 # 2b- Solved problems for quadratic interpolation

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## Solved problems for quadratic interpolation.

Two solved problems for quadratic interpolation are introduced by using the given three points and the corresponding y values to get the coefficients a0,a1, and a2. And then, we can write the quadratic polynomial expression as P(x)=a0+a1x+a2x2.

The last step is the multiply the inverse matrix V-1 by the X- matrix to find the value of the factor column vector. In the end, we have obtained these values, as shown in the last slide image.

These equations can be written in a matrix form. The Form of V*X=Y to be used, where V is the Vandermonde matrix, the column vector for the coefficients of a0,a1, and a2.y is the column vector of y values for the three points.

This is the final value for the coefficients, for more details please refer to the previous post to learn about how to derive this expression. The matrix b is formed with 3×3 and the different factors for all the rows and columns are shown in the next slide image.

### Solved problem 1/2 of the solved problems for quadratic interpolation.

It is required to derive an expression for the quadratic polynomial for a given three points x and y values and to check the P(x) value of x=2.70.

The first step is to find the value of coefficients a0,a1, and a2 by substituting the corresponding elements of the B matrix, considering x0=1, y0=3, x1=2, y1=5, and x2=3 and y2=8. All the rows have different x0,x1,x2, and fractions values.

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Once we have written the three rows of matrix B, we will multiply matrix b by the vector-matrix( y0,y1,y2). The product will give the corresponding value of the coefficients as follows:

a0=2, a1=1/2 and a2=1/2. The quadratic polynomial can be written in the form of P(x)=a0+a1x+a2*x2. The last step is to find P(2.70) which will be=6.995.

### Solved problem 2/2 of the solved problems for quadratic interpolation.

It is required to derive an expression for the quadratic polynomial for a given three points x and y values and to check the P(x) value of x=pi/12.

The given function is (sin x+cos x). the given x values are x0=10 degrees, x1=20 degrees and x2=30 degrees. The first step is to find the value of coefficients a0,a1, and a2 by substituting in the corresponding elements of the B matrix.

After converting the values in degrees to the corresponding values in radians.consider x0=0.1745, y0=1.1585, x1=0.3491, y1=1.2817, and x2=0.5236 and y2=1.3660.

The next slide image explains the values of the first column of matrix B and the values of the different elements.

The next slide image explains the values of the second column of matrix B and the values of the different elements.

The next slide image are explaining the values of the third column of matrix B and the values of the different elements.

Once we have written the three rows of matrix B, we will multiply matrix b by the vector-matrix( y0,y1,y2). the product will give the corresponding value of the coefficients as follows:

a0=0.9968, a1=1.0176 and a2=-0.6009. The quadratic polynomial can be written in the form of P(x)=a0+a1x+a2*x2. The last step is to find P(oi/12) which will be=1.222.

The corresponding value of Pi/12 for the original function is obtained by writing sin (pi/12)+ cos(pi/12)=1.2247, which is having little difference from the quadratic polynomial value.

This is a link to a post that has the title How to Use a matrix for the quadratic function?

This is a link to the pdf file used to illustrate this post.

The next post is Introduction to Newton-divided difference interpolation.

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