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C4 Arghhhh!!!
Can anyone help me with this TORTUREEEE!!
1) (1+kx)(1+4x)^½ - in the expansion of this, the coefficient of x is 7. Find the value of the constant k and hencee the coefficient of x².
2) Expand - (1+3x)^-1 |x|<1/3
i get hw to expand but thennnn..
Hence show that for small x..
(1+x)/(1+3x) ≈ 1-2x+6x²-18x³
THEN..
take a suitable value for x which should be stated..use the series expansion in part b to find an approximate value for 101/103.
cheers
1) (1+kx)(1+4x)^½ - in the expansion of this, the coefficient of x is 7. Find the value of the constant k and hencee the coefficient of x².
2) Expand - (1+3x)^-1 |x|<1/3
i get hw to expand but thennnn..
Hence show that for small x..
(1+x)/(1+3x) ≈ 1-2x+6x²-18x³
THEN..
take a suitable value for x which should be stated..use the series expansion in part b to find an approximate value for 101/103.
cheersCan anyone help me with this TORTUREEEE!!
1) (1+kx)(1+4x)^½ - in the expansion of this, the coefficient of x is 7. Find the value of the constant k and hencee the coefficient of x².
2) Expand - (1+3x)^-1 |x|<1/3
i get hw to expand but thennnn..
Hence show that for small x..
(1+x)/(1+3x) ≈ 1-2x+6x²-18x³
THEN..
take a suitable value for x which should be stated..use the series expansion in part b to find an approximate value for 101/103.
cheers
1) (1+kx)(1+4x)^½ - in the expansion of this, the coefficient of x is 7. Find the value of the constant k and hencee the coefficient of x².
2) Expand - (1+3x)^-1 |x|<1/3
i get hw to expand but thennnn..
Hence show that for small x..
(1+x)/(1+3x) ≈ 1-2x+6x²-18x³
THEN..
take a suitable value for x which should be stated..use the series expansion in part b to find an approximate value for 101/103.
cheersFizzle
Can anyone help me with this TORTUREEEE!!
1) (1+kx)(1+4x)^½ - in the expansion of this, the coefficient of x is 7. Find the value of the constant k and hencee the coefficient of x².
2) Expand - (1+3x)^-1 |x|<1/3
i get hw to expand but thennnn..
Hence show that for small x..
(1+x)/(1+3x) ≈ 1-2x+6x²-18x³
THEN..
take a suitable value for x which should be stated..use the series expansion in part b to find an approximate value for 101/103.
cheers
1) (1+kx)(1+4x)^½ - in the expansion of this, the coefficient of x is 7. Find the value of the constant k and hencee the coefficient of x².
2) Expand - (1+3x)^-1 |x|<1/3
i get hw to expand but thennnn..
Hence show that for small x..
(1+x)/(1+3x) ≈ 1-2x+6x²-18x³
THEN..
take a suitable value for x which should be stated..use the series expansion in part b to find an approximate value for 101/103.
cheers(1+kx)(1+4x)^½ = (1+kx)(1 + (0.5)4x + (0.5*-0.5/2)(4x)^2...)
equation x terms, 7 = 2 + k, so k = 5
so x^2 terms = 4*2 + (0.5*-0.5/2)4^2 = 8 - 2 = 6
probably wrong.
Fizzle
Can anyone help me with this TORTUREEEE!!
1) (1+kx)(1+4x)^½ - in the expansion of this, the coefficient of x is 7. Find the value of the constant k and hencee the coefficient of x².
2) Expand - (1+3x)^-1 |x|<1/3
i get hw to expand but thennnn..
Hence show that for small x..
(1+x)/(1+3x) ≈ 1-2x+6x²-18x³
THEN..
take a suitable value for x which should be stated..use the series expansion in part b to find an approximate value for 101/103.
cheers
1) (1+kx)(1+4x)^½ - in the expansion of this, the coefficient of x is 7. Find the value of the constant k and hencee the coefficient of x².
2) Expand - (1+3x)^-1 |x|<1/3
i get hw to expand but thennnn..
Hence show that for small x..
(1+x)/(1+3x) ≈ 1-2x+6x²-18x³
THEN..
take a suitable value for x which should be stated..use the series expansion in part b to find an approximate value for 101/103.
cheers(1+kx)(1+4x)^½ = (1+kx)(1 + (0.5)4x + (0.5*-0.5/2)(4x)^2...)
equation x terms, 7 = 2 + k, so k = 5
so x^2 terms = 4*2 + (0.5*-0.5/2)4^2 = 8 - 2 = 6
probably wrong.
1. (1+kx)(1+4x)1/2
=(1+kx)(1+2x+½(½)(-½)(4x)²+...)
=(1+kx)(1+2x-2x²+...)
=1+(2+k)x+(2k-2)x²+...
2+k=7 so k=5
so coefficient of x² = 2(5)-2=8
2. (1+3x)-1 = 1-3x+9x²-27x³+...
=∑(-3x)n from n=0 to n=infinity
(1+x)/(1+3x)=(1+x)(1+3x)-1
=(1+x)(1-3x+9x²-27x³+...)
=1-2x+6x²-18x³+...
let (1+x)/(1+3x)=101/103=1.01/1.03
so x=0.01=1/100
then 101/103≈1-2/100+6/10000-18/1000000
then calculate what this is.
=(1+kx)(1+2x+½(½)(-½)(4x)²+...)
=(1+kx)(1+2x-2x²+...)
=1+(2+k)x+(2k-2)x²+...
2+k=7 so k=5
so coefficient of x² = 2(5)-2=8
2. (1+3x)-1 = 1-3x+9x²-27x³+...
=∑(-3x)n from n=0 to n=infinity
(1+x)/(1+3x)=(1+x)(1+3x)-1
=(1+x)(1-3x+9x²-27x³+...)
=1-2x+6x²-18x³+...
let (1+x)/(1+3x)=101/103=1.01/1.03
so x=0.01=1/100
then 101/103≈1-2/100+6/10000-18/1000000
then calculate what this is.
1. (1+kx)(1+4x)1/2
=(1+kx)(1+2x+½(½)(-½)(4x)²+...)
=(1+kx)(1+2x-2x²+...)
=1+(2+k)x+(2k-2)x²+...
2+k=7 so k=5
so coefficient of x² = 2(5)-2=8
2. (1+3x)-1 = 1-3x+9x²-27x³+...
=∑(-3x)n from n=0 to n=infinity
(1+x)/(1+3x)=(1+x)(1+3x)-1
=(1+x)(1-3x+9x²-27x³+...)
=1-2x+6x²-18x³+...
let (1+x)/(1+3x)=101/103=1.01/1.03
so x=0.01=1/100
then 101/103≈1-2/100+6/10000-18/1000000
then calculate what this is.
=(1+kx)(1+2x+½(½)(-½)(4x)²+...)
=(1+kx)(1+2x-2x²+...)
=1+(2+k)x+(2k-2)x²+...
2+k=7 so k=5
so coefficient of x² = 2(5)-2=8
2. (1+3x)-1 = 1-3x+9x²-27x³+...
=∑(-3x)n from n=0 to n=infinity
(1+x)/(1+3x)=(1+x)(1+3x)-1
=(1+x)(1-3x+9x²-27x³+...)
=1-2x+6x²-18x³+...
let (1+x)/(1+3x)=101/103=1.01/1.03
so x=0.01=1/100
then 101/103≈1-2/100+6/10000-18/1000000
then calculate what this is.
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