Linear Interpolation for Given Function

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In summary, the task at hand is to determine the linear interpolant for the given function with the following values: f(-pi)=4, f(-pi/2)=5/4, f(0)=1, f(pi/2)=-3/4, f(pi)=0. The function itself is (1/pi^2)*(x-pi)^2 - cos^2(x-pi/2). While the initial attempt using Lagrange basis functions did not yield the correct result, it was later discovered that the answer for the first interpolant (interval -pi to -pi/2) is given by the equation 4-11(x+pi)/(2pi). The task ultimately involves finding the equations that connect the
  • #1
ForTheGreater
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Decide the linear interpolant

f(-pi)=4 f(-pi/2)=5/4 f(0)=1 f(pi/2)=-3/4 f(pi)=0

the function is (1/pi2 ) (x-pi)2 - cos2 (x-pi/2)

Don't know how to do this. I tried lagrange basis functions f(x0)(x1-x)/(x1-x0)+f(x1)(x-x0)/(x1-x0)

But it doesn't turn out right.

The answer for the first interpolant (interval -pi to -pi/2) is: 4-11(x+pi)/(2pi)
 
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  • #2
I sense a contradiction between the problem statement "Decide the interpolant" and the book answer "The answer for the first interpolant ... is...".
Could it be that all you are asked to provide is the four equations "connecting the dots" ?

(The book answer for the first section is the equation of the straight line through ##(-\pi, 4)## and ##(-\pi/2, 5/4)##. )

--
 
  • #3
I have the answer here, it wasn't in the problem.

I'm suppose to give the interpolant in all intervals, or as you put it connect the dots.

Thank you!
 

Related to Linear Interpolation for Given Function

1. What is a linear interpolant?

A linear interpolant is a method for estimating values between two known points by creating a straight line that connects the two points. This technique is commonly used in mathematics and science to approximate data points that fall between two measured values.

2. How is the linear interpolant calculated?

The linear interpolant is calculated by finding the slope of the line between two known points and using that slope to determine the value of the unknown point. This can be done using the formula y = mx + b, where m is the slope and b is the y-intercept of the line.

3. What is the purpose of using a linear interpolant?

The purpose of using a linear interpolant is to estimate values that fall between two known data points. This can be useful in situations where only a limited amount of data is available or when precise measurements are not necessary.

4. What are the limitations of using a linear interpolant?

One limitation of using a linear interpolant is that it assumes a linear relationship between the known data points. This may not always be the case and can lead to inaccurate estimations. Additionally, linear interpolation is not suitable for predicting values outside of the range of the known data points.

5. Can the linear interpolant be used for non-linear data?

No, the linear interpolant is only applicable for data that follows a linear trend. For non-linear data, other interpolation methods such as polynomial or spline interpolation may be more appropriate.

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