Consider the function f(\mu)=\sum_{i=1}(x_{i}-\mu)^{2}, where x_[i}=i,\ i=1,\ 2,\ \cdots,\ n What is the

Bobbie Comstock 2021-12-16 Answered
Consider the function
\(\displaystyle{f{{\left(\mu\right)}}}=\sum_{{{i}={1}}}{\left({x}_{{{i}}}-\mu\right)}^{{{2}}},\) where
\(\displaystyle{x}_{{{i}}}={i},\ {i}={1},\ {2},\ \cdots,\ {n}\)
What is the first and second derivative os \(\displaystyle{f{{\left(\mu\right)}}}\)?

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Expert Answer

Janet Young
Answered 2021-12-17 Author has 1413 answers
Step 1
\(\displaystyle{\frac{{{d}}}{{{d}\mu}}}{f{{\left(\mu\right)}}}=-{2}{\sum_{{{i}={1}}}^{{{n}}}}{\left({x}_{{{1}}}-\mu\right)}=-{2}{\sum_{{{i}={1}}}^{{{n}}}}{x}_{{{i}}}+{2}{n}\mu\)
\(\displaystyle{\frac{{{d}^{{{2}}}}}{{{d}\mu^{{{2}}}}}}{f{{\left(\mu\right)}}}={2}{n}\)
In the 1st step on line 1, the chain rule is applied. The term within the brackets is differentiated, producing a -1.
Then the bracket itself is differentiated, producing the 2 at the front. The 2nd step on line 1 involves no differentiation. Instead, the bracket is split into two terms.
The second term has an n because it is simply the summation from \(\displaystyle{i}={1}\) to \(\displaystyle{i}={n}\) of a constant.
The summation of a constant is equal to n multiplied by the constant. Then for the second line, there are no extra rules.
The first term becomes 0 because it's a constant and the second term loses \(\displaystyle\mu\).
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accimaroyalde
Answered 2021-12-18 Author has 3123 answers
Step 1
Adding an answer here to further clarify the other ones which are simply answers without steps. To get the first derivative, this can be re-written as:
\(\displaystyle{\frac{{{d}}}{{{d}\mu}}}{s}{u}{n}{\left({x}-\mu\right)}^{{{2}}}\)
\(\displaystyle=\sum{\frac{{{d}}}{{{d}\mu}}}{\left({x}-\mu\right)}^{{{2}}}\)
After that it's standard fare chain rule
\(\displaystyle=\sum-{1}\times{2}{\left({x}-\mu\right)}\)
\(\displaystyle=-{2}\sum{\left({x}-\mu\right)}\)
Second derivative: you can observe the same property of linear summation:
\(\displaystyle{\frac{{{d}}}{{{d}\mu}}}-{2}\sum{\left({x}-\mu\right)}\)
\(\displaystyle=-{2}\sum{\frac{{{d}}}{{{d}\mu}}}{\left({x}-\mu\right)}\)
\(\displaystyle=-{2}\sum{\left(-{1}\right)}\)
\(\displaystyle={2}{n}\)
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nick1337
Answered 2021-12-28 Author has 9672 answers

Step 1
Here we have one function \((x_{i}-\mu)\) that is part of a larger function in that it is raised to a power of 2, \((\mu-i)^{2}\)
According to the extended power rule, we multiply the derivative of the outer function
\((\mu-i)^{2}x\)
the derivative of the inner function \((x_{i}-\mu)\)
The derivative of the outer function brings the 2 down in front as
\(2\times(x_{i}-\mu)\)
and the derivative of the inner function
\((x_{i}-\mu)\) is -1.
So the -2 comes from multiplying the two derivatives according to the extend power rule:
\(2\times(x_{i}-\mu)\times-1=-2(x_{i}-\mu)\)
Answer:
\(f'(\mu)=-2\sum_{i=1}^{n}(x_{i}-\mu)\) and \(f''(\mu)=2n\)

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