Simple Question: gradient function Formula

Question

Santino M 2015-2-2

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https://ww2.mathworks.cn/matlabcentral/answers/174620-simple-question-gradient-function-formula

编辑： Santino M 2015-2-3

采纳的回答： Roger Stafford

in the following line, x is a 200 element column vector, h is a scalar eg.say 20 or 30

a = gradient(x,h)

can you please tell how 'a' is computed from x and h? Thanks in advance

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Answer 1

Roger Stafford 2015-2-2

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https://ww2.mathworks.cn/matlabcentral/answers/174620-simple-question-gradient-function-formula#answer_166765

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For 2 <= k <= 199, the computation is a central difference:

a(k) = (x(k+1)-x(k-1))/(2*h)

However, for the two endpoints it is:

   a(1) = (x(2)-x(1))/h
   a(200) = (x(200)-x(199))/h

In case h is a vector of the same length as x, then it becomes a divided difference:

a(k) = (x(k+1)-x(k-1))/(h(k+1)-h(k-1))

and

   a(1) = (x(2)-x(1))/(h(2)-h(1))
   a(200) = (x(200)-x(199))/(h(200)-h(199))

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Santino M 2015-2-3

编辑：Santino M 2015-2-3

Thanks Star Strider I did try this by plotting. Thanks Roger for your reply. I did not understand the need of dividing by the scalar value h in a gradient. I initially assumed that the h denoted spacing between points eg. for h=20; grad(k) = (x(k-10) - x(k+10))/h. apparently it is not so.Can you please tell me how to use gradient to achieve the above. I want to set the interval for calculating the difference while calculating gradient. I can as well make a small function to do the same. If there is a way to get it done through an in built function it would be faster..especially if it is used a large no. of times (say > 10000000 times or more)

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