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Evgesh-ka [11]
3 years ago
10

Estimate the maximum error made in approximating e^x by the polynomial 1 + x + {1}/{2}x^2 over the interval x of [-0.4,0.4].

Mathematics
1 answer:
kolezko [41]3 years ago
7 0
E^x = 1 + x + x² / 2 + x³/ 3! + x^4 / 4! + .....
       = (1 + x + x²/2 ) + x³   [ 1/6 + x /4! + x² / 5! + ....  ]
 Error =  e^x - (1+ x + x² )   =  x³ [  1/6 + x /4! + x² / 5! + ....  ]
          x / 4!  < x / 6            x² / 5!   <  x² / 6  and so on
         So if we replace all factorials by 1/6 ..
     error  <  x² [ 1/6 + x/6 + x²/6 + ... ]
             <  x² / 6    [ 1 + x + x² ..... ]
             < x² / 6    * 1 / (1 -x)    = x² / 6 (1-x)        if   x < 1
 maximum error  =  x² /6(1-x)  occurs at 0.4 or -0.4 in the given interval.
             =  0.0444444


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What multiplies to 16 and adds to negative 12
Romashka-Z-Leto [24]
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hmm


we will solve with math
xy=16
x+y=-12
minus  x both sides
y=-12-x
sub for y in other equation
x(-12-x)=16
-12x-x^2=16
add 12x+x^2 both sides
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use quadratic formula
if you have
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}


0=x^2+12x+16
a=1
b=12
c=16

x=\frac{-12+/- \sqrt{12^{2}-4(1)(16) }}{2(1)}
x=\frac{-12+/- \sqrt{144-64 }}{2}
x=\frac{-12+/- \sqrt{80}}{2}
x=\frac{-12+/- 4\sqrt{5}}{2}
x=\frac{-6+/- 2\sqrt{5}
x=\frac{-6+ 2\sqrt{5} or \frac{-6- 2\sqrt{5}

aprox
x=-1.52786 or -10.4721

those are the numbers

the numbes are -1.52786 or -10.4721

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