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Spline interpolation
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<h2 id="introduction">Introduction</h2> <p>Originally, <i><a href="/facts/Flat_spline/gWl5ceBu">spline</a></i> was a term for elastic <a href="/facts/Ruler/jKWAdmso">rulers</a> that were bent to pass through a number of predefined points, or <i>knots</i>. These were used to make <a href="/facts/Technical_drawing/WtTmaUwL">technical drawings</a> for <a href="/facts/Shipbuilding/XLy9lj1W">shipbuilding</a> and construction by hand, as illustrated in the figure. </p><p>We wish to model similar kinds of curves using a set of mathematical equations. Assume we have a sequence of n + 1 {\displaystyle n+1} knots, ( x 0 , y 0 ) {\displaystyle (x_{0},y_{0})} through ( x n , y n ) {\displaystyle (x_{n},y_{n})} . There will be a cubic polynomial q i ( x ) = y {\displaystyle q_{i}(x)=y} between each successive pair of knots ( x i − 1 , y i − 1 ) {\displaystyle (x_{i-1},y_{i-1})} and ( x i , y i ) {\displaystyle (x_{i},y_{i})} connecting to both of them, where i = 1 , 2 , … , n {\displaystyle i=1,2,\dots ,n} . So there will be n {\displaystyle n} polynomials, with the first polynomial starting at ( x 0 , y 0 ) {\displaystyle (x_{0},y_{0})} , and the last polynomial ending at ( x n , y n ) {\displaystyle (x_{n},y_{n})} . </p><p>The <a href="/facts/Curvature/RFCAEsj8">curvature</a> of any curve y = y ( x ) {\displaystyle y=y(x)} is defined as </p> κ = y ″ ( 1 + y ′ 2 ) 3 / 2 , {\displaystyle \kappa ={\frac {y''}{(1+y'^{2})^{3/2}}},} <p>where y ′ {\displaystyle y'} and y ″ {\displaystyle y''} are the first and second derivatives of y ( x ) {\displaystyle y(x)} with respect to x {\displaystyle x} . To make the spline take a shape that minimizes the bending (under the constraint of passing through all knots), we will define both y ′ {\displaystyle y'} and y ″ {\displaystyle y''} to be continuous everywhere, including at the knots. Each successive polynomial must have equal values (which are equal to the y-value of the corresponding datapoint), derivatives, and second derivatives at their joining knots, which is to say that </p> { q i ( x i ) = q i + 1 ( x i ) = y i q i ′ ( x i ) = q i + 1 ′ ( x i ) q i ″ ( x i ) = q i + 1 ″ ( x i ) 1 ≤ i ≤ n − 1. {\displaystyle {\begin{cases}q_{i}(x_{i})=q_{i+1}(x_{i})=y_{i}\\q'_{i}(x_{i})=q'_{i+1}(x_{i})\\q''_{i}(x_{i})=q''_{i+1}(x_{i})\end{cases}}\qquad 1\leq i\leq n-1.} <p>This can only be achieved if polynomials of degree 3 (cubic polynomials) or higher are used. The classical approach is to use polynomials of exactly degree 3 — <a href="/facts/Cubic_spline/S29nwRUF">cubic splines</a>. </p><p>In addition to the three conditions above, a <i>natural cubic spline</i> has the condition that q 1 ″ ( x 0 ) = q n ″ ( x n ) = 0 {\displaystyle q''_{1}(x_{0})=q''_{n}(x_{n})=0} . </p><p>In addition to the three main conditions above, a <i>clamped cubic spline</i> has the conditions that q 1 ′ ( x 0 ) = f ′ ( x 0 ) {\displaystyle q'_{1}(x_{0})=f'(x_{0})} and q n ′ ( x n ) = f ′ ( x n ) {\displaystyle q'_{n}(x_{n})=f'(x_{n})} where f ′ ( x ) {\displaystyle f'(x)} is the derivative of the interpolated function. </p><p>In addition to the three main conditions above, a <i>not-a-knot spline</i> has the conditions that q 1 ‴ ( x 1 ) = q 2 ‴ ( x 1 ) {\displaystyle q'''_{1}(x_{1})=q'''_{2}(x_{1})} and q n − 1 ‴ ( x n − 1 ) = q n ‴ ( x n − 1 ) {\displaystyle q'''_{n-1}(x_{n-1})=q'''_{n}(x_{n-1})} .<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> </p> <h2 id="algorithm-to-find-the-interpolating-cubic-spline">Algorithm to find the interpolating cubic spline</h2> <p>We wish to find each polynomial q i ( x ) {\displaystyle q_{i}(x)} given the points ( x 0 , y 0 ) {\displaystyle (x_{0},y_{0})} through ( x n , y n ) {\displaystyle (x_{n},y_{n})} . To do this, we will consider just a single piece of the curve, q ( x ) {\displaystyle q(x)} , which will interpolate from ( x 1 , y 1 ) {\displaystyle (x_{1},y_{1})} to ( x 2 , y 2 ) {\displaystyle (x_{2},y_{2})} . This piece will have slopes k 1 {\displaystyle k_{1}} and k 2 {\displaystyle k_{2}} at its endpoints. Or, more precisely, </p> q ( x 1 ) = y 1 , {\displaystyle q(x_{1})=y_{1},} q ( x 2 ) = y 2 , {\displaystyle q(x_{2})=y_{2},} q ′ ( x 1 ) = k 1 , {\displaystyle q'(x_{1})=k_{1},} q ′ ( x 2 ) = k 2 . {\displaystyle q'(x_{2})=k_{2}.} <p>The full equation q ( x ) {\displaystyle q(x)} can be written in the symmetrical form </p> <table><tbody><tr><td> q ( x ) = ( 1 − t ( x ) ) y 1 + t ( x ) y 2 + t ( x ) ( 1 − t ( x ) ) ( ( 1 − t ( x ) ) a + t ( x ) b ) , {\displaystyle q(x)={\big (}1-t(x){\big )}\,y_{1}+t(x)\,y_{2}+t(x){\big (}1-t(x){\big )}{\Big (}{\big (}1-t(x){\big )}\,a+t(x)\,b{\Big )},} </td> <td></td> <td>1</td></tr></tbody></table> <p>where </p> <table><tbody><tr><td> t ( x ) = x − x 1 x 2 − x 1 , {\displaystyle t(x)={\frac {x-x_{1}}{x_{2}-x_{1}}},} </td> <td></td> <td>2</td></tr></tbody></table> <table><tbody><tr><td> a = k 1 ( x 2 − x 1 ) − ( y 2 − y 1 ) , {\displaystyle a=k_{1}(x_{2}-x_{1})-(y_{2}-y_{1}),} </td> <td></td> <td>3</td></tr></tbody></table> <table><tbody><tr><td> b = − k 2 ( x 2 − x 1 ) + ( y 2 − y 1 ) . {\displaystyle b=-k_{2}(x_{2}-x_{1})+(y_{2}-y_{1}).} </td> <td></td> <td>4</td></tr></tbody></table> <p>But what are k 1 {\displaystyle k_{1}} and k 2 {\displaystyle k_{2}} ? To derive these critical values, we must consider that </p> q ′ = d q d x = d q d t d t d x = d q d t 1 x 2 − x 1 . {\displaystyle q'={\frac {dq}{dx}}={\frac {dq}{dt}}{\frac {dt}{dx}}={\frac {dq}{dt}}{\frac {1}{x_{2}-x_{1}}}.} <p>It then follows that </p> <table><tbody><tr><td> q ′ = y 2 − y 1 x 2 − x 1 + ( 1 − 2 t ) a ( 1 − t ) + b t x 2 − x 1 + t ( 1 − t ) b − a x 2 − x 1 , {\displaystyle q'={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}+(1-2t){\frac {a(1-t)+bt}{x_{2}-x_{1}}}+t(1-t){\frac {b-a}{x_{2}-x_{1}}},} </td> <td></td> <td>5</td></tr></tbody></table> <table><tbody><tr><td> q ″ = 2 b − 2 a + ( a − b ) 3 t ( x 2 − x 1 ) 2 . {\displaystyle q''=2{\frac {b-2a+(a-b)3t}{{(x_{2}-x_{1})}^{2}}}.} </td> <td></td> <td>6</td></tr></tbody></table> <p>Setting <i>t</i> = <i>0</i> and <i>t</i> = <i>1</i> respectively in equations (5) and (6), one gets from (2) that indeed first derivatives <i>q′</i>(<i>x</i>1) = <i>k</i>1 and <i>q′</i>(<i>x</i>2) = <i>k</i>2, and also second derivatives </p> <table><tbody><tr><td> q ″ ( x 1 ) = 2 b − 2 a ( x 2 − x 1 ) 2 , {\displaystyle q''(x_{1})=2{\frac {b-2a}{{(x_{2}-x_{1})}^{2}}},} </td> <td></td> <td>7</td></tr></tbody></table> <table><tbody><tr><td> q ″ ( x 2 ) = 2 a − 2 b ( x 2 − x 1 ) 2 . {\displaystyle q''(x_{2})=2{\frac {a-2b}{{(x_{2}-x_{1})}^{2}}}.} </td> <td></td> <td>8</td></tr></tbody></table> <p>If now (<i>xi</i>, <i>yi</i>), <i>i</i> = 0, 1, ..., <i>n</i> are <i>n</i> + 1 points, and </p> <table><tbody><tr><td> q i = ( 1 − t ) y i − 1 + t y i + t ( 1 − t ) ( ( 1 − t ) a i + t b i ) , {\displaystyle q_{i}=(1-t)\,y_{i-1}+t\,y_{i}+t(1-t){\big (}(1-t)\,a_{i}+t\,b_{i}{\big )},} </td> <td></td> <td>9</td></tr></tbody></table> <p>where <i>i</i> = 1, 2, ..., <i>n</i>, and t = x − x i − 1 x i − x i − 1 {\displaystyle t={\tfrac {x-x_{i-1}}{x_{i}-x_{i-1}}}} are <i>n</i> third-degree polynomials interpolating y in the interval <i>x</i><i>i</i>−1 ≤ <i>x</i> ≤ <i>xi</i> for <i>i</i> = 1, ..., <i>n</i> such that <i>q′i</i> (<i>xi</i>) = <i>q′</i><i>i</i>+1(<i>xi</i>) for <i>i</i> = 1, ..., <i>n</i> − 1, then the <i>n</i> polynomials together define a differentiable function in the interval <i>x</i>0 ≤ <i>x</i> ≤ <i>xn</i>, and </p> <table><tbody><tr><td> a i = k i − 1 ( x i − x i − 1 ) − ( y i − y i − 1 ) , {\displaystyle a_{i}=k_{i-1}(x_{i}-x_{i-1})-(y_{i}-y_{i-1}),} </td> <td></td> <td>10</td></tr></tbody></table> <table><tbody><tr><td> b i = − k i ( x i − x i − 1 ) + ( y i − y i − 1 ) {\displaystyle b_{i}=-k_{i}(x_{i}-x_{i-1})+(y_{i}-y_{i-1})} </td> <td></td> <td>11</td></tr></tbody></table> <p>for <i>i</i> = 1, ..., <i>n</i>, where </p> <table><tbody><tr><td> k 0 = q 1 ′ ( x 0 ) , {\displaystyle k_{0}=q_{1}'(x_{0}),} </td> <td></td> <td>12</td></tr></tbody></table> <table><tbody><tr><td> k i = q i ′ ( x i ) = q i + 1 ′ ( x i ) , i = 1 , … , n − 1 , {\displaystyle k_{i}=q_{i}'(x_{i})=q_{i+1}'(x_{i}),\qquad i=1,\dots ,n-1,} </td> <td></td> <td>13</td></tr></tbody></table> <table><tbody><tr><td> k n = q n ′ ( x n ) . {\displaystyle k_{n}=q_{n}'(x_{n}).} </td> <td></td> <td>14</td></tr></tbody></table> <p>If the sequence <i>k</i>0, <i>k</i>1, ..., <i>kn</i> is such that, in addition, <i>q′′i</i>(<i>xi</i>) = <i>q′′</i><i>i</i>+1(<i>xi</i>) holds for <i>i</i> = 1, ..., <i>n</i> − 1, then the resulting function will even have a continuous second derivative. </p><p>From (7), (8), (10) and (11) follows that this is the case if and only if </p> <table><tbody><tr><td> k i − 1 x i − x i − 1 + ( 1 x i − x i − 1 + 1 x i + 1 − x i ) 2 k i + k i + 1 x i + 1 − x i = 3 ( y i − y i − 1 ( x i − x i − 1 ) 2 + y i + 1 − y i ( x i + 1 − x i ) 2 ) {\displaystyle {\frac {k_{i-1}}{x_{i}-x_{i-1}}}+\left({\frac {1}{x_{i}-x_{i-1}}}+{\frac {1}{x_{i+1}-x_{i}}}\right)2k_{i}+{\frac {k_{i+1}}{x_{i+1}-x_{i}}}=3\left({\frac {y_{i}-y_{i-1}}{{(x_{i}-x_{i-1})}^{2}}}+{\frac {y_{i+1}-y_{i}}{{(x_{i+1}-x_{i})}^{2}}}\right)} </td> <td></td> <td>15</td></tr></tbody></table> <p>for <i>i</i> = 1, ..., <i>n</i> − 1. The relations (15) are <i>n</i> − 1 linear equations for the <i>n</i> + 1 values <i>k</i>0, <i>k</i>1, ..., <i>kn</i>. </p><p>For the elastic rulers being the model for the spline interpolation, one has that to the left of the left-most "knot" and to the right of the right-most "knot" the ruler can move freely and will therefore take the form of a straight line with <i>q′′</i> = 0. As q′′ should be a continuous function of x, "natural splines" in addition to the <i>n</i> − 1 linear equations (15) should have </p> q 1 ″ ( x 0 ) = 2 3 ( y 1 − y 0 ) − ( k 1 + 2 k 0 ) ( x 1 − x 0 ) ( x 1 − x 0 ) 2 = 0 , {\displaystyle q''_{1}(x_{0})=2{\frac {3(y_{1}-y_{0})-(k_{1}+2k_{0})(x_{1}-x_{0})}{{(x_{1}-x_{0})}^{2}}}=0,} q n ″ ( x n ) = − 2 3 ( y n − y n − 1 ) − ( 2 k n + k n − 1 ) ( x n − x n − 1 ) ( x n − x n − 1 ) 2 = 0 , {\displaystyle q''_{n}(x_{n})=-2{\frac {3(y_{n}-y_{n-1})-(2k_{n}+k_{n-1})(x_{n}-x_{n-1})}{{(x_{n}-x_{n-1})}^{2}}}=0,} <p>i.e. that </p> <table><tbody><tr><td> 2 x 1 − x 0 k 0 + 1 x 1 − x 0 k 1 = 3 y 1 − y 0 ( x 1 − x 0 ) 2 , {\displaystyle {\frac {2}{x_{1}-x_{0}}}k_{0}+{\frac {1}{x_{1}-x_{0}}}k_{1}=3{\frac {y_{1}-y_{0}}{(x_{1}-x_{0})^{2}}},} </td> <td></td> <td>16</td></tr></tbody></table> <table><tbody><tr><td> 1 x n − x n − 1 k n − 1 + 2 x n − x n − 1 k n = 3 y n − y n − 1 ( x n − x n − 1 ) 2 . {\displaystyle {\frac {1}{x_{n}-x_{n-1}}}k_{n-1}+{\frac {2}{x_{n}-x_{n-1}}}k_{n}=3{\frac {y_{n}-y_{n-1}}{(x_{n}-x_{n-1})^{2}}}.} </td> <td></td> <td>17</td></tr></tbody></table> <p>Eventually, (15) together with (16) and (17) constitute <i>n</i> + 1 linear equations that uniquely define the <i>n</i> + 1 parameters <i>k</i>0, <i>k</i>1, ..., <i>kn</i>. </p><p>There exist other end conditions, "clamped spline", which specifies the slope at the ends of the spline, and the popular "not-a-knot spline", which requires that the third derivative is also continuous at the <i>x</i>1 and <i>x</i><i>n</i>−1 points. For the "not-a-knot" spline, the additional equations will read: </p> q 1 ‴ ( x 1 ) = q 2 ‴ ( x 1 ) ⇒ 1 Δ x 1 2 k 0 + ( 1 Δ x 1 2 − 1 Δ x 2 2 ) k 1 − 1 Δ x 2 2 k 2 = 2 ( Δ y 1 Δ x 1 3 − Δ y 2 Δ x 2 3 ) , {\displaystyle q'''_{1}(x_{1})=q'''_{2}(x_{1})\Rightarrow {\frac {1}{\Delta x_{1}^{2}}}k_{0}+\left({\frac {1}{\Delta x_{1}^{2}}}-{\frac {1}{\Delta x_{2}^{2}}}\right)k_{1}-{\frac {1}{\Delta x_{2}^{2}}}k_{2}=2\left({\frac {\Delta y_{1}}{\Delta x_{1}^{3}}}-{\frac {\Delta y_{2}}{\Delta x_{2}^{3}}}\right),} q n − 1 ‴ ( x n − 1 ) = q n ‴ ( x n − 1 ) ⇒ 1 Δ x n − 1 2 k n − 2 + ( 1 Δ x n − 1 2 − 1 Δ x n 2 ) k n − 1 − 1 Δ x n 2 k n = 2 ( Δ y n − 1 Δ x n − 1 3 − Δ y n Δ x n 3 ) , {\displaystyle q'''_{n-1}(x_{n-1})=q'''_{n}(x_{n-1})\Rightarrow {\frac {1}{\Delta x_{n-1}^{2}}}k_{n-2}+\left({\frac {1}{\Delta x_{n-1}^{2}}}-{\frac {1}{\Delta x_{n}^{2}}}\right)k_{n-1}-{\frac {1}{\Delta x_{n}^{2}}}k_{n}=2\left({\frac {\Delta y_{n-1}}{\Delta x_{n-1}^{3}}}-{\frac {\Delta y_{n}}{\Delta x_{n}^{3}}}\right),} <p>where Δ x i = x i − x i − 1 , Δ y i = y i − y i − 1 {\displaystyle \Delta x_{i}=x_{i}-x_{i-1},\ \Delta y_{i}=y_{i}-y_{i-1}} . </p> <h2 id="example">Example</h2> <p>In case of three points the values for k 0 , k 1 , k 2 {\displaystyle k_{0},k_{1},k_{2}} are found by solving the <a href="/facts/Tridiagonal_matrix/1SK5A0rP">tridiagonal linear equation system</a> </p> [ a 11 a 12 0 a 21 a 22 a 23 0 a 32 a 33 ] [ k 0 k 1 k 2 ] = [ b 1 b 2 b 3 ] {\displaystyle {\begin{bmatrix}a_{11}&a_{12}&0\\a_{21}&a_{22}&a_{23}\\0&a_{32}&a_{33}\\\end{bmatrix}}{\begin{bmatrix}k_{0}\\k_{1}\\k_{2}\\\end{bmatrix}}={\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\\\end{bmatrix}}} <p>with </p> a 11 = 2 x 1 − x 0 , {\displaystyle a_{11}={\frac {2}{x_{1}-x_{0}}},} a 12 = 1 x 1 − x 0 , {\displaystyle a_{12}={\frac {1}{x_{1}-x_{0}}},} a 21 = 1 x 1 − x 0 , {\displaystyle a_{21}={\frac {1}{x_{1}-x_{0}}},} a 22 = 2 ( 1 x 1 − x 0 + 1 x 2 − x 1 ) , {\displaystyle a_{22}=2\left({\frac {1}{x_{1}-x_{0}}}+{\frac {1}{x_{2}-x_{1}}}\right),} a 23 = 1 x 2 − x 1 , {\displaystyle a_{23}={\frac {1}{x_{2}-x_{1}}},} a 32 = 1 x 2 − x 1 , {\displaystyle a_{32}={\frac {1}{x_{2}-x_{1}}},} a 33 = 2 x 2 − x 1 , {\displaystyle a_{33}={\frac {2}{x_{2}-x_{1}}},} b 1 = 3 y 1 − y 0 ( x 1 − x 0 ) 2 , {\displaystyle b_{1}=3{\frac {y_{1}-y_{0}}{(x_{1}-x_{0})^{2}}},} b 2 = 3 ( y 1 − y 0 ( x 1 − x 0 ) 2 + y 2 − y 1 ( x 2 − x 1 ) 2 ) , {\displaystyle b_{2}=3\left({\frac {y_{1}-y_{0}}{{(x_{1}-x_{0})}^{2}}}+{\frac {y_{2}-y_{1}}{{(x_{2}-x_{1})}^{2}}}\right),} b 3 = 3 y 2 − y 1 ( x 2 − x 1 ) 2 . {\displaystyle b_{3}=3{\frac {y_{2}-y_{1}}{(x_{2}-x_{1})^{2}}}.} <p>For the three points </p> ( − 1 , 0.5 ) , ( 0 , 0 ) , ( 3 , 3 ) , {\displaystyle (-1,0.5),\ (0,0),\ (3,3),} <p>one gets that </p> k 0 = − 0.6875 , k 1 = − 0.1250 , k 2 = 1.5625 , {\displaystyle k_{0}=-0.6875,\ k_{1}=-0.1250,\ k_{2}=1.5625,} <p>and from (10) and (11) that </p> a 1 = k 0 ( x 1 − x 0 ) − ( y 1 − y 0 ) = − 0.1875 , {\displaystyle a_{1}=k_{0}(x_{1}-x_{0})-(y_{1}-y_{0})=-0.1875,} b 1 = − k 1 ( x 1 − x 0 ) + ( y 1 − y 0 ) = − 0.3750 , {\displaystyle b_{1}=-k_{1}(x_{1}-x_{0})+(y_{1}-y_{0})=-0.3750,} a 2 = k 1 ( x 2 − x 1 ) − ( y 2 − y 1 ) = − 3.3750 , {\displaystyle a_{2}=k_{1}(x_{2}-x_{1})-(y_{2}-y_{1})=-3.3750,} b 2 = − k 2 ( x 2 − x 1 ) + ( y 2 − y 1 ) = − 1.6875. {\displaystyle b_{2}=-k_{2}(x_{2}-x_{1})+(y_{2}-y_{1})=-1.6875.} <p>In the figure, the spline function consisting of the two cubic polynomials q 1 ( x ) {\displaystyle q_{1}(x)} and q 2 ( x ) {\displaystyle q_{2}(x)} given by (9) is displayed. </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Akima_spline/G1UXSX2d">Akima spline</a></li> <li><a href="/facts/Circular_interpolation/6CSNdb6t">Circular interpolation</a></li> <li><a href="/facts/Cubic_Hermite_spline/S29nwRUF">Cubic Hermite spline</a></li> <li><a href="/facts/Centripetal_Catmull%25E2%2580%2593Rom_spline/SfHV0CD3">Centripetal Catmull–Rom spline</a></li> <li><a href="/facts/Discrete_spline_interpolation/gwMJdb5b">Discrete spline interpolation</a></li> <li><a href="/facts/Monotone_cubic_interpolation/3LlrJ032">Monotone cubic interpolation</a></li> <li><a href="/facts/Non-uniform_rational_B-spline/MA0NjmeC">Non-uniform rational B-spline</a></li> <li><a href="/facts/Multivariate_interpolation/qj0F4Jze">Multivariate interpolation</a></li> <li><a href="/facts/Polynomial_interpolation/DmnuxaKT">Polynomial interpolation</a></li> <li><a href="/facts/Smoothing_spline/oHeBXmE8">Smoothing spline</a></li> <li><a href="/facts/Spline_wavelet/b6k6wH6e">Spline wavelet</a></li> <li><a href="/facts/Thin_plate_spline/FIalQnKv">Thin plate spline</a></li> <li><a href="/facts/Polyharmonic_spline/ge8mYmnU">Polyharmonic spline</a></li></ul> <h2 id="further-reading">Further reading</h2> <ul><li>Schoenberg, Isaac J. (1946). <a href="https://doi.org/10.1090%2Fqam%2F15914">"Contributions to the Problem of Approximation of Equidistant Data by Analytic Functions:Part A.—On the Problem of Smoothing or Graduation. A First Class of Analytic Approximation Formulae"</a>. <i>Quarterly of Applied Mathematics</i>. 4 (2): 45–99. <a href="/facts/Doi_(identifier)/muM9Etpq">doi</a>:<a href="https://doi.org/10.1090%2Fqam%2F15914">10.1090/qam/15914</a>.</li> <li>Schoenberg, Isaac J. (1946). <a href="https://doi.org/10.1090%2Fqam%2F16705">"Contributions to the Problem of Approximation of Equidistant Data by Analytic Functions:Part B.—On the Problem of Osculatory Interpolation. A Second Class of Analytic Approximation Formulae"</a>. <i>Quarterly of Applied Mathematics</i>. 4 (2): 112–141. <a href="/facts/Doi_(identifier)/muM9Etpq">doi</a>:<a href="https://doi.org/10.1090%2Fqam%2F16705">10.1090/qam/16705</a>.</li></ul> <h2 id="external-links">External links</h2> <ul><li><a href="http://tools.timodenk.com/?p=cubic-spline-interpolation">Cubic Spline Interpolation Online Calculation and Visualization Tool (with JavaScript source code)</a></li> <li><a href="https://www.encyclopediaofmath.org/index.php?title=Spline_interpolation">"Spline interpolation"</a>, <i><a href="/facts/Encyclopedia_of_Mathematics/WC6mGtPm">Encyclopedia of Mathematics</a></i>, <a href="/facts/European_Mathematical_Society/B3h7b672">EMS Press</a>, 2001 [1994]</li> <li><a href="http://jsxgraph.uni-bayreuth.de/wiki/index.php/Cubic_spline_interpolation">Dynamic cubic splines with JSXGraph</a></li> <li><a href="https://www.youtube.com/view_play_list?p=DAB608CD1A9A0D55">Lectures on the theory and practice of spline interpolation</a></li> <li><a href="https://web.archive.org/web/20090408054627/http://online.redwoods.cc.ca.us/instruct/darnold/laproj/Fall98/SkyMeg/Proj.PDF">Paper which explains step by step how cubic spline interpolation is done, but only for equidistant knots.</a></li> <li><a href="http://apps.nrbook.com/c/index.html">Numerical Recipes in C, Go to Chapter 3 Section 3-3</a></li> <li><a href="http://www.cs.tau.ac.il/~turkel/notes/numeng/spline_note.pdf">A note on cubic splines</a></li> <li><a href="https://websites.pmc.ucsc.edu/~fnimmo/eart290c_17/NumericalRecipesinF77.pdf">Information about spline interpolation (including code in Fortran 77)</a></li> <li><a href="https://github.com/msteinbeck/tinyspline">TinySpline:Open source C-library for splines which implements cubic spline interpolation</a></li> <li><a href="https://docs.scipy.org/doc/scipy/tutorial/interpolate.html">SciPy Spline Interpolation:a Python package that implements interpolation</a></li> <li><a href="https://github.com/ValexCorp/Cubic-Interpolation">Cubic Interpolation:Open source C#-library for cubic spline interpolation</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Hall, Charles A.; Meyer, Weston W. (1976). "Optimal Error Bounds for Cubic Spline Interpolation". Journal of Approximation Theory. 16 (2): 105–122. doi:10.1016/0021-9045(76)90040-X. <a href="https://doi.org/10.1016%2F0021-9045%2876%2990040-X" target="_blank">https://doi.org/10.1016%2F0021-9045%2876%2990040-X</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Burden, Richard; Faires, Douglas (2015). Numerical Analysis (10th ed.). Cengage Learning. pp. 142–157. ISBN 9781305253667. <a href="9781305253667" target="_blank">9781305253667</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> </ol>