Interpolation Formula Online

Interpolation Formula Online: Interpolation is the development of finding a value between two stops on a line or curve. To help us commemorate what it means, we should think of the first part of the word, ‘inter,’ as meaning ‘enter,’ which reminds us to look ‘inside’ the data we basically had. This tool, interpolation, is not only profitable in statistics, but is also useful in science, business or any time there is a need to predict profits that fall within two existing data points.

Here’s an illustration that will illustrate the concept of interpolation. A gardener planted a tomato plant and she computed and kept track of its prosperity every other day. This gardener is an inquisitive person, and she would like to estimate how tall her plant was on the fourth day.

Interpolation Formula Online

Interpolation Formula

In the analytical field of numerical analysis, interpolation is a type of estimation, a method of manufacturing new data points within the range of a disconnected set of known data points.

In engineering and science, one often has a number of data points, obtained by sampling or experiment, which represent the values of a function for a limited number of values of the independent variable. It is often required to interpolate, i.e., estimate the value of that function for an intermediate value of the independent variable.

A closely related problem is the approximation of a troublesome function by a simple function. Suppose the formula for some given function is known, but too troublesome to evaluate efficiently. A few data points from the original function can be interpolated to produce a simpler function which is still fairly close to the original. The resulting gain in simplicity may counterbalance the loss from interpolation error.

Linear Interpolation Formula

Linear Interpolation Formula

The method of finding new values for any gathering using the set of values is done by interpolation. The unknown value on a point is found out using this formula. If linear interpolation formula is dealt with then it should be used to find the new value from the two given points. If compared to Lagrange’s interpolation creed, the “n” set of numbers should be available and Lagrange’s technique is to be used to find the new value.

The following is  Linear Interpolation Formula

y=y1+(xx1)(x2x1)×(y2y1)

The La-grange’s Interpolation Rubric is given as,

y=(xx1)(xx2)(xxn)(xox1)(x0x2).
(x0xn)y+(xx0)(xxn).(xxn)(x1x0)(x1x2.
((x1xn)y1+(xx1)(xx2).(xxn)(xox1)(x0x2)..(xoxn)yn

 Some solved problems on interpolation are given below:

Solved Examples

Question 1: Using the interpolation custom, find the value of y at x = 8 given some set of values (2, 6), (5, 9) ?
Solution:

The known values are,x0=8,
x1=2,
x2=5,
y1=6,

y2=9y=y1+(xx1)(x2x1)×(y2y1)

y=6+((82)(52)×(96)

y = 6 + 6
y = 12

Interpolation Formula Excel

If the two known points are given by the coordinates {\displaystyle (x_{0},y_{0})} and {\displaystyle (x_{1},y_{1})}, the linear interpolant is the straight line between these points. For a value x in the interval {\displaystyle (x_{0},x_{1})}, the value y along the straightforward line is given from the equation of slopes

{\displaystyle {\frac {y-y_{0}}{x-x_{0}}}={\frac {y_{1}-y_{0}}{x_{1}-x_{0}}},}

which can be derived geometrically from the intelligence on the right. It is a special case of polynomial interpolation with n = 1.

Determining this equation for y, which is the unknown value at x, gives

{\displaystyle y=y_{0}+(x-x_{0}){\frac {y_{1}-y_{0}}{x_{1}-x_{0}}}={\frac {y_{0}(x_{1}-x)+y_{1}(x-x_{0})}{x_{1}-x_{0}}},}

which is the creed for linear interpolation in the interval {\displaystyle (x_{0},x_{1})}. Outside this interval, the formula is interchangeable to linear extrapolation.

This formula can also be understood as a weighted average. The weights are inversely related to the length from the end points to the unknown point; the closer point has more influence than the farther point. Thus, the weights are {\textstyle {\frac {x-x_{0}}{x_{1}-x_{0}}}} and {\textstyle {\frac {x_{1}-x}{x_{1}-x_{0}}}}, which are normalized distances between the unknown point and each of the end points. Because these sum to 1,

{\displaystyle y=y_{0}\left(1-{\frac {x-x_{0}}{x_{1}-x_{0}}}\right)+y_{1}\left(1-{\frac {x_{1}-x}{x_{1}-x_{0}}}\right)=y_{0}\left(1-{\frac {x-x_{0}}{x_{1}-x_{0}}}\right)+y_{1}\left({\frac {x-x_{0}}{x_{1}-x_{0}}}\right),}

which yields the formula for linear interpolation given above.

Lagrange Interpolation Formula

Interpolation is a method for estimating the value of a function between two known values. Often some relationship is measured experimentally or traced with Dagra at a range of values. Interpolation can be used to estimate the function for untabulated points.

For example, suppose we have tabulated data for the thermal resistance of a transistor tabulated for air velocity from 0 to 1800 FPM in 200 FPM steps. Interpolation can be used to estimate the thermal resistance at non-tabulated values such as 485 FPM.

The linear interpolation equation above can be implemented directly in Microsoft Excel provided the tabulated values are monotonic in x, that is the x-values are sorted and no two are equal. The online Microwave Encyclopedia has the full 6 line implementation along with a good explanation of how it works.

However, here is a simpler implementation for Excel:
=FORECAST(NewX,OFFSET(KnownY,MATCH(NewX,KnownX,1)-1,0,2), OFFSET(KnownX,MATCH(NewX,KnownX,1)-1,0,2))
To use it either:

  1. Copy the formula above into Excel and replace KnownX and KnownY with the cell reference for the tabulated x and y values and NewX with the x-value to interpolate, OR
  2. Define names for the KnownX and KnownY ranges (Insert→Name→Define… in Excel 2003) and replace NewX with the x-value to interpolate.

Interpolation Formula Calculator

The simple implementation is easiest to understand by dissecting from the outside and working in. Here’s the full equation:
=FORECAST(NewX,OFFSET(KnownY,MATCH(NewX,KnownX,1)-1,0,2), OFFSET(KnownX,MATCH(NewX,KnownX,1)-1,0,2))
In brief, the equation consists of 3 parts:

  1. the FORECAST function to calculate the linear interpolation,
  2. two calls to the MATCH function to find the tabulated x-value closest too, but less than the new-x value, and
  3. two calls to the OFFSET function to reference the tabulated x-values and y-values just above and just below the new-x value.

In more detail, the FORECAST function performs the actual interpolation using the linear interpolation equation shown above. Its syntax is: FORECAST(NewXknown_y_pairknown_x_pair).

The first parameter, NewX is simply the value to interpolate. The next two parameters, known_y_pair and known_x_pair are the values either side of NewX. That is, {x1, x2} and {y1, y2} in the diagram above.

The MATCH function is used to find the tabulated x-value just below NewX. Its syntax is: MATCH(lookup_valuelookup_table,match_type)MATCH returns the relative position of an item in a sorted array. So, lookup_value is the value to interpolate, lookup_table is the array of KnownX values, and match_type is 1 to find the largest value in the array that is less than or equal to NewX.

The MATCH function returns an index, but the FORECAST function requires two cell ranges: one for the known_x_pair and one for the known_y_pair. So, the OFFSET function is used twice to create these ranges. Its syntax is OFFSET(reference,row_offsetcolumn_offsetrow_count,column_count). It takes a starting point, the reference, and creates a cell reference with the given offset and size. To obtain the known_y_pair range, the reference is set to the table of KnownYvalues; for the known_x_pair range, reference is set to the array of KnownX values. If the tabulated values are arranged vertically, the row_offset is the result from the MATCH function less 1 and row_count is 2; column_offset is 0 andcolumn_count is 1. This gives us a cell array reference 2 cells high and 1 cell wide. If the tabulated values are arranged horizontally, row and column are switched in the OFFSET function.

How do you calculate the interpolation rate?

Divide the result from Step 1 by the difference between the lengths of the two time periods. For example, the difference between the 60-day time period and the 30-day time period is 30 days. Divide 0.2613 percent by 30 days and the result is 0.00871 percent.

What is linear interpolation method?

In mathematics, linear interpolation is a method of curve fitting using linear polynomials to construct new data points within the range of a discrete set of known data points.

What do you mean by interpolation?

Extrapolation is an estimation of a value based on extending a known sequence of values or facts beyond the area that is certainly known. … Interpolation is an estimation of a value within two known values in a sequence of values. Polynomial interpolation is a method of estimating values between known data points.