For the given x_0=0, x1=1 and x2=2,
1)Construct the quadratic Lagrange Interpolation Polinomial for f(x)=3^x
2)Find the error for each x€[0,2]
1) I found the quadratic Lagrange Interpolation Polinomial as,
y_(x)= (2*(x)^2) + 1
So y'(x)=4x
y''(x)=4
y'''(x)=0
I know the error bound(formula) for Lagrange Interpolation Polinomial is,
|error| <= [(1/(n+1)!]*| [(f^(n+1))(c)|*| ((pi) (i=0-->n)(x - x_i))|
Since y'''(x)=0, how do we find the error?
What is meant by find error each x€[0,2]?
Thank you
If Y"' = 0 then the error is zero.
In fact, at x = 0,1,2 the error is zero.
3^0 = 2*0^2 + 1
3^1 = 2*1^2 + 1
3^2 = 2*2^2 + 1
Thank you very much
To find the error for each x ∈ [0,2], we need to calculate the difference between the actual function value at each x and the corresponding interpolated function value.
In this case, the interpolated function \(y(x) = 2x^2 + 1\) is a quadratic polynomial that we obtained using the quadratic Lagrange interpolation method.
To find the error at a specific x value, we can use the formula:
\[ \text{error}_x = |f(x) - y(x)| \]
where \(f(x) = 3^x\) is the actual function.
For example, let's calculate the error at x = 0:
\[ \text{error}_{x=0} = |f(0) - y(0)| = |3^0 - (2(0)^2 + 1)| = |1 - 1| = 0 \]
Similarly, let's calculate the error at x = 1:
\[ \text{error}_{x=1} = |f(1) - y(1)| = |3^1 - (2(1)^2 + 1)| = |3 - 3| = 0 \]
Finally, let's calculate the error at x = 2:
\[ \text{error}_{x=2} = |f(2) - y(2)| = |3^2 - (2(2)^2 + 1)| = |9 - 9| = 0 \]
Hence, for each \(x\) ∈ [0,2], the error in the quadratic Lagrange interpolation polynomial is 0, indicating that the interpolation perfectly matches the actual function in this case.