Newton's method of approximation. AGAIN. (1 Viewer)

Smilebuffalo · Aug 6, 2009

Use x = 0.5 to find an approximation to the root of cosx = x, correct to two decimal places, using 2 applications of Newton's method.

Answer: x = 0.76

cutemouse · Aug 7, 2009

Smilebuffalo said:
Use x = 0.5 to find an approximation to the root of cosx = x, correct to two decimal places, using 2 applications of Newton's method.

Answer: x = 0.76

$cosx=x\\cosx-x=0\\Let\ P(x)=cosx-x\\ P'(x)=-sinx-1\\By\ Newton's\ Method\ z_2=z_1-\frac{P(z_1)}{P'(z_1)}\ Let\ z_1=0.5\\z_2=0.5-\frac{P(0.5)}{P'(0.5)}\\=0.5-\frac{0.3775...}{-1.479...}\\=0.755222417...\\=0.76\ (to\ 2\ dp)$

Trebla · Aug 8, 2009

Did you remember to use RADIANS on your calculator when subbing in x = 0.5 as opposed to degrees?

Smilebuffalo · Aug 9, 2009

yea i did. The question is weird because i get the answer right using 1 application of newton's method. But using a 2nd application results in a less accurate answer that is not equal to 0.76

Timothy.Siu · Aug 9, 2009

Smilebuffalo said:
yea i did. The question is weird because i get the answer right using 1 application of newton's method. But using a 2nd application results in a less accurate answer that is not equal to 0.76

i got a different answer too, but it isn't less accurate.

Newton's method of approximation. AGAIN. (1 Viewer)

Smilebuffalo

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cutemouse

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Trebla

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Timothy.Siu

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