Let's do some calculus! (32)

Calculus Level 4

If $a_i \in \mathbb{R}$ , where $i \in \left\{ 1,2,3,4,5,\cdots,n \right\}$ , then find the value of $x$ for which the expression $\displaystyle \sum_{i=1}^n {(x-{a_i})}^2$ assumes its minimum value.

Bonus: Can this be solved this without calculus?

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No such value exists for

x \in \mathbb{R}

\dfrac{\displaystyle \sum_{i=1}^n {\left(-1\right)}^{i + 1} \cdot a_i}{n}

\dfrac{\displaystyle \sum_{i=1}^n a_i}{n}

0

1 solution

Kushal Bose
Sep 30, 2016

Using calculus it can be done easily by differentiating w.r.t. $x$ and make first derivative equal to zero.But without using calculus I have approached in the following way.

. $\displaystyle \sum_{i=1}^n{(x-a_i)^2} \\ =n x^2 - 2 \sum_{i=1}^n a_i x + \sum_{i=1}^n {a_i}^2 \\ =n(x^2 - \frac{2 \sum_{i=1}^n a_i}{n}x) +\sum_{i=1}^n {a_i}^2 \\ =n(x^2 - \frac{2 \sum_{i=1}^n a_i}{n}x + (\frac{\sum_{i=1}^n a_i}{n})^2) - \frac{(\sum_{i=1}^n a_i)^2}{n} + \sum_{i=1}^n {a_i}^2 \\ =n (x-\frac{\sum_{i=1}^n a_i}{n})^2 + \sum_{i=1}^n {a_i}^2 - \frac{(\sum_{i=1}^n a_i)^2}{n}$

So from above expression if the perfect square attains minimum value then total value will be minimum. The minimum value of x is = $\frac{\sum_{i=1}^n a_i}{n}$

Perfect!

Exactly the way I had thought. Actually, the reason I said not to use calculus was because even though calculus makes it very easy but there are still many methods to do this problem. One of them being the famous 'completing squares' method that you solved this with. For higher powers though, calculus is the best option but for quadratics, you don't need the knowledge of calculus to solve this. Great work!

P.S. It's okay that you have took the starting value of $i$ and zero, but in my problem, I've stated the starting value as $i=1$ . Just to avoid confusion for the reader, could you modify it?

Tapas Mazumdar - 4 years, 8 months ago

Updated...

Kushal Bose - 4 years, 8 months ago

Let's do some calculus! (32)

For more problems on calculus, click here .

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