Binomial expansion involving partial fractions
$begingroup$
Sorry I do not know how to use the formatting will try my best.
Q. Find the binomial expansion up to $x^2$ of:
$$frac{3+2x^2}{(2x+1)(x-3)^2}$$
For the partial fraction I get:
$$frac{2}{7}frac{1}{2x+1} + frac{6}{7}frac{1}{x-3} + frac{3}{(x-3)^2}$$
Then I did the following:
$$(2x+1)^{-1} = 1 - 2x + 4x^2$$
$$(x-3)^{-1}= frac{1}{3} + frac{x}{9} + frac{x^2}{27}$$
$$(x-3)^{-2} = frac{1}{3} + frac{2x}{9} + frac{x^2}{9}$$
When I add them I get completely the wrong answer:
Correct answer is
$$frac{1}{3} + frac{4x}{9} + frac{11x^2}{9}.$$
partial-fractions
$endgroup$
add a comment |
$begingroup$
Sorry I do not know how to use the formatting will try my best.
Q. Find the binomial expansion up to $x^2$ of:
$$frac{3+2x^2}{(2x+1)(x-3)^2}$$
For the partial fraction I get:
$$frac{2}{7}frac{1}{2x+1} + frac{6}{7}frac{1}{x-3} + frac{3}{(x-3)^2}$$
Then I did the following:
$$(2x+1)^{-1} = 1 - 2x + 4x^2$$
$$(x-3)^{-1}= frac{1}{3} + frac{x}{9} + frac{x^2}{27}$$
$$(x-3)^{-2} = frac{1}{3} + frac{2x}{9} + frac{x^2}{9}$$
When I add them I get completely the wrong answer:
Correct answer is
$$frac{1}{3} + frac{4x}{9} + frac{11x^2}{9}.$$
partial-fractions
$endgroup$
$begingroup$
Do you polynomial division by increasing powers?
$endgroup$
– Bernard
Nov 21 '17 at 15:32
add a comment |
$begingroup$
Sorry I do not know how to use the formatting will try my best.
Q. Find the binomial expansion up to $x^2$ of:
$$frac{3+2x^2}{(2x+1)(x-3)^2}$$
For the partial fraction I get:
$$frac{2}{7}frac{1}{2x+1} + frac{6}{7}frac{1}{x-3} + frac{3}{(x-3)^2}$$
Then I did the following:
$$(2x+1)^{-1} = 1 - 2x + 4x^2$$
$$(x-3)^{-1}= frac{1}{3} + frac{x}{9} + frac{x^2}{27}$$
$$(x-3)^{-2} = frac{1}{3} + frac{2x}{9} + frac{x^2}{9}$$
When I add them I get completely the wrong answer:
Correct answer is
$$frac{1}{3} + frac{4x}{9} + frac{11x^2}{9}.$$
partial-fractions
$endgroup$
Sorry I do not know how to use the formatting will try my best.
Q. Find the binomial expansion up to $x^2$ of:
$$frac{3+2x^2}{(2x+1)(x-3)^2}$$
For the partial fraction I get:
$$frac{2}{7}frac{1}{2x+1} + frac{6}{7}frac{1}{x-3} + frac{3}{(x-3)^2}$$
Then I did the following:
$$(2x+1)^{-1} = 1 - 2x + 4x^2$$
$$(x-3)^{-1}= frac{1}{3} + frac{x}{9} + frac{x^2}{27}$$
$$(x-3)^{-2} = frac{1}{3} + frac{2x}{9} + frac{x^2}{9}$$
When I add them I get completely the wrong answer:
Correct answer is
$$frac{1}{3} + frac{4x}{9} + frac{11x^2}{9}.$$
partial-fractions
partial-fractions
edited Nov 21 '17 at 15:44
boi Shift
asked Nov 21 '17 at 15:19
boi Shiftboi Shift
266
266
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Do you polynomial division by increasing powers?
$endgroup$
– Bernard
Nov 21 '17 at 15:32
add a comment |
$begingroup$
Do you polynomial division by increasing powers?
$endgroup$
– Bernard
Nov 21 '17 at 15:32
$begingroup$
Do you polynomial division by increasing powers?
$endgroup$
– Bernard
Nov 21 '17 at 15:32
$begingroup$
Do you polynomial division by increasing powers?
$endgroup$
– Bernard
Nov 21 '17 at 15:32
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator:
$$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$
We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):
begin{alignat}{6}
& & color{red}{dfrac13 -dfrac49x}&color{red}{{}+dfrac{11}{9}x^2} & color{red}{{}-dfrac{182}{81}x^3}&+dotsm\%
9 + 12x-11x^2+ 2x^3 & enspacebiggl(enspace & 3+2x^2 \[-18mu]
& & -3 - 4x& +dfrac{11}{3}x^2 &{} -dfrac23 x^3 \
& & -4x&+ dfrac{17}3x^2&{} -dfrac{2}{3}x^3 \
&&4x&+dfrac{16}3x^2&{} -dfrac{44}{9}x^3&+dotsm\
&&&hskip30mu 11x^2&{}-dfrac{50}{9}x^3&+dotsm\
&&&hskip12mu - 11x^2&{}-dfrac{132}{9}x^3&+dotsm\
&&&&-dfrac{182}{9}x^3&+dotsm
end{alignat}
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add a comment |
$begingroup$
Hint:
How about finding $a,b,c$ in
$$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$
by comparing the coefficients of different powers of $x$
For example, $(-3)^2a=3$
$endgroup$
$begingroup$
The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
$endgroup$
– boi Shift
Nov 21 '17 at 15:45
add a comment |
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$$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^mleft(1-dfrac{mx}3+dfrac{m(m-1)}2left(-dfrac x3right)^2+cdotsright)$$ for $|—x/3|<1$
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator:
$$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$
We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):
begin{alignat}{6}
& & color{red}{dfrac13 -dfrac49x}&color{red}{{}+dfrac{11}{9}x^2} & color{red}{{}-dfrac{182}{81}x^3}&+dotsm\%
9 + 12x-11x^2+ 2x^3 & enspacebiggl(enspace & 3+2x^2 \[-18mu]
& & -3 - 4x& +dfrac{11}{3}x^2 &{} -dfrac23 x^3 \
& & -4x&+ dfrac{17}3x^2&{} -dfrac{2}{3}x^3 \
&&4x&+dfrac{16}3x^2&{} -dfrac{44}{9}x^3&+dotsm\
&&&hskip30mu 11x^2&{}-dfrac{50}{9}x^3&+dotsm\
&&&hskip12mu - 11x^2&{}-dfrac{132}{9}x^3&+dotsm\
&&&&-dfrac{182}{9}x^3&+dotsm
end{alignat}
$endgroup$
add a comment |
$begingroup$
You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator:
$$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$
We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):
begin{alignat}{6}
& & color{red}{dfrac13 -dfrac49x}&color{red}{{}+dfrac{11}{9}x^2} & color{red}{{}-dfrac{182}{81}x^3}&+dotsm\%
9 + 12x-11x^2+ 2x^3 & enspacebiggl(enspace & 3+2x^2 \[-18mu]
& & -3 - 4x& +dfrac{11}{3}x^2 &{} -dfrac23 x^3 \
& & -4x&+ dfrac{17}3x^2&{} -dfrac{2}{3}x^3 \
&&4x&+dfrac{16}3x^2&{} -dfrac{44}{9}x^3&+dotsm\
&&&hskip30mu 11x^2&{}-dfrac{50}{9}x^3&+dotsm\
&&&hskip12mu - 11x^2&{}-dfrac{132}{9}x^3&+dotsm\
&&&&-dfrac{182}{9}x^3&+dotsm
end{alignat}
$endgroup$
add a comment |
$begingroup$
You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator:
$$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$
We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):
begin{alignat}{6}
& & color{red}{dfrac13 -dfrac49x}&color{red}{{}+dfrac{11}{9}x^2} & color{red}{{}-dfrac{182}{81}x^3}&+dotsm\%
9 + 12x-11x^2+ 2x^3 & enspacebiggl(enspace & 3+2x^2 \[-18mu]
& & -3 - 4x& +dfrac{11}{3}x^2 &{} -dfrac23 x^3 \
& & -4x&+ dfrac{17}3x^2&{} -dfrac{2}{3}x^3 \
&&4x&+dfrac{16}3x^2&{} -dfrac{44}{9}x^3&+dotsm\
&&&hskip30mu 11x^2&{}-dfrac{50}{9}x^3&+dotsm\
&&&hskip12mu - 11x^2&{}-dfrac{132}{9}x^3&+dotsm\
&&&&-dfrac{182}{9}x^3&+dotsm
end{alignat}
$endgroup$
You can mechanically obtain the expansion with a simple division by increasing powers of the numerator by the denominator. First expand the denominator:
$$(1+2x)(3-x)^2=(1+2x)(9-6x+x^2)=9+12x-11x^2+2x^3$$
We'll expand up to order $3$, dividing $3+2x^2$ by $9+12x-11x^2+2x^3$ (for an expansion up to order $2$, you may truncate the divisor at order $2$ first):
begin{alignat}{6}
& & color{red}{dfrac13 -dfrac49x}&color{red}{{}+dfrac{11}{9}x^2} & color{red}{{}-dfrac{182}{81}x^3}&+dotsm\%
9 + 12x-11x^2+ 2x^3 & enspacebiggl(enspace & 3+2x^2 \[-18mu]
& & -3 - 4x& +dfrac{11}{3}x^2 &{} -dfrac23 x^3 \
& & -4x&+ dfrac{17}3x^2&{} -dfrac{2}{3}x^3 \
&&4x&+dfrac{16}3x^2&{} -dfrac{44}{9}x^3&+dotsm\
&&&hskip30mu 11x^2&{}-dfrac{50}{9}x^3&+dotsm\
&&&hskip12mu - 11x^2&{}-dfrac{132}{9}x^3&+dotsm\
&&&&-dfrac{182}{9}x^3&+dotsm
end{alignat}
answered Nov 21 '17 at 17:15
BernardBernard
119k740113
119k740113
add a comment |
add a comment |
$begingroup$
Hint:
How about finding $a,b,c$ in
$$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$
by comparing the coefficients of different powers of $x$
For example, $(-3)^2a=3$
$endgroup$
$begingroup$
The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
$endgroup$
– boi Shift
Nov 21 '17 at 15:45
add a comment |
$begingroup$
Hint:
How about finding $a,b,c$ in
$$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$
by comparing the coefficients of different powers of $x$
For example, $(-3)^2a=3$
$endgroup$
$begingroup$
The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
$endgroup$
– boi Shift
Nov 21 '17 at 15:45
add a comment |
$begingroup$
Hint:
How about finding $a,b,c$ in
$$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$
by comparing the coefficients of different powers of $x$
For example, $(-3)^2a=3$
$endgroup$
Hint:
How about finding $a,b,c$ in
$$(a+bx+cx^2)(2x+1)(x-3)^2=3+2x^2$$
by comparing the coefficients of different powers of $x$
For example, $(-3)^2a=3$
answered Nov 21 '17 at 15:33
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
$begingroup$
The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
$endgroup$
– boi Shift
Nov 21 '17 at 15:45
add a comment |
$begingroup$
The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
$endgroup$
– boi Shift
Nov 21 '17 at 15:45
$begingroup$
The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
$endgroup$
– boi Shift
Nov 21 '17 at 15:45
$begingroup$
The way I've tried to do this is the way it's supposed to be done - I cannot deviate from this method really. There is just somewhere I am going wrong.
$endgroup$
– boi Shift
Nov 21 '17 at 15:45
add a comment |
$begingroup$
$$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^mleft(1-dfrac{mx}3+dfrac{m(m-1)}2left(-dfrac x3right)^2+cdotsright)$$ for $|—x/3|<1$
$endgroup$
add a comment |
$begingroup$
$$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^mleft(1-dfrac{mx}3+dfrac{m(m-1)}2left(-dfrac x3right)^2+cdotsright)$$ for $|—x/3|<1$
$endgroup$
add a comment |
$begingroup$
$$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^mleft(1-dfrac{mx}3+dfrac{m(m-1)}2left(-dfrac x3right)^2+cdotsright)$$ for $|—x/3|<1$
$endgroup$
$$(x-3)^m=(-3)^m(1-x/3)^m=(-3)^mleft(1-dfrac{mx}3+dfrac{m(m-1)}2left(-dfrac x3right)^2+cdotsright)$$ for $|—x/3|<1$
answered Nov 21 '17 at 15:50
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
add a comment |
add a comment |
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$begingroup$
Do you polynomial division by increasing powers?
$endgroup$
– Bernard
Nov 21 '17 at 15:32