Calculate the inner product of X=(−4,2,3) and Y=(4,4,−1)
时间: 2024-05-22 16:10:53 浏览: 14
The inner product of two vectors X and Y is defined as:
X · Y = (X1 * Y1) + (X2 * Y2) + (X3 * Y3)
where X1, X2, X3 are the components of vector X, and Y1, Y2, Y3 are the components of vector Y.
Using this formula, we can calculate the inner product of X=(-4,2,3) and Y=(4,4,-1) as follows:
X · Y = (-4 * 4) + (2 * 4) + (3 * -1) = -16 + 8 - 3 = -11
Therefore, the inner product of X and Y is -11.
相关问题
Consider the cubic spline Y=9+8x+8x2+4x3+4h(x,1)+7h(x,2) Calculate the value of Y for x=−8
To calculate the value of Y for x = -8, we first need to determine the values of h(x,1) and h(x,2) at x = -8.
Using the cubic spline interpolation formula, we have:
h(x,1) = [3(-8-(-2))^2 - 2(-8-(-2))^3](8-(-4))/6h^2 + [3(-8-(-4))^2 - 2(-8-(-4))^3](4-(-8))/6h^2
= (-3/2)h + (9/2)h
= 3h
h(x,2) = [3(-8-(-4))^2 - 2(-8-(-4))^3](4-(-8))/6h^2 + [3(-8-0)^2 - 2(-8-0)^3](9-4)/6h^2
= (-1/2)h + (5/2)h
= 2h
Substituting these values into the expression for Y, we get:
Y = 9 + 8(-8) + 8(-8)^2 + 4(-8)^3 + 4(3h) + 7(2h)
= -1695 + 22h
Therefore, the value of Y for x = -8 is -1695 + 22h, where h is the interval between the data points used to construct the cubic spline.
Please calculate the value of for x= by the step of 0.1 (accurate to 10-4). Calculate the approximation of by cubic times Newton forward and backward interpolation formula.
I'm sorry, but it seems like there is missing information in the question. It is not clear what function we are trying to approximate using Newton forward and backward interpolation formulas. Without this information, I cannot provide a complete answer. Could you please provide more context or details about the problem?
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