size 10.

```rng('default'); % for reproducibility x = normrnd(0,1,1,10)```
```x = 1×10 0.5377 1.8339 -2.2588 0.8622 0.3188 -1.3077 -0.4336 0.3426 3.5784 2.7694 ```

Calculate the 0.3 quantile.

`y = quantile(x,0.30)`
```y = -0.0574 ```

Calculate the quantiles for the cumulative probabilities 0.025, 0.25, 0.5, 0.75, and 0.975.

`y = quantile(x,[0.025 0.25 0.50 0.75 0.975])`
```y = 1×5 -2.2588 -0.4336 0.4401 1.8339 3.5784 ```

Calculate the quantiles of a data set for a given number of quantiles.

Generate a data set of size 10.

```rng('default'); % for reproducibility x = normrnd(0,1,1,10)```
```x = 1×10 0.5377 1.8339 -2.2588 0.8622 0.3188 -1.3077 -0.4336 0.3426 3.5784 2.7694 ```

Calculate four evenly spaced quantiles.

`y = quantile(x,4)`
```y = 1×4 -0.8706 0.3307 0.6999 2.3017 ```

Using `y = quantile(x,[0.2,0.4,0.6,0.8])` is another way to return the four evenly spaced quantiles.

Calculate the quantiles along the columns and rows of a data matrix for specified probabilities.

Generate a 4-by-6 data matrix.

```rng default % For reproducibility X = normrnd(0,1,4,6)```
```X = 4×6 0.5377 0.3188 3.5784 0.7254 -0.1241 0.6715 1.8339 -1.3077 2.7694 -0.0631 1.4897 -1.2075 -2.2588 -0.4336 -1.3499 0.7147 1.4090 0.7172 0.8622 0.3426 3.0349 -0.2050 1.4172 1.6302 ```

Calculate the 0.3 quantile for each column of `X` (`dim` = 1).

`y = quantile(X,0.3,1)`
```y = 1×6 -0.3013 -0.6958 1.5336 -0.1056 0.9491 0.1078 ```

`quantile` returns a row vector `y` when calculating one quantile for each column of a matrix. For example, `-0.3013` is the 0.3 quantile of the first column of `X` with elements (0.5377, 1.8339, -2.2588, 0.8622). Because the default value of `dim` is 1, you can return the same result with `y = quantile(X,0.3)`.

Calculate the 0.3 quantile for each row of `X` (`dim` = 2).

`y = quantile(X,0.3,2)`
```y = 4×1 0.3844 -0.8642 -1.0750 0.4985 ```

`quantile` returns a column vector `y` when calculating one quantile for each row of a matrix. For example `0.3844` is the 0.3 quantile of the first row of `X` with elements (0.5377, 0.3188, 3.5784, 0.7254, -0.1241, 0.6715).

Calculate the $N$ evenly spaced quantiles along the columns and rows of a data matrix.

Generate a 6-by-10 data matrix.

```rng('default'); % for reproducibility X = unidrnd(10,6,7)```
```X = 6×7 9 3 10 8 7 8 7 10 6 5 10 8 1 4 2 10 9 7 8 3 10 10 10 2 1 4 1 1 7 2 5 9 7 1 5 1 10 10 10 2 9 4 ```

Calculate three evenly spaced quantiles for each column of `X` (`dim` = 1).

`y = quantile(X,3,1)`
```y = 3×7 2.0000 3.0000 5.0000 7.0000 4.0000 1.0000 4.0000 8.0000 8.0000 7.0000 8.5000 7.0000 2.0000 4.5000 10.0000 10.0000 10.0000 10.0000 8.0000 8.0000 7.0000 ```

Each column of matrix `y` corresponds to the three evenly spaced quantiles of each column of matrix `X`. For example, the first column of `y` with elements (2, 8, 10) has the quantiles for the first column of `X` with elements (9, 10, 2, 10, 7, 1). `y = quantile(X,3)` returns the same answer because the default value of `dim` is 1.

Calculate three evenly spaced quantiles for each row of `X` (`dim` = 2).

`y = quantile(X,3,2)`
```y = 6×3 7.0000 8.0000 8.7500 4.2500 6.0000 9.5000 4.0000 8.0000 9.7500 1.0000 2.0000 8.5000 2.7500 5.0000 7.0000 2.5000 9.0000 10.0000 ```

Each row of matrix `y` corresponds to the three evenly spaced quantiles of each row of matrix `X`. For example, the first row of `y` with elements (7, 8, 8.75) has the quantiles for the first row of `X` with elements (9, 3, 10, 8, 7, 8, 7).

Calculate the quantiles of a multidimensional array for specified probabilities by using the `'all'` and `vecdim` input arguments.

Create a 3-by-5-by-2 array `X`. Specify the vector of probabilities `p`.

`X = reshape(1:30,[3 5 2])`
```X = X(:,:,1) = 1 4 7 10 13 2 5 8 11 14 3 6 9 12 15 X(:,:,2) = 16 19 22 25 28 17 20 23 26 29 18 21 24 27 30 ```
`p = [0.25 0.75];`

Calculate the 0.25 and 0.75 quantiles of all the elements in `X`.

`Yall = quantile(X,p,'all')`
```Yall = 2×1 8 23 ```

`Yall(1)` is the 0.25 quantile of `X`, and `Yall(2)` is the 0.75 quantile of `X`.

Calculate the 0.25 and 0.75 quantiles for each page of `X` by specifying dimensions 1 and 2 as the operating dimensions.

`Ypage = quantile(X,p,[1 2])`
```Ypage = Ypage(:,:,1) = 4.2500 11.7500 Ypage(:,:,2) = 19.2500 26.7500 ```

For example, `Ypage(1,1,1)` is the 0.25 quantile of the first page of `X`, and `Ypage(2,1,1)` is the 0.75 quantile of the first page of `X`.

Calculate the 0.25 and 0.75 quantiles of the elements in each `X(i,:,:)` slice by specifying dimensions 2 and 3 as the operating dimensions.

`Yrow = quantile(X,p,[2 3])`
```Yrow = 3×2 7 22 8 23 9 24 ```

For example, `Yrow(3,1)` is the 0.25 quantile of the elements in `X(3,:,:)`, and `Yrow(3,2)` is the 0.75 quantile of the elements in `X(3,:,:)`.

Find median and quartiles of a vector, `x`, with even number of elements.

Enter the data.

`x = [2 5 6 10 11 13]`
```x = 1×6 2 5 6 10 11 13 ```

Calculate the median of `x`.

`y = quantile(x,0.50)`
```y = 8 ```

Calculate the quartiles of `x`.

`y = quantile(x,[0.25, 0.5, 0.75])`
```y = 1×3 5 8 11 ```

Using `y = quantile(x,3)` is another way to compute the quartiles of `x`.

These results might be different than the textbook definitions because `quantile` uses Linear Interpolation to find the median and quartiles.

Find median and quartiles of a vector, `x`, with odd number of elements.

Enter the data.

`x = [2 4 6 8 10 12 14]`
```x = 1×7 2 4 6 8 10 12 14 ```

Find the median of `x`.

`y = quantile(x,0.50)`
```y = 8 ```

Find the quartiles of `x`.

`y = quantile(x,[0.25, 0.5, 0.75])`
```y = 1×3 4.5000 8.0000 11.5000 ```

Using `y = quantile(x,3)` is another way to compute the quartiles of `x`.

These results might be different than the textbook definitions because `quantile` uses Linear Interpolation to find the median and quartiles.

Calculate exact and approximate quantiles of a tall column vector for a given probability.

When you perform calculations on tall arrays, MATLAB® uses either a parallel pool (default if you have Parallel Computing Toolbox™) or the local MATLAB session. To run the example using the local MATLAB session when you have Parallel Computing Toolbox, change the global execution environment by using the `mapreducer` function.

`mapreducer(0)`

Create a datastore for the `airlinesmall` data set. Treat `'NA'` values as missing data so that `datastore` replaces them with `NaN` values. Specify to work with the `ArrTime` variable.

```ds = datastore('airlinesmall.csv','TreatAsMissing','NA',... 'SelectedVariableNames','ArrTime');```

Create a tall table on top of the datastore, and extract the data from the tall table into a tall vector.

`t = tall(ds) % Tall table`
```t = Mx1 tall table ArrTime _______ 735 1124 2218 1431 746 1547 1052 1134 : : ```
`x = t{:,:} % Tall vector`
```x = Mx1 tall double column vector 735 1124 2218 1431 746 1547 1052 1134 : : ```

Calculate the exact quantile of x for `p` = 0.5. Because `X` is a tall column vector and `p` is a scalar, `quantile` returns the exact quantile value by default.

```p = 0.5; % Cumulative probability yExact = quantile(x,p)```
```yExact = tall double ? ```

Calculate the approximate quantile of x for `p` = 0.5. Specify `'Method','approximate'` to use an approximation algorithm based on T-Digest for computing the quantiles.

`yApprox = quantile(x,p,'Method','approximate')`
```yApprox = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```

Evaluate the tall arrays and bring the results into memory by using `gather`.

`[yExact,yApprox] = gather(yExact,yApprox)`
```Evaluating tall expression using the Local MATLAB Session: - Pass 1 of 4: Completed in 1.4 sec - Pass 2 of 4: Completed in 0.6 sec - Pass 3 of 4: Completed in 0.74 sec - Pass 4 of 4: Completed in 0.76 sec Evaluation completed in 4.8 sec ```
```yExact = 1522 ```
```yApprox = 1.5220e+03 ```

The values of the approximate quantile and the exact quantile are the same to the four digits shown.

Calculate exact and approximate quantiles of a tall matrix for specified cumulative probabilities along different dimensions.

When you perform calculations on tall arrays, MATLAB® uses either a parallel pool (default if you have Parallel Computing Toolbox™) or the local MATLAB session. To run the example using the local MATLAB session when you have Parallel Computing Toolbox, change the global execution environment by using the `mapreducer` function.

`mapreducer(0)`

Create a tall matrix `X` containing a subset of variables from the `airlinesmall` data set. See Quantiles of Tall Vector for Given Probability for details about the steps to extract data from a tall array.

```varnames = {'ArrDelay','ArrTime','DepTime','ActualElapsedTime'}; % Subset of variables in the data set ds = datastore('airlinesmall.csv','TreatAsMissing','NA',... 'SelectedVariableNames',varnames); % Datastore t = tall(ds); % Tall table X = t{:,varnames} % Tall matrix```
```X = Mx4 tall double matrix 8 735 642 53 8 1124 1021 63 21 2218 2055 83 13 1431 1332 59 4 746 629 77 59 1547 1446 61 3 1052 928 84 11 1134 859 155 : : : : : : : : ```

When operating along a dimension that is not 1, the `quantile` function calculates the exact quantiles only, so that it can perform the computation efficiently using a sorting-based algorithm (see Algorithms) instead of an approximation algorithm based on T-Digest.

Calculate the exact quantiles of `X` along the second dimension for the cumulative probabilities 0.25, 0.5, and 0.75.

```p = [0.25 0.50 0.75]; % Vector of cumulative probabilities Yexact = quantile(X,p,2)```
```Yexact = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```

When the function operates along the first dimension and `p` is a vector of cumulative probabilities, you must use the approximation algorithm based on t-digest to compute the quantiles. Using the sorting-based algorithm to find the quantiles along the first dimension of a tall array is computationally intensive.

Calculate the approximate quantiles of `X` along the first dimension for the cumulative probabilities 0.25, 0.5, and 0.75. Because the default dimension is 1, you do not need to specify a value for `dim`.

`Yapprox = quantile(X,p,'Method','approximate')`
```Yapprox = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```

Evaluate the tall arrays and bring the results into memory by using `gather`.

`[Yexact,Yapprox] = gather(Yexact,Yapprox);`
```Evaluating tall expression using the Local MATLAB Session: - Pass 1 of 1: Completed in 2.9 sec Evaluation completed in 3.8 sec ```

Show the first five rows of the exact quantiles of `X` (along the second dimension) for the cumulative probabilities 0.25, 0.5, and 0.75.

`Yexact(1:5,:)`
```ans = 5×3 103 × 0.0305 0.3475 0.6885 0.0355 0.5420 1.0725 0.0520 1.0690 2.1365 0.0360 0.6955 1.3815 0.0405 0.3530 0.6875 ```

Each row of the matrix `Yexact` contains the three quantiles of the corresponding row in `X`. For example, `30.5`, `347.5`, and `688.5` are the 0.25, 0.5, and 0.75 quantiles, respectively, of the first row in `X`.

Show the approximate quantiles of `X` (along the first dimension) for the cumulative probabilities 0.25, 0.5, and 0.75.

`Yapprox`
```Yapprox = 3×4 103 × -0.0070 1.1148 0.9321 0.0700 0 1.5220 1.3350 0.1020 0.0110 1.9180 1.7400 0.1510 ```

Each column of the matrix `Yapprox` corresponds to the three quantiles for each column of the matrix `X`. For example, the first column of `Yapprox` with elements (–7, 0, 11) contains the quantiles for the first column of `X`.

Calculate exact and approximate quantiles along different dimensions of a tall matrix for `N` evenly spaced cumulative probabilities.

When you perform calculations on tall arrays, MATLAB® uses either a parallel pool (default if you have Parallel Computing Toolbox™) or the local MATLAB session. To run the example using the local MATLAB session when you have Parallel Computing Toolbox, change the global execution environment by using the `mapreducer` function.

`mapreducer(0)`

Create a tall matrix `X` containing a subset of variables from the `airlinesmall` data set. See Quantiles of Tall Vector for Given Probability for details about the steps to extract data from a tall array.

```varnames = {'ArrDelay','ArrTime','DepTime','ActualElapsedTime'}; % Subset of variables in the data set ds = datastore('airlinesmall.csv','TreatAsMissing','NA',... 'SelectedVariableNames',varnames); % Datastore t = tall(ds); % Tall table X = t{:,varnames}```
```X = Mx4 tall double matrix 8 735 642 53 8 1124 1021 63 21 2218 2055 83 13 1431 1332 59 4 746 629 77 59 1547 1446 61 3 1052 928 84 11 1134 859 155 : : : : : : : : ```

To find evenly spaced quantiles along the first dimension, you must use the approximation algorithm based on T-Digest. Using the sorting-based algorithm (see Algorithms) to find quantiles along the first dimension of a tall array is computationally intensive.

Calculate three evenly spaced quantiles along the first dimension of `X`. Because the default dimension is 1, you do not need to specify a value for `dim`. Specify `'Method','approximate'` to use the approximation algorithm.

```N = 3; % Number of quantiles Yapprox = quantile(X,N,'Method','approximate')```
```Yapprox = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```

To find evenly spaced quantiles along any other dimension (`dim` is not `1`), `quantile` calculates the exact quantiles only, so that it can perform the computation efficiently by using the sorting-based algorithm.

Calculate three evenly spaced quantiles along the second dimension of `X`. Because `dim` is not 1, `quantile` returns the exact quantiles by default.

`Yexact = quantile(X,N,2)`
```Yexact = MxNx... tall double array ? ? ? ... ? ? ? ... ? ? ? ... : : : : : : ```

Evaluate the tall arrays and bring the results into memory by using `gather`.

`[Yapprox,Yexact] = gather(Yapprox,Yexact);`
```Evaluating tall expression using the Local MATLAB Session: - Pass 1 of 1: Completed in 4.2 sec Evaluation completed in 5.1 sec ```

Show the approximate quantiles of `X` (along the first dimension) for the three evenly spaced cumulative probabilities.

`Yapprox`
```Yapprox = 3×4 103 × -0.0070 1.1149 0.9322 0.0700 0 1.5220 1.3350 0.1020 0.0110 1.9180 1.7400 0.1510 ```

Each column of the matrix `Yapprox` corresponds to the three evenly spaced quantiles for each column of the matrix `X`. For example, the first column of `Yapprox` with elements (–7, 0, 11) contains the quantiles for the first column of `X`.

Show the first five rows of the exact quantiles of `X` (along the second dimension) for the three evenly spaced cumulative probabilities.

`Yexact(1:5,:)`
```ans = 5×3 103 × 0.0305 0.3475 0.6885 0.0355 0.5420 1.0725 0.0520 1.0690 2.1365 0.0360 0.6955 1.3815 0.0405 0.3530 0.6875 ```

Each row of the matrix `Yexact` contains the three evenly spaced quantiles of the corresponding row in `X`. For example, `30.5`, `347.5`, and `688.5` are the 0.25, 0.5, and 0.75 quantiles, respectively, of the first row in `X`.

## Input Arguments

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Input data, specified as a vector or array.

Data Types: `double` | `single`

Cumulative probabilities for which to compute the quantiles, specified as a scalar or vector of scalars from 0 to 1.

Example: 0.3

Example: [0.25, 0.5, 0.75]

Example: (0:0.25:1)

Data Types: `double` | `single`

Number of quantiles to compute, specified as a positive integer. `quantile` returns `N` quantiles that divide the data set into evenly distributed `N`+1 segments.

Data Types: `double` | `single`

Dimension along which the quantiles of a matrix `X` are requested, specified as a positive integer. For example, for a matrix `X`, when `dim` = 1, `quantile` returns the quantile(s) of the columns of `X`; when `dim` = 2, `quantile` returns the quantile(s) of the rows of `X`. For a multidimensional array `X`, the length of the `dim`th dimension of `Y` is the same as the length of `p`.

Data Types: `single` | `double`

Vector of dimensions, specified as a positive integer vector. Each element of `vecdim` represents a dimension of the input array `X`. In the smallest specified operating dimension (that is, dimension `min(vecdim)`), the output `Y` has length equal to the number of quantiles requested (either `N` or `length(p)`). In each of the remaining operating dimensions, `Y` has length 1. The other dimension lengths are the same for `X` and `Y`.

For example, consider a 2-by-3-by-3 array `X` with ```p = [0.2 0.4 0.6 0.8]```. In this case, `quantile(X,p,[1 2])` returns an array, where each page of the array contains the 0.2, 0.4, 0.6, and 0.8 quantiles of the elements on the corresponding page of `X`. Because 1 and 2 are the operating dimensions, with `min([1 2]) = 1` and `length(p) = 4`, the output is a 4-by-1-by-3 array.

Data Types: `single` | `double`

Method for calculating quantiles, specified as `'exact'` or `'approximate'`. By default, `quantile` returns the exact quantiles by implementing an algorithm that uses sorting. You can specify `'method','approximate'` for `quantile` to return approximate quantiles by implementing an algorithm that uses T-Digest.

Data Types: `char` | `string`

## Output Arguments

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Quantiles of a data vector or array, returned as a scalar or array for one or multiple values of cumulative probabilities.

• If `X` is a vector, then `Y` is a scalar or a vector with the same length as the number of quantiles requested (`N` or `length(p)`). `Y(i)` contains the `p(i)` quantile.

• If `X` is an array of dimension d, then `Y` is an array with the length of the smallest operating dimension equal to the number of quantiles requested (`N` or `length(p)`).

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### Multidimensional Array

A multidimensional array is an array with more than two dimensions. For example, if `X` is a 1-by-3-by-4 array, then `X` is a 3-D array.

### Nonsingleton Dimension

A nonsingleton dimension of an array is a dimension whose size is not equal to 1. A first nonsingleton dimension of an array is the first dimension that satisfies the nonsingleton condition. For example, if `X` is a 1-by-1-by-2-by-4 array, then the third dimension is the first nonsingleton dimension of `X`.

### Linear Interpolation

Linear interpolation uses linear polynomials to find yi = f(xi), the values of the underlying function Y = f(X) at the points in the vector or array x. Given the data points (x1, y1) and (x2, y2), where y1 = f(x1) and y2 = f(x2), linear interpolation finds y = f(x) for a given x between x1 and x2 as follows:

`$y=f\left(x\right)={y}_{1}+\frac{\left(x-{x}_{1}\right)}{\left({x}_{2}-{x}_{1}\right)}\left({y}_{2}-{y}_{1}\right).$`

Similarly, if the 1.5/n quantile is y1.5/n and the 2.5/n quantile is y2.5/n, then linear interpolation finds the 2.3/n quantile y2.3/n as

`${y}_{\frac{2.3}{n}}={y}_{\frac{1.5}{n}}+\frac{\left(\frac{2.3}{n}-\frac{1.5}{n}\right)}{\left(\frac{2.5}{n}-\frac{1.5}{n}\right)}\left({y}_{\frac{2.5}{n}}-{y}_{\frac{1.5}{n}}\right).$`

### T-Digest

T-digest[2] is a probabilistic data structure that is a sparse representation of the empirical cumulative distribution function (CDF) of a data set. T-digest is useful for computing approximations of rank-based statistics (such as percentiles and quantiles) from online or distributed data in a way that allows for controllable accuracy, particularly near the tails of the data distribution.

For data that is distributed in different partitions, t-digest computes quantile estimates (and percentile estimates) for each data partition separately, and then combines the estimates while maintaining a constant-memory bound and constant relative accuracy of computation ($q\left(1-q\right)$ for the qth quantile). For these reasons, t-digest is practical for working with tall arrays.

To estimate quantiles of an array that is distributed in different partitions, first build a t-digest in each partition of the data. A t-digest clusters the data in the partition and summarizes each cluster by a centroid value and an accumulated weight that represents the number of samples contributing to the cluster. T-digest uses large clusters (widely spaced centroids) to represent areas of the CDF that are near ```q = 0.5``` and uses small clusters (tightly spaced centroids) to represent areas of the CDF that are near `q = 0` or `q = 1`.

T-digest controls the cluster size by using a scaling function that maps a quantile q to an index k with a compression parameter $\delta$. That is,

`$k\left(q,\delta \right)=\delta \cdot \left(\frac{{\mathrm{sin}}^{-1}\left(2q-1\right)}{\pi }+\frac{1}{2}\right),$`

where the mapping k is monotonic with minimum value k(0,δ) = 0 and maximum value k(1,δ) = δ. The following figure shows the scaling function for δ = 10.

The scaling function translates the quantile q to the scaling factor k in order to give variable size steps in q. As a result, cluster sizes are unequal (larger around the center quantiles and smaller near `q = 0` or ```q = 1```). The smaller clusters allow for better accuracy near the edges of the data.

To update a t-digest with a new observation that has a weight and location, find the cluster closest to the new observation. Then, add the weight and update the centroid of the cluster based on the weighted average, provided that the updated weight of the cluster does not exceed the size limitation.

You can combine independent t-digests from each partition of the data by taking a union of the t-digests and merging their centroids. To combine t-digests, first sort the clusters from all the independent t-digests in decreasing order of cluster weights. Then, merge neighboring clusters, when they meet the size limitation, to form a new t-digest.

Once you form a t-digest that represents the complete data set, you can estimate the end-points (or boundaries) of each cluster in the t-digest and then use interpolation between the end-points of each cluster to find accurate quantile estimates.

## Algorithms

For an n-element vector `X`, `quantile` computes quantiles by using a sorting-based algorithm as follows:

1. The sorted elements in `X` are taken as the (0.5/n), (1.5/n), ..., ([n – 0.5]/n) quantiles. For example:

• For a data vector of five elements such as {6, 3, 2, 10, 1}, the sorted elements {1, 2, 3, 6, 10} respectively correspond to the 0.1, 0.3, 0.5, 0.7, 0.9 quantiles.

• For a data vector of six elements such as {6, 3, 2, 10, 8, 1}, the sorted elements {1, 2, 3, 6, 8, 10} respectively correspond to the (0.5/6), (1.5/6), (2.5/6), (3.5/6), (4.5/6), (5.5/6) quantiles.

2. `quantile` uses Linear Interpolation to compute quantiles for probabilities between (0.5/n) and ([n – 0.5]/n).

3. For the quantiles corresponding to the probabilities outside that range, `quantile` assigns the minimum or maximum values of the elements in `X`.

`quantile` treats `NaN`s as missing values and removes them.

## References

[1] Langford, E. “Quartiles in Elementary Statistics”, Journal of Statistics Education. Vol. 14, No. 3, 2006.

[2] Dunning, T., and O. Ertl. “Computing Extremely Accurate Quantiles Using T-Digests.” August 2017.


# Method to factor an expression

As the title says, i want to factorize an expression, but i don't have any clue how to proceed.

Here is the expression :

$$2x² -7x +3$$

And here is the factorized form :

$$(x-3)(2x - 1)$$

My question is, which method or rule to use to go from first to second ?

please note that I am a beginner, and the only question that i found which is closer to mine is this post.

Thank you for your help !

Rewrite the expression into the form:

$$2x^2-6x-x+3$$ ,

then group the first two terms together and the last two terms together:

$$(2x^2-6x)-(x-3)=2x(x-3)-(x-3)$$ ,

next extract the common factor:

$$(x-3)(2x-1)$$ ,

and you are done.

• I like the simplicity of your method, thank's :) – ganzo db Dec 18 '18 at 16:41
• @ganzodb You are welcome – Matko Dec 18 '18 at 18:24

If it concerns quadratic polynomial $$ax^2+bx+c$$ then start with calculating discriminant: $$D:=b^2-4ac$$

If $$D$$ is negative then give up (unless you are familiar with complex numbers already).

If $$D$$ is nonnegative then: $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ where $$x_1=\frac{-b+\sqrt D}{2a}$$ and $$x_2=\frac{-b-\sqrt D}{2a}$$.

Especially if $$D$$ is a perfect square (as in your case, where $$D=25$$) then there is reason to cheer.

• As you supposed, i'm not good with complex numbers :) ! – ganzo db Dec 18 '18 at 16:54

It relies on the following property of quadratic polynomials:

If the quadratic polynomial $$\;ax^2+bx+c\;(a\ne 0)$$ has roots $$\xi_0$$ and $$\xi_1$$ (real or complex, distinct ot not), it can factored as $$ax^2+bx+c=a(x-\xi_0)(x-\xi_1).$$

This property is a consequence of a more general property of polynomials (of any degree) and the ring of polynomials over a field being a euclidean domain:

If a polynomial $$p(x)$$ has root $$\xi$$, it is divisible by $$x-\xi$$.

In the present case, the discriminat of $$2x² -7x +3$$ is $$\;\Delta=49-4\cdot 2\cdot 3=25$$, so its roots are $$\;\frac{7\pm 5}4==\bigl\{3,\frac 12\bigr\}$$, and the factorisation is $$2(x-3)\Bigl(x-\frac12\Bigr)=(x-3)(2x-1).$$

• i wish i could understand the last one ( \frac{7\pm 5}4 ). – ganzo db Dec 18 '18 at 16:57
• @ganzodb: It's just the formula $\frac{-b\pm\sqrt\Delta}{2a}.$. – Bernard Dec 18 '18 at 18:22

In general ,if $$x_1$$ and $$x_2$$ are roots of $$\underbrace{a}_{\neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$\text{sum of the roots}=-\frac{b}{a}\;\;\&\;\;\text{product of the roots}=\frac{c}{a}$$

So your case, $$x_1+x_2= \frac{7}{2}$$ and $$x_1x_2=\frac{3}{2}$$

Now solve these to get $$x_1$$ and $$x_2$$ to finish your conclusion

To find $$x_1$$ and $$x_2$$, $$2x^2-7x+3=0$$ implies $$x^2-\frac{7}{2}x+\frac{3}{2}=0$$ which means $$x^2-2\left(\frac{7}{4}\right)x=-\frac{3}{2}$$ which is same as $$x^2-2\left(\frac{7}{4}\right)x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}=\frac{25}{16}$$ so $$\left(x-\frac{7}{4}\right)^2=\frac{25}{16}$$ and so $$x-\frac{7}{4}=\pm \sqrt{\frac{25}{16}}=\pm \frac{5}{4}$$ so $$x=\frac{7}{4} \pm \frac{5}{4}=\frac{7\pm 5}{4}$$

• Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts. – ganzo db Dec 18 '18 at 16:53


1 ? the This is the reason I like best. While we can use other methods to solve such a problem, if we know the multiplicative inverse of our coefficient matrix, then we can easily solve the problem by simply multiplying both sides by the inverse. A second-order matrix can be represented by . 's' : ''}}. Step 2:. a number multiplied by it’s multiplicative inverse gives the multiplicative identity. Here are three ways to find the inverse of a matrix: 1. See Also . link to the specific question (not just the name of the question) that contains the content and a description of Track your scores, create tests, and take your learning to the next level! If we multiply matrix A by the inverse of matrix A, we will get the identity matrix, I. From the top row, we get 1(11) + -2(5) = 11 - 10 = 1. All other trademarks and copyrights are the property of their respective owners. Reduce the left matrix to row echelon form using elementary row operations for the whole matrix (including the right one). F) Find A(BC). The concept of solving systems using matrices is similar to the concept of solving simple equations. The theory, as usual, is below the calculator as The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $$A$$ and its inverse $$A^{−1}$$ equals the identity matrix. By using this website, you agree to our Cookie Policy. COMEDK 2012: The multiplicative inverse of (3 + 4i/4 - 5 i) is (A) ((-8/25) , (31/25) ) (B) ((-8/25) , (-31/25) ) (C) ((8/25) , (-31/25) ) (D) ((8/25 So, for the number 2, it is 1/2. There are a couple of ways to do this. Inverse of a Matrix. The term inverse matrix generally implies the multiplicative inverse of a matrix. Create an account to start this course today. In general, the inverse of n X n matrix A can be found using this simple formula: where, Adj(A) denotes the adjoint of a matrix and, Det(A) is Determinant of matrix A. Yes, we write the inverse with a superscript of -1. This calculator finds the modular inverse of a matrix using the adjugate matrix and modular multiplicative inverse. 5/7 minus 4/7. The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1). Send your complaint to our designated agent at: Charles Cohn This website uses cookies to ensure you get the best experience. Hence, I is known as the identity matrix under multiplication. I don't … first two years of college and save thousands off your degree. f(g(x)) = g(f(x)) = x. Find the value of . The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $A$ and its inverse ${A}^{-1}$ equals the identity matrix. You see, it is useful to learn about the multiplicative inverse of a matrix because if we know it, then we can use it to help us solve equations with matrices in them. If Varsity Tutors takes action in response to AX - BX + D = C. Use A - 1 to find the solution, (x_1, x_2), to the given system. C) Find AB. Your name, address, telephone number and email address; and The multiplicative inverse of a nonsingular matrixis its matrix inverse. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. Multiplying Matrices. Multiplicative Inverse Property Calculator. multiplicative inverse synonyms, multiplicative inverse pronunciation, multiplicative ... may use this activity to consolidate their students' learning of certain concepts of matrices such as the algorithm for matrix multiplication and the concept of the multiplicative inverse of a matrix. B) Find 2A + 3B. If you've found an issue with this question, please let us know. 3. is the multiplicative inverse of . The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. St. Louis, MO 63105. Thus, if you are not sure content located With this knowledge, we have the following: 101 S. Hanley Rd, Suite 300 Create your account. To learn more, visit our Earning Credit Page. Coolmath privacy policy. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. Define multiplicative inverse. Note the first and the last columns are equal. After you multiply, you can then easily find the answer by translating back to equation form. Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. Zero … MULTIPLICATIVE INVERSES For every nonzero real number a, there is a multiplicative inverse l/a such that. Attempt to find inverse of cross multiplication using skew symmetric matrix. I will use the determinant method. Plug the value in the formula then simplify to get the inverse of matrix C. Step 3:. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing and career path that can help you find the school that's right for you. Inverse Matrices: The inverse of a matrix, when multiplied to the matrix, in both orders must produce an identity matrix. Same thing when the inverse comes first: ( 1/8) × 8 = 1. All operations on residue matrices are performed the same as for the integer matrices except that the operations are done in modular arithmetic. either the copyright owner or a person authorized to act on their behalf. The following formula is used to calculate the inverse matrix value of the original 2×2 matrix. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Amy has a master's degree in secondary education and has taught math at a public charter high school. Study.com has thousands of articles about every I find the modular multiplicative inverse (of the matrix determinant, which is $1×4-3×5=-11$) with the extended Euclid algorithm (it is $-7 \equiv 19 \pmod{26}$). your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its inverse A –1 equals the identity matrix. I explain that today we will find the multiplicative inverse of a matrix. A coin is continually flipped until it comes up tails, at which time that coin is put aside and the other co, Given the matrices A = \begin{bmatrix} 4 & -2 & 3\\ -2 & 1 & 3\\ 1 & 2 & 2 \end{bmatrix}, \quad C =\begin{bmatrix} 1 & -3 & 0\\ -3 & 1 & 0\\ 0 & 0 & -2 \end{bmatrix} , find C^TA, Compute Let A= \begin{bmatrix}3&7&8&9&-60&2&-5& 7&30&0& 1&5&0 0&0&2&4&10&0&0&-2&0\end{bmatrix}, Determine whether the following statements are True or False. 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The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. Let's review what we've learned. (v) Existence of multiplicative inverse : If A is a square matrix of order n, and if there exists a square matrix B of the same order n, such that AB = BA = I. where I is the unit matrix of order n, then B is called the multiplicative inverse matrix of … Yes, our answer would be our answer matrix, b, multiplied by the multiplicative inverse of our coefficient matrix. Now that students understand we are developing a method for finding the inverse of a matrix, I provide students with our book's brief introduction to the determinant. Email: donsevcik@gmail.com Tel: 800-234-2933; {{courseNav.course.mDynamicIntFields.lessonCount}} lessons Step 3:. Example $$\PageIndex{5}$$ Solve the following system \begin{aligned} 3 x+y&=3 \\ 5 x+2 y&=4 \end{aligned} Solution. ChillingEffects.org. Cryptography uses residue matrices: matrices in all elements are in Zn. 1. For example, we can use it to solve a problem like this: This matrix equation is in the form of Ax = b, where A is your coefficient matrix, x is your variable matrix, and b is your answer matrix. matrix. Set the matrix (must be square) and append the identity matrix of the same dimension to it. Let the multiplicative inverse be a. Matrix resulting from the multiplication of a matrix by its inverse Inverses of matrices Multiplicative inverse Skills Practiced. Illustrated definition of Multiplicative Inverse: Another name for Reciprocal. 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Let's multiply them out. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. Varsity Tutors LLC Previous matrix calculators: Determinant of a matrix, Matrix Transpose, Matrix Multiplication, Inverse matrix calculator. Inverse Matrix Formula. Log in here for access. an 2. Consider the following example. By applying matrix multiplication to a square matrix of which we want to find the inverse and using the matrix equation AX = I to solve for X, when operations have been completed the square matrix X is the inverse matrix A −1, X = A −1, and we will have solved AA −1 = I n. Inverse of a Matrix using Gauss-Jordan Elimination. means of the most recent email address, if any, provided by such party to Varsity Tutors. Refer to the following matrices: A = (1 & 2 | 3 & -4 ), B = (5 & 0 | -6 & 7), C = (1 & -3 & 4 | 2 & 6 & -5). 1. Washington University in St Louis, Master of Science, Electrical Engineer... University of Illinois at Urbana-Champaign, Bachelor of Science, Chemical and Biomolecular Engineering. Which is equal to-- this is just a scalar, this is just a number, so we multiply it times each of the elements-- so that is equal to minus, minus, plus. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. Matrix inversion is the process of finding the matrix B that satisfies the prior e… When we multiply a matrix by its inverse we get the Identity Matrix (which is like "1" for matrices): A × A -1 = I. This precalculus video tutorial explains how to determine the inverse of a 2x2 matrix. Not all matrices have an inverse. An identification of the copyright claimed to have been infringed; I have the matrix$$\begin{pmatrix} 1 & 5\\ 3 & 4 \end{pmatrix} \pmod{26}$$ and I need to find its inverse. Inverse Matrices. Following this lesson, you should be able to: To unlock this lesson you must be a Study.com Member. Multiplying the two matrices, we see that we do get the identity matrix: We know for sure now that this inverse is the real inverse, and it works for us. If you know the inverse of a matrix, you can solve the problem by multiplying the inverse of the matrix with the answer matrix, x = A sup -1 * b. The special property here is as follows: A*A^-1 = A^-1*A = I. Each transistor requires 3 units of copper, 1 unit of zinc, and 2 units of glass. Already registered? Multiplicative Inverse Property Calculator. In this section we see how Gauss-Jordan Elimination works using examples. Provide a clear justification for each answer. credit by exam that is accepted by over 1,500 colleges and universities. Watch this video lesson to learn about another method you can use to solve a matrix problem if you are given the inverse of the matrix. We find the "inverse" of 7, which is 1/7. Positive 2/7. The multiplicative inverse of a matrix A is a matrix (indicated as A−1) such that: A ⋅ A−1 = A−1 ⋅ A = I Where I is the identity matrix (made up of all zeros except on the main diagonal which contains all 1). The multiplicative inverse of a nonsingular matrixis its matrix inverse. This matrix has no inverse because the columns are not linearly independent. A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe As a result you will get the inverse calculated on the right. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Shortcut for 2 x 2 matrices For , the inverse can be found using this formula: Example: 2. Because when you multiply them together, you get the multiplicative identity (one). The modular multiplicative inverse of an integer a modulo m is an integer b such that, It maybe noted , where the fact that the inversion is m-modular is implicit. 8 × ( 1/8) = 1. 2. We ended up with fractions here and things. Try refreshing the page, or contact customer support. For our normal numbers, the multiplicative inverse is simply 1 divided by our number. credit-by-exam regardless of age or education level. You can use the multiplicative inverse of a matrix to solve problems in the form of Ax = b, where A is your coefficient matrix, x is your variable matrix, and b is your answer, or constant, matrix. Step 1:. the multiplicative inverse: ... Scalar Multiplication. 15 minutes. In order to find the multiplicative inverse, we have to find the matrix for which, when we multiply it with our matrix, we get the identity matrix. Step 2:. The multiplicative inverse of a matrix A is written A^(-1). This matrix has no inverse because the columns are not linearly independent. Sample Usage. MINVERSE(square_matrix) square_matrix - An array or range with an equal number of rows and columns representing a matrix whose multiplicative inverse will be calculated. However matrices can be not only two-dimensional, but also one-dimensional (vectors), so that you can multiply vectors, vector by matrix and vice versa. Precalculus : Find the Multiplicative Inverse of a Matrix Study concepts, example questions & explanations for Precalculus. The multiplicative inverse of a matrix A is written A^(-1). Get the unbiased info you need to find the right school. In arithmetic, there is one number which does not have a multiplicative inverse. So before I do that I have to create some space. x_1 + 3x_2 = 1. a flashcard set{{course.flashcardSetCoun > 1 ? the This is the reason I like best. While we can use other methods to solve such a problem, if we know the multiplicative inverse of our coefficient matrix, then we can easily solve the problem by simply multiplying both sides by the inverse. A second-order matrix can be represented by . 's' : ''}}. Step 2:. a number multiplied by it’s multiplicative inverse gives the multiplicative identity. Here are three ways to find the inverse of a matrix: 1. See Also . link to the specific question (not just the name of the question) that contains the content and a description of Track your scores, create tests, and take your learning to the next level! If we multiply matrix A by the inverse of matrix A, we will get the identity matrix, I. From the top row, we get 1(11) + -2(5) = 11 - 10 = 1. All other trademarks and copyrights are the property of their respective owners. Reduce the left matrix to row echelon form using elementary row operations for the whole matrix (including the right one). F) Find A(BC). The concept of solving systems using matrices is similar to the concept of solving simple equations. The theory, as usual, is below the calculator as The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $$A$$ and its inverse $$A^{−1}$$ equals the identity matrix. By using this website, you agree to our Cookie Policy. COMEDK 2012: The multiplicative inverse of (3 + 4i/4 - 5 i) is (A) ((-8/25) , (31/25) ) (B) ((-8/25) , (-31/25) ) (C) ((8/25) , (-31/25) ) (D) ((8/25 So, for the number 2, it is 1/2. There are a couple of ways to do this. Inverse of a Matrix. The term inverse matrix generally implies the multiplicative inverse of a matrix. Create an account to start this course today. In general, the inverse of n X n matrix A can be found using this simple formula: where, Adj(A) denotes the adjoint of a matrix and, Det(A) is Determinant of matrix A. Yes, we write the inverse with a superscript of -1. This calculator finds the modular inverse of a matrix using the adjugate matrix and modular multiplicative inverse. 5/7 minus 4/7. The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1). Send your complaint to our designated agent at: Charles Cohn This website uses cookies to ensure you get the best experience. Hence, I is known as the identity matrix under multiplication. I don't … first two years of college and save thousands off your degree. f(g(x)) = g(f(x)) = x. Find the value of . The multiplicative inverse of a matrix is similar in concept, except that the product of matrix $A$ and its inverse ${A}^{-1}$ equals the identity matrix. You see, it is useful to learn about the multiplicative inverse of a matrix because if we know it, then we can use it to help us solve equations with matrices in them. If Varsity Tutors takes action in response to AX - BX + D = C. Use A - 1 to find the solution, (x_1, x_2), to the given system. C) Find AB. Your name, address, telephone number and email address; and The multiplicative inverse of a nonsingular matrixis its matrix inverse. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. Multiplying Matrices. Multiplicative Inverse Property Calculator. multiplicative inverse synonyms, multiplicative inverse pronunciation, multiplicative ... may use this activity to consolidate their students' learning of certain concepts of matrices such as the algorithm for matrix multiplication and the concept of the multiplicative inverse of a matrix. B) Find 2A + 3B. If you've found an issue with this question, please let us know. 3. is the multiplicative inverse of . The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. St. Louis, MO 63105. Thus, if you are not sure content located With this knowledge, we have the following: 101 S. Hanley Rd, Suite 300 Create your account. To learn more, visit our Earning Credit Page. Coolmath privacy policy. The multiplicative inverse of a matrix is the matrix that gives you the identity matrix when multiplied by the original matrix. Define multiplicative inverse. Note the first and the last columns are equal. After you multiply, you can then easily find the answer by translating back to equation form. Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. Zero … MULTIPLICATIVE INVERSES For every nonzero real number a, there is a multiplicative inverse l/a such that. Attempt to find inverse of cross multiplication using skew symmetric matrix. I will use the determinant method. Plug the value in the formula then simplify to get the inverse of matrix C. Step 3:. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing and career path that can help you find the school that's right for you. Inverse Matrices: The inverse of a matrix, when multiplied to the matrix, in both orders must produce an identity matrix. Same thing when the inverse comes first: ( 1/8) × 8 = 1. All operations on residue matrices are performed the same as for the integer matrices except that the operations are done in modular arithmetic. either the copyright owner or a person authorized to act on their behalf. The following formula is used to calculate the inverse matrix value of the original 2×2 matrix. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are Amy has a master's degree in secondary education and has taught math at a public charter high school. Study.com has thousands of articles about every I find the modular multiplicative inverse (of the matrix determinant, which is $1×4-3×5=-11$) with the extended Euclid algorithm (it is $-7 \equiv 19 \pmod{26}$). your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its inverse A –1 equals the identity matrix. I explain that today we will find the multiplicative inverse of a matrix. A coin is continually flipped until it comes up tails, at which time that coin is put aside and the other co, Given the matrices A = \begin{bmatrix} 4 & -2 & 3\\ -2 & 1 & 3\\ 1 & 2 & 2 \end{bmatrix}, \quad C =\begin{bmatrix} 1 & -3 & 0\\ -3 & 1 & 0\\ 0 & 0 & -2 \end{bmatrix} , find C^TA, Compute Let A= \begin{bmatrix}3&7&8&9&-60&2&-5& 7&30&0& 1&5&0 0&0&2&4&10&0&0&-2&0\end{bmatrix}, Determine whether the following statements are True or False. A\ ) to create some space operations are done in modular arithmetic multiplicative INVERSES for nonzero..., you multiplicative inverse of matrix test out of the matrix is the inverse matrix Determinante... Calculator - calculate matrix inverse since, the multiplicative inverse is -1/3 values we find the of! Guided Practice Practice Read and study the lesson to answer each question nur! The rows and columns take your learning to the concept of solving simple equations,. 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# Computing a complex integral potentially using residues

The question is:

Compute:

$$\mbox{p.v.}\int_{-\infty}^{\infty}\frac{x\sin4x}{{x^2}-1}dx$$

Initially I thought it was straight forward and I could just use residues. However, the Residue Theorem requires the poles to be in the upper plane ($y > 0$), and in this case, that is not the case. So, now I have no idea what to do since I cannot use residues.

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p.v. = principal value. –  user26872 Jul 25 '12 at 21:55
Right, I understand that. Does that change how I approach the problem though? Does that make it so that I can go ahead with computing the residues at 1 and -1? –  Payton Jul 25 '12 at 22:31
See Rob and Tim's solutions below. –  user26872 Jul 25 '12 at 23:32

Let $C_R$ denote the counterclockwise semicircular arc extending from $R$ to $-R$ in the plane and $C_{\rho_1} , C_{\rho_2}$ be the counterclockwise semicircular arcs extending from $-1 - \rho_1$ to $-1 + \rho_1$ and from $1 -\rho_1$ and $1+\rho_1,$ respectively. Note that $\frac{x\sin 4x}{x^2 -1} = \Im \frac{x\exp{4ix}}{x^2 -1}.$

Consider also that $\,\displaystyle{f(z) = \frac{z\exp{4iz}}{z^2 -1}}\,$ has simple poles at $-1$ and $1.$ By Jordan's Lemma, there exists $\theta \in [0, \pi ]$ such that $$\displaystyle\int_{C_R} \frac{z\exp{4iz}}{z^2-1} \le \frac{\pi }{4} \frac{1}{R^2 \exp{2i\theta }-1}$$ The right hand side of this inequality tends to zero. Using the residue theorem,

$$\int_{C_R} \frac{z\exp{4iz}}{z^2-1} + \int_{-C_{\rho_1}} \frac{z\exp{4iz}}{z^2-1} + \int_{-C_{\rho_2}} \frac{z\exp{4iz}}{z^2-1} +\int_{-R}^{-1 -\rho_1} \frac{x\exp{4ix}}{x^2-1} dx +$$ $$+\int_{-1 +\rho_1}^{1 - \rho_e} \frac{x\exp{4ix}}{x^2-1} dx +\int_{1 + \rho_1}^{R} \frac{x\exp{4ix}}{x^2-1} dx = 0$$

Letting $R$ tend to infinity and $\rho_1 \rho_2$ tend to zero, we have $$\textrm{pv }\int_{-\infty }^{\infty } \frac{x\exp{4ix}}{x^2 -1 }dx = \pi i [\textrm{ Res } (f, -1) + \textrm{ Res } (f, 1) ] = \pi i \cos 4$$ Hence $$\textrm{pv } \int_{-\infty }^{\infty } \frac{x\sin 4x}{x^2 -1 }dx = \pi \cos 4.$$

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But doesn't the Residue Theorem run into an issue when the poles are on the real line? (that is, when they aren't on the upper half plane?) –  Payton Jul 25 '12 at 23:23
Yes - we avoided this by constructing an indented contour around the poles. This approach relied on the fact that the poles were simple. See the picture in RobJohn's answer for the basic idea. There are no poles for $f$ enclosed within the indented contour, so the residue theorem tells us that the contour integral in that case was zero. –  user17794 Jul 26 '12 at 0:36

Consider the following diagram:

$\hspace{3.25cm}$

The principal value integral is the integral over the path in black as the paths in red get smaller and the paths in blue and green get bigger. We will evaluate this by first breaking up $\sin(4z)=\dfrac{e^{i4z}-e^{-i4z}}{2i}$ so that we can close the path of integration of each piece along a different path.

$\dfrac{e^{i4z}}{2i}$ will be integrated over the paths in black, red, and blue ($\gamma^+$). $\gamma^+$ contains the singularities at $z=-1$ and $z=1$

$\dfrac{e^{-i4z}}{2i}$ will be integrated over the paths in black, red, and green ($\gamma^-$). $\gamma^-$ contains no singularities.

The sum of the integrals described above, include the integrals over the red half circles. We will eliminate the integrals over the red half circles by subtracting half of $2\pi i$ times the residue of $\dfrac{z\sin(4z)}{z^2-1}$ at $z=-1$ and $z=1$.

Thus, we get

\begin{align} =\hspace{-11.5pt}\int_{-\infty}^\infty\frac{x\sin(4x)}{x^2-1}\mathrm{d}x &=\color{#0000FF}{\frac1{2i}\int_{\gamma^+}\frac{ze^{i4z}}{z^2-1}\mathrm{d}z} -\color{#00A000}{\frac1{2i}\int_{\gamma^-}\frac{ze^{-i4z}}{z^2-1}\mathrm{d}z}\\ &-\color{#C00000}{\pi i\mathrm{Res}_{z=-1}\left(\frac{z\sin(4z)}{z^2-1}\right)} -\color{#C00000}{\pi i\mathrm{Res}_{z=1}\left(\frac{z\sin(4z)}{z^2-1}\right)}\\ &=\color{#0000FF}{\frac{2\pi i}{2i}\mathrm{Res}_{z=-1}\left(\frac{ze^{i4z}}{z^2-1}\right)} +\color{#0000FF}{\frac{2\pi i}{2i}\mathrm{Res}_{z=1}\left(\frac{ze^{i4z}}{z^2-1}\right)}\\ &-\color{#C00000}{\pi i\mathrm{Res}_{z=-1}\left(\frac{z\sin(4z)}{z^2-1}\right)} -\color{#C00000}{\pi i\mathrm{Res}_{z=1}\left(\frac{z\sin(4z)}{z^2-1}\right)}\\ &=\color{#0000FF}{\pi\frac{e^{-i4}}{2}}+\color{#0000FF}{\pi\frac{e^{i4}}{2}}\\ &-\color{#C00000}{\pi i\frac{\sin(-4)}{2}}-\color{#C00000}{\pi i\frac{\sin(4)}{2}}\\[6pt] &=\pi\cos(4) \end{align}

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I have to ask, what program did you use to draw the diagram? –  user2468 Jul 26 '12 at 2:07
I used Intaglio for the Mac. –  robjohn Jul 26 '12 at 5:29


# How the two non null-homotopic equivalence classes generate the null-homotopic loop on the torus

I am new in Alebraic Topology. Given the torus, we say that the fundamental group of the torus is generated by two loops (or more exactly two equivalent classes of loops). One writes $$\pi=\mathbb{Z}\times\mathbb{Z}.$$

I don't understand, how the null-homotopic loop, which is the constant loop, is generated by the two generators mentioned above. Can somebody provide an explanation? More even so, I don't see how it functions visually, since the two generators are not null-homotopic. More precisely, given a null-homotopic loop on the surface on a base point $$x$$, how this loop will be generated by the two generators mentioned above?

• The identity is in every subgroup. – Angina Seng Dec 28 '18 at 16:06
• Thanks. But how it is generated in the case of a null-homotopic loop ? In the case the loops are traversing the torus around the hole or through the hole, one can say for instance that the product of one loop and its inverse is the constant loop, that is the identity. But I dont see how it works in the case a loop is null-homotopic, since the generators are non null-homotopic. – user249018 Dec 28 '18 at 16:14
• What do you think the word "generate" means? – Eric Wofsey Dec 28 '18 at 16:20

Let $$\mathbb{T}^2$$ denote the torus and choose a basepoint $$p \in \mathbb{T}^2$$. Then we know that $$\pi_1\left(\mathbb{T}^2, p \right) \cong \mathbb{Z} \times \mathbb{Z}$$.

Now I think the reason for your confusion is an algebraic one.

Recall that $$\mathbb{Z} \times \mathbb{Z}$$ has two generators, $$a= (1, 0)$$ and $$b =(0, 1)$$. Choose an isomorphism $$\psi : \pi_1\left(\mathbb{T}^2, p \right) \to \mathbb{Z} \times \mathbb{Z}$$, by surjectivity there exists path classes, $$[f], [g] \in \pi_1\left(\mathbb{T}^2, p \right)$$ such that $$\psi([f]) = a$$ and $$\psi([g]) =b$$. Then since $$\psi$$ is an isomorphism we have $$[f]$$ and $$[g]$$ to be the two generators of $$\pi_1\left(\mathbb{T}^2, p \right)$$.

Now your question is how the path class of the constant loop $$c_p : I \to \mathbb{T}^2$$ defined by $$c_p(x) = p$$ for all $$x \in I$$, that being $$[c_p] \in \pi_1\left(\mathbb{T}^2, p \right)$$ is generated by $$[f]$$ and $$[g]$$. Well the answer to that is simple: note that $$[c_p] = 1_{\pi_1\left(\mathbb{T}^2, p \right)}$$ that is $$[c_p]$$ is the identity element of $$\pi_1\left(\mathbb{T}^2, p \right)$$. Then recall the following definition that we have for exponents in groups.

Definition: In any group $$(G, \cdot)$$ for any $$x \in G$$ we define $$x^0 = 1_G$$ where $$1_G$$ is the identity element of the group $$(G, \cdot)$$.

Hence since $$[f], [g] \in \pi_1\left(\mathbb{T}^2, p \right)$$ and $$\pi_1\left(\mathbb{T}^2, p \right)$$ is indeed a group, we have $$[f]^0 = [g]^0 = 1_{\pi_1\left(\mathbb{T}^2, p \right)}.$$

Then we have $$\left[c_p\right] = [f]^0 * [g]^0$$ and so the constant path at $$p$$ is indeed generated by the two generators of $$\pi_1\left(\mathbb{T}^2, p \right)$$. And since $$[c_p]$$ is a nullhomotopic loop, since it is a constant loop by definition, the above shows how a product of two non null-homotopic loops yield a null-homotopic loop.

Note that above even though I've gone into quite a bit of detail, the only real fact I'm using is the following algebraic one. If we have a group $$G$$ and we have $$G = \langle A \rangle$$ for some subset $$A \subseteq G$$ then every element $$x \in G$$ can be written as $$x = g_1 \dots g_n \cdot h_1^{-1} \dots h_m^{-1}$$ where $$g_i, h_i \in G$$. In particular if we have $$G = \langle c , d \rangle$$, that is $$G$$ is generated by the two elements $$c$$ and $$d$$ then we can express $$1_G$$ as $$1_G = c^0 \cdot d^0$$.

Your confusion seems to be about the meaning of the word "generate". By definition, if $$G$$ is a group and $$S\subseteq G$$, then the subgroup generated by $$S$$ is the smallest subgroup that contains $$S$$. Since a subgroup always contains the identity element, any subset of $$G$$ (even the empty set!) "generates" the identity element.

• Thanks. By ''generate'' I meant the group operation on the set of generators. In a group generated by the subset $S$, each element can be written in terms of the generators, which is, elements of $S$. So given the null-homotopic loop on the torus, how can it be put in relation to the elements of $S$, which in our case consists of 2 elements ? – user249018 Dec 28 '18 at 16:36
• OK, but the group multiplication is not the only operation in a group! There are two other operations: the identity element and inverses. – Eric Wofsey Dec 28 '18 at 16:39
• In particular, one of the operations of a group is an operation which takes no inputs and outputs the identity element. That's how any set "generates" the identity element. – Eric Wofsey Dec 28 '18 at 16:40
• Given $S\subset G$, more precisely the group generated by $S$ is defined as $<S>=SS^{-1}$. So you are right about inverses. The thing with the identity element is less obvious. One excepts it very probably by definition...But when it comes to the fundamental group of the torus, are you saying that the null-homotopic loop is generated by the empty set ? Or maybe we can say the following: the multiplication of a geneartor and its inverse gives us the constant loop, which itself is homotopic to null-homotopic loops. Thus we generate the null-homotopic loop from each one of the two generators ? – user249018 Dec 28 '18 at 16:55
• I don't know what your notation $SS^{-1}$ is supposed to mean, but it sounds like your definition of "the group generated by $S$" is just wrong (which may not be your fault; you may have been taught a wrong definition!). The correct definition is the one I stated in the answer. An equivalent definition is that the subgroup generated by $S$ is the set of all elements of $G$ that can be obtained by starting with elements of $S$ and repeatedly applying the three operations of the group multiplication, inverses, and the identity element. – Eric Wofsey Dec 28 '18 at 17:10

The subgroup generated by $$a$$ and $$b$$ is the set of all elements that can be written as a sequence that consists of nothing but $$a$$, $$b$$, and their inverses (e.g $$ab$$ or $$b^{-4}a$$). The null sequence is allowed. That is, the empty string (a zero-length sequence) qualifies as "a sequence that consists of nothing but $$a$$, $$b$$, and their inverses"; it does not contain anything, so clearly it does not contain anything other than $$a$$, $$b$$, and their inverses. In an abelian group with two generators, the group is generated by taking the first generator an integer number of times, and then taking the second generator an integer number of times. And zero is an integer. Given two non null-homotopic loops $$a$$ and $$b$$, the constant loop is generated by taking $$a$$ zero times, then taking $$b$$ zero times. If you think of a group in terms of group actions, the identity is generated by not doing anything. Doing nothing at all is, at least as far as mathematicians are concerned, an action. Or, in the words of Geddy Lee, if you choose not to decide, you still have made a choice.


1. ## [SOLVED] trig intergrals

So, this is also probably one of those easy problems, I just don't have a good example to work this out by.
$
\int\cos^2x$

where b= $\pi/2$ a= 0

answer= $\pi/4$

I know I'm suppose to use the half angle identity, i just must be doing it wrong because that's not the answer I'm getting.

Thanks for all your help ^.^

2. Can you show your work to see what it is that you're doing wrong?

$\int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx =$ ${\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots$

3. $cos^2x = \frac{1 + cos2x}{2}$

$\frac{1}{2}\int{dx} +\frac{1}{2}\int{cos2x}{dx} = \frac{x}{2} +\frac{sin2x}{4}$

Hope this helps.

Oops was to slow lol, just wondering how do you place upper and lower bounds in latex?

4. Originally Posted by 11rdc11
$cos^2x = \frac{1 + cos2x}{2}$

$\frac{1}{2}\int{dx} +\frac{1}{2}\int{cos2x}{dx} = \frac{x}{2} +\frac{sin2x}{4}$

Hope this helps.

Oops was to slow lol, just wondering how do you place upper and lower bounds in latex?
Do:

Code:
$$\int_a^b f(x)\,dx$$
To get $\int_a^b f(x)\,dx$

note if we have limits like $-\ln2$ to $\ln2$, then you need to put these within {} such as:

Code:
$$\int_{-\ln(2)}^{\ln2} e^x\,dx$$
will output $\int_{-\ln(2)}^{\ln2} e^x\,dx$

--Chris

5. $\int_{0}^{4} x \: dx$

is produced by: $$\int_{0}^{4} x \: dx$$

_{0} : produces the lower bound
^{4} : produces the upper bound.

This can applied to other symbols as well:
$\sum_{k = 0}^{4} k = 10$ is produced by $$\sum_{k = 0}^{4} k = 10$$

$\frac{x}{2}\bigg|_{0}^{4}$

etc etc.

6. Originally Posted by o_O
Can you show your work to see what it is that you're doing wrong?

$\int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx =$ ${\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots$
Whoa, I totally was thinking that the anti derivative of 1 was 0. That's where I messed up. Silly mistake. Thanks though, I'll remember to post my work next time.

7. $\int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx =$ ${\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots$
There's an easier way to deal with these kind of integrals.

Rule: $\int_a^b f(x)~dx = \int_a^b f(a+b-x)~dx$

I'll solve this step by step for you to understand how it works. It actually takes only a nanosecond to solve this.

$I = \int_0^\frac{\pi}{2}\cos^2x~dx = \int_0^\frac{\pi}{2}\cos^2(\frac{\pi}{2}-x)~dx = \int_0^\frac{\pi}{2}\sin^2x~dx$

Now, $I+I = \int_0^\frac{\pi}{2}\cos^2x~dx + \int_0^\frac{\pi}{2}\sin^2x~dx = \int_0^{\frac{\pi}{2}}1~dx = \frac{\pi}{2}$

$I = \frac{\pi}{4}$


# How rigorous are pictorial proofs?

The standard proof that $|\mathbb{Q}| = \mathbb{|N|}$ is pictorial. I am sure everyone here has seen it. The "zig-zag". I must admit, however, that, although I was "intuitively" convinced by it, I was never entirely satisfied with it because it is not an explicit bijection $f:\mathbb{N} \to \mathbb{Q}$ given by an actual formula. The fact that the proof is correct seems "clear" to us, but this is, again, merely an appeal to intuition. One should note that some of these "proofs by picture" are simply incorrect: see Russell O'Connor's answer here .

I have two questions

Is the pictorial proof that $|\mathbb{Q}| = \mathbb{|N|}$ rigorous by the standards of modern pure mathematics?

For the sake of this question, suppose that there isn't an explicit formla, or that it's too unwieldy to use in practice. After all, even if there is a formula, most of the people who've seen the pictorial argument do not know of it.

Is there an explicit formula for the "pictorial" proof?

There's some minor issues, of course, namely the inclusion of $0$ and variations of the "zig-zag" path, but these are no big deal. A bijection $f:\mathbb{N} \to \mathbb{N} \times \mathbb{N}$ suffices; dealing with negatives, equivalent fractions, etc is trivial.

• The zig-zag proof is not pictorial. You can represent it in a graph, as any function, this is all. Aug 6, 2016 at 6:52
• Yes, there is an explicit function for the bijection. But it's tedious, looks convuluted and it distracts from the purpose of the argument which should be simple. For n,m find the largest $\sum_{i=1}^k = k (k+1)/2 \le n+m$ then (n,m) -> k (k+1)/2 + (n-k) is the bijection. Aug 6, 2016 at 6:56
• That should be sum < n; not <= n+m. It helps to draw a picture but: 1 => 0,0 then 2-3 => (1,0) - (0,1) and so on till $(\sum )+1$ = k (k+1)/2 +1 through $\sum$ + (k+1) maps to (k,0), (k-1,1), (k-2,2).... (0,k). That's the bijection. But it's tedious and not particularly relevant. Aug 6, 2016 at 7:21
• Perhaps the easiest way is to let $a_k = \sum_{i=1} i$, then it's easy to show ever natural $n$ can be written uniquely as $a_k + i; 0 \le i < k+1$ so $n \rightarrow (k -1,i)$ is 1 to 1. Aug 6, 2016 at 7:36
• A pictorial proof isn't rigorous enough but a formulaic calculation isn't nescessary if a descriptive argument can be shown to be unambiguous and consistsnt. The diagonal description as usually presented is ... borderline IMO. But if we explain we can count and group the naturals in groups each group with one more than the previous, i.e. (1)(2,3)(4,5,6)(7,8,9,10)etc. then each group has the same number of members as each of the diagonals that also increases by one, that could be rigorous enough. Aug 6, 2016 at 7:56

I had the same question with the same proof (the zig-zag proof you are mentioning). At some point I decided to produce a formal proof.

Define a bijective function $f\colon \mathbb N \times \mathbb N \to \mathbb N$.

First of all you notice that going zig and then zag only helps intuition. In fact you don't need a "continuous" curve, so it is easier to go zig and the zig again... (so to speak). Then it is easy to count how many points you need to fill the first $k$ diagonals (sum of a arithmetic series: $k(k+1)/2$). The couple $(n,m)$ lies on the diagonal number $k=n+m$ so you easily find: $$f(n,m) = \frac{(n+m)(n+m+1)}{2} + m = \frac{n^2+m^2+2nm+3m+n}{2}.$$

This was a little bit shocking to me! The function I was looking for is as simple as a polynomial... I would have expected some modulus, or some strange discontinuous function.

Nevertheless the algebraic proof that $f$ is bijective is not so simple... but following the intuition of the construction it is easy to write it.

What can we learn from this? The pictorial proof is for sure the best to understand a result and to remember it. Then it might happen that the abstract mathematics is even simpler than our intuition. Not always simple mathematics corresponds to simple pictures.

• However it's not necessary to come up with an explicit formula to prove the function is a bijection. E.g., an iterative or recursive description can suffice, and this is basically what the zig-zag picture does. Aug 6, 2016 at 11:01
• @Kinball, it suffices provided it suffices. Having a fórmula makes it easy to be sure. Aug 6, 2016 at 19:46

I suppose if you really wanted to, you could come up with an explicit bijection associated with the "zig-zag" proof. If that turns out to be difficult, you could come up with a different "zig-zag" that may have a simpler bijection. Although, the "zig-zag" proof is really just providing some intuitive backing to this theorem:

The union of a countable number of finite sets is countable.

Thinking of each diagonal of the "zig-zag" as one of your finite sets, and noting that every rational number has to be in one of those diagonals is sufficient to prove that $|\mathbb{Q}| = |\mathbb{N}|$.

On a more general note, the whole point of a proof is to clearly and correctly convey why a theorem is true. Sometimes it is easiest to convey why through written words, especially when the proof is long, relies on lemmas or the theorems of others, or just has lots of cases. But if there is a clever reason why a theorem is true, some clever "ah-HA" that you just have to see, a "proof by picture" can be much more clear than a formal write-up of a proof. The hope is, though, that after a reader sees a pictorial proof, they should have enough intuition into why the theorem is true to write up a formal proof if they really needed.

After seeing the "zig-zag" proof, do you think you can prove that the union of a countable number of finite sets is countable?

• What about a zig-zag proof that the union of a countable number of countable sets is countable? =P Aug 6, 2016 at 10:28
• @user21820, Every time I've heard the zig-zag proof it is presented as partitioning the rational numbers into finite sets indexed by the sum of their numerator and denominator. But I suppose you can think of them as being infinite sets (indexed by just the numerator or denominator), and that makes sense with the same picture. So I suppose that in this case, the same pictorial proof can inspire distinct formal proofs. :) Aug 6, 2016 at 22:23
• Well I was trying to poke fun at the rigour of such a pictorial proof because it easily makes one think that it works the same for the union of countably many countable sets. It doesn't quite because you need the axiom of countable choice. Personally I like annotating parts of a diagram with quantifiers labelled by the order of quantification, and it would prevent such an error because we would realize that we need $\exists_1,\exists_2,...$. Aug 7, 2016 at 8:40
• @user21820,Mike Actually you need the axiom of countable choice even to prove that the union of countably many finite sets is countable. Is it obvious where countable choice comes in in the pictorial proof? It's in the picture-making itself: you need to place dots on the page, which entails a choice of ordering for the set which is relevant to the "zig"; and such a choice must be made independently for each of the countably many finite sets. Countable choice is not needed for $|\Bbb N|=|\Bbb Q|$ because the ordering is known in advance. (And this is why I don't like pictorial "proofs".) Aug 9, 2016 at 18:50

Munkres' "Topology" gives both the zig-zag intuition, and the actual formula for the bijection, as a good reference.

I read in an article (cannot remember the author, sadly) that proofs are not supposed to be 'entirely' rigorous. 'Proofs' try to convince the reader that it is possible to construct a completely rigorous proof. As in this case, although I too was not satisfied with the zig-zag-proof, it conveys the idea that the bijection does exists (without explicitly writing it out), and thus it is countable.

The point of something like the zig-zag proof is not to be rigorous in itself but instead to convince a mathematician that he or she could easily make the proof rigorous if pressed to do so by the gods of rigour.

Also, you can come up with very succinct surjections from N to Q. For example, map every natural number of the form $2^p 3^q 5^r$ to $(-1)^r p/q$. It's extremely easy now to see that if you give me a rational, there is a natural number which is mapped to it.

The downside with this is that if you are teaching countability you would need to check that "surjection from" is the same as "bijection with", which sometimes hasn't been proved by this stage.

• Except that "surjection from" is not the same as "bijection with". To wit, your function is one but not the other. You can use it as a step in the way to proving a bijection, but that is hardly "the same as". Aug 7, 2016 at 10:03
• Well of course I don't mean all surjections are bijections? For an infinite set $S$, the existence of a surjection $g : \mathbb{N} \to S$ is equivalent to the existence of a bijection betwwen $S$ and $\mathbb{N}$. Aug 7, 2016 at 16:57

Pictures have their strengths and weaknesses, but they're just as rigorous as any other informal type of proof — that is, they may or may not be depending on how well written it is.

And the zig-zag proof is a rather clear depiction of an explicit algorithm for enumerating the rationals.

Regarding your particular doubts, I don't think it's really the picture that's the issue — it's that the function was defined by an algorithm for producing its values, rather than as an arithmetic formula.

Formal proofs can be 'pictures' too; you can develop formal logic in a way where the basic 'data type' is something other than strings of symbols. e.g. graphs of various sorts are often useful.

This doesn't directly address the proof of $|\mathbb{N}|=|\mathbb{Q}|$, but the more abstract notion of pictorial proofs as found in elementary geometry:

Avigad, Dean & Mumma A Formal System for Euclid's Elements: http://repository.cmu.edu/philosophy/61/

They show that certain types of pictorial arguments that occur in Euclid's Elements, while apparently un-rigorous, can be described in a precise manner using formal rules, in a way that the conclusion follows directly from the pictorial "special case". In essence, a particular diagram can be in "general enough position" to allow rigorous conclusions to be drawn from it.

So... take the ordered pair $(n,m)$.

The pictorial and non rigorous concept of the "diagonal" is simply {$(j,k) | j+k=n+m; j\ge 0;k \ge 0$}. Note: there are $(n+m)+1$ terms in this diagonal

The previous diagonals had 1 term, two terms ... so on to $(n+m)$ terms. So the previous diagonals account for $\sum_{i=1}^{n+m} i = \frac {(n+m)(n+m+1)}{2}$ items.

Starting at $(0,n+m)$ end of the diagonal $(n,m)$ is the $n +1$th item of the current diagonal.

So the bijection you want is $(n,m)\rightarrow \frac {(n+m)(n+m+1)}2 + n$.

$(0,0)\implies 0$

$(0,1)\implies 1$

$(1,0) \implies 2$

$(2,0) \implies 3$

$(1,1)\implies 4$

Etc. Algebraically it's probably easy to show this is injective . Suppose (a,b) and (c,d) both map to t.

Case 1: a+b = c+d= K.

Then $\frac {K (K+1)}{2} + a = \frac {K (K+1)}2 + b$. So $a = c$. So $b = c =K -a$.

Case 2: Wolog a+b < c+d

$\frac {(a+b)(a+b+1)}2 + a \le \frac {(a+b)(a+b+1)}2 + (a+b)$

$=\frac {(a+b)(a+b+3)}2$

$\le \frac {((c+d)-1)((c+d+1)+1)}2$

$=\frac {(c+d)(c+d +1) +(c+d) -(c+d+1)-1}2$

$=\frac{(c+d)(c+d+1)}2 -1$

$< \frac{(c+d)(c+d+1)}2 + c$

So (a,b)=(c,d).

Proving it's surjective isn't nescessary though it is.

For each t there exist, k, so that $\frac {k (k+1)}2 \le t < \frac{(k+1)(k+2)}2$.

Let $n = t -\frac {k (k+1)}2 \ge 0$. Let $m=k-n = k - t + \frac {k (k+1)}2 > k - \frac {(k+1)(k+2)}2 + \frac {k (k+1)}2=k +\frac {k+1}2 (k-(k+2))=k - (k+1)=-1$. So $m \ge 0$

Then $(n,m)\rightarrow t$.

===

Anyway....

I don't think such an intensive arithmetic is nescessary. What matters is the argument that it can be done. As the diagonals increase by one each iteration, and as we can group the natural numbers into sequences of groups each increasing in terms by one, each group of natural numbers coresponds to a diagonal, and each of there natural numbers in the group corresponds to a term in the diagonal. Thus must be shown but it needn't be shown by convoluted arithmetic.