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DENIUS [597]
3 years ago
7

How Do You Do This ? Please show work .

Mathematics
1 answer:
snow_tiger [21]3 years ago
5 0
So basically you are going to line up your equations (already done). Next, you will just add each term. So x + 3x = 4x, then 2y - 2y = 0 (we wouldn't put anything) then 7 - 3 = 4
So then we have all of our terms figured out and will have 4x = 4 then divide 4x by 4 to get x alone and then you will also divide the four that is by itself by four to equal one. So, your answer is x = 1. I hope that helps!
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In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T
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Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

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x_{i+1} = x_{i} + \frac{1}{f''x_{i}}  [-f'(x_{i}) ± \sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]

Part B.

It is evident that Newton's method fails in two cases, as:

1.  if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)    

Part C.

In case  x_{i+1} is close to x_{i}, the choice that shouldbe made instead of ± in part A is:

f'(x) = \sqrt{f'(x)^{2} - 2f(x)f''(x)}  ⇔ x_{i+1} = x_{i}

Part D.

As given x_{i+1} = x_{i} = h

or                 h = x_{i+1} - x_{i}

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also,             (x_{i+1}-x_{i})² = -(x_{i+1}-x_{i})(f(x_{i})/f'(x_{i}))

So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also,             x_{i+1} = x_{i} -f(x_{i})/f'(x_{i}) + [(x_{i+1} - x_{i})f''(x_{i})f(x_{i})]/[2(f'(x_{i}))²]

6 0
3 years ago
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