Topics:
Calculus

Analysis,
Pages 2 (339 words)

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The equations we have to solve are as follows:

<=>{sum_(i=1)^n (X_i^2 a+X_i b-X_iY_i)=0

{sum_(i=1)^n (X_i a+b-y_1)=0

So the first equation can be simplified by looking at the coefficients of a and b. You'll

see that there are actually linear equations in a and b, so there's a lot of clutter with

all these x's and y's all over the place.

Let's simplify the formula by dividing out the factors of two.

We can eliminate a and b

from the variable terms. After that, we can focus on the coefficient of a. So when we

do this, we get xi squared times a plus xi times b minus xi yi. And we set this equal to

0.

Let us proceed to the next example. We multiply axi by minus 1 and then add b to

yield xia plus b minus yi. Let us rearrange this equation so that all the a's are on one

side. This means that the sum of all the xi squared plus the sum of all xi times b

minus the sum of xiy equals zero.

So, in the case of y = x 2 + x, if we rewrite this as y = (×2) +x, then we have y = (sum

of ×2) times a plus sum of x. So that equals sum of xi times a, and that equals (sum

of x times a. Plus, how many b's do we get from this one? Well, we get one for each

data point. When we sum them together, we will get n. So n times b equals the sum

of vi.

Now, these quantities will look scary, but they're really just numbers. So for example,

for this one, you simply sum all the data points. And you get a two-by-two linear

system in x and y. Now, solving that system is easy. Just plug in the numbers from

your data and solve the linear system that you get.

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