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attashe74 [19]
3 years ago
11

Let f (x) = (1 − x)−1 and x0 = 0.

Mathematics
1 answer:
Papessa [141]3 years ago
5 0

Answer:

n = 7

Step-by-step explanation:

Recalling Taylor nth polynimial, we have,

Pₙ (x) = f (a) f ¹ (a) (x −a) + f ¹¹ (a) (x −a)²/2 + f ¹¹¹ (a) (x −a)³/3x2 + .......+ f ⁽ⁿ⁾ (a) (x-a)ⁿ / n (n - 1) (n - 2) .......3x2

Therefore,

If f(x) = ( 1 - x) ⁻¹ ,

Then the Lagrange form of the remainder hold which sates that,

( 1 - x) ⁻¹ - Pₙ (x) = f ⁿ⁺¹ (c) x ⁿ⁺¹ / (n + 1) !

Noting that,

where C is between 0 and x,

Hence therefore, the error is bounded by,

/ (0.5) ⁿ⁺¹ ( 1 - x) ⁻¹ ⁽⁰⁵⁾  ÷   (n + 1) !/ ≤ 10⁻⁶

These then equates to the fact that, that inequality first equates to 7.

Therefore,  n = 7.

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If √2 = 1.414 then the value of 5+√2 / 5-√2 is
Gala2k [10]

Answer:

1.787

Step-by-step explanation:

\\  \sf \frac{5 +  \sqrt{2} }{5 -  \sqrt{2} }  \times  \frac{5 +  \sqrt{2} }{5  +   \sqrt{2} }

\\   \sf =  \frac{(5 +  \sqrt{2}) }{5 -  \sqrt{2} }  \times  \frac{(5 +  \sqrt{2}) }{5 +  \sqrt{2} }

\\  \sf =  \frac{25 + 5 \sqrt{2 } + 5 \sqrt{2} + 2  }{( {5})^{2}  - ( \sqrt{ {2}})^{2} }

\\  \sf =  \frac{27 + 10 \sqrt{2} }{25 - 2}

\\  \sf =  \frac{27 - 10 \sqrt{2} }{23}

Now putting the value of√2

\\  \sf =  \frac{27 + 10 \times 1.414}{23}

\\  \sf =  \frac{27 + 14.14}{23}

\\  \sf =  \frac{41.14}{23}

\\ \sf = 1.787

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2 years ago
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Serga [27]

The answer COULD possibly be letter B

4 0
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Alchen [17]
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If x = 64, what is the value of x(x - 6)?
Fiesta28 [93]

Answer:

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Step-by-step explanation:

64(64 -6)

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64(58) = 3712

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