Mixing
What is the partial fraction of $frac{x}{((x)^2+1)^2}$
$begingroup$
I was trying to find the partial fraction of
$$frac{x}{(x^2+1)^2}$$
By the method of assuming
$$frac{x}{(x^2+1)^2}=frac{(Ax+B)}{(x^2+1)} + frac{(Cx+D)}{(x^2+1)^2} $$
But, my values for $A, B$ and $D$ are coming $0$.
i.e. $$A=B=D=0$$ and $$C=1$$
Which is directly equal to
$$frac{x}{(x^2+1)^2}$$
So, technically I got NO solution or Partial Fraction
I am afraid if I'm doing any foolish mistake but, please help me out in this issue. I haven't practised Partial Fractions since a long time.
Thank you so much in advance!
Great Day Ahead!
partial-fractions
edited Jan 22 at 17:53
Thomas Shelby
3,9042625
asked Jan 22 at 17:46
Naved THE SheikhNaved THE Sheikh
186
$endgroup$
|
show 4 more comments
$begingroup$
I was trying to find the partial fraction of
$$frac{x}{(x^2+1)^2}$$
By the method of assuming
$$frac{x}{(x^2+1)^2}=frac{(Ax+B)}{(x^2+1)} + frac{(Cx+D)}{(x^2+1)^2} $$
But, my values for $A, B$ and $D$ are coming $0$.
i.e. $$A=B=D=0$$ and $$C=1$$
Which is directly equal to
$$frac{x}{(x^2+1)^2}$$
So, technically I got NO solution or Partial Fraction
I am afraid if I'm doing any foolish mistake but, please help me out in this issue. I haven't practised Partial Fractions since a long time.
Thank you so much in advance!
Great Day Ahead!
partial-fractions
edited Jan 22 at 17:53
Thomas Shelby
3,9042625
asked Jan 22 at 17:46
Naved THE SheikhNaved THE Sheikh
186
$endgroup$
1
$begingroup$
The ansatz that you are using only works when your rational function has a denominator that is a $textbf{linear term}$ squared. The correct ansatz here is $frac{x}{(x^{2} + 1)^{2}} = frac{A}{x + i} + frac{B}{x-i} + frac{Cx + D}{(x+i)^{2}} + frac{Ex + F}{(x - i)^{2}}$. I'm not sure if you wanted your partial fractions to have complex coefficients, but this will give you the complete partial fraction expansion of $frac{x}{(x^{2} + 1)^{2}}$
$endgroup$
– Adam Higgins
Jan 22 at 17:55
1
$begingroup$
This is right. You start off with a fraction in the form you require, and when you do the calculations they show that this is the right decomposition. You didn't get "no solution", just an unexpected solution.
$endgroup$
– Mark Bennet
Jan 22 at 17:56
1
$begingroup$
@AdamHiggins You can decompose into partial fractions in the way suggested in the question. For example if integrating a rational function over the reals the factorisation of the denominator into linear and quadratic factors with real coefficients does make sense.
$endgroup$
– Mark Bennet
Jan 22 at 17:58
1
$begingroup$
@MarkBennet I'm not sure I understand your point
$endgroup$
– Adam Higgins
Jan 22 at 18:00
1
$begingroup$
@AdamHiggins You say this only works with linear factors. It can be done perfectly well with irreducible quadratic factors.
$endgroup$
– Mark Bennet
Jan 22 at 18:01
|
show 4 more comments
$begingroup$
I was trying to find the partial fraction of
$$frac{x}{(x^2+1)^2}$$
By the method of assuming
$$frac{x}{(x^2+1)^2}=frac{(Ax+B)}{(x^2+1)} + frac{(Cx+D)}{(x^2+1)^2} $$
But, my values for $A, B$ and $D$ are coming $0$.
i.e. $$A=B=D=0$$ and $$C=1$$
Which is directly equal to
$$frac{x}{(x^2+1)^2}$$
So, technically I got NO solution or Partial Fraction
I am afraid if I'm doing any foolish mistake but, please help me out in this issue. I haven't practised Partial Fractions since a long time.
Thank you so much in advance!
Great Day Ahead!
partial-fractions
edited Jan 22 at 17:53
Thomas Shelby
3,9042625
asked Jan 22 at 17:46
Naved THE SheikhNaved THE Sheikh
186
$endgroup$
I was trying to find the partial fraction of
$$frac{x}{(x^2+1)^2}$$
By the method of assuming
$$frac{x}{(x^2+1)^2}=frac{(Ax+B)}{(x^2+1)} + frac{(Cx+D)}{(x^2+1)^2} $$
But, my values for $A, B$ and $D$ are coming $0$.
i.e. $$A=B=D=0$$ and $$C=1$$
Which is directly equal to
$$frac{x}{(x^2+1)^2}$$
So, technically I got NO solution or Partial Fraction
I am afraid if I'm doing any foolish mistake but, please help me out in this issue. I haven't practised Partial Fractions since a long time.
Thank you so much in advance!
Great Day Ahead!
partial-fractions
partial-fractions
edited Jan 22 at 17:53
Thomas Shelby
3,9042625
asked Jan 22 at 17:46
Naved THE SheikhNaved THE Sheikh
186
edited Jan 22 at 17:53
Thomas Shelby
3,9042625
asked Jan 22 at 17:46
Naved THE SheikhNaved THE Sheikh
186
edited Jan 22 at 17:53
Thomas Shelby
3,9042625
edited Jan 22 at 17:53
Thomas Shelby
3,9042625
edited Jan 22 at 17:53
Thomas Shelby
3,9042625
3,9042625
asked Jan 22 at 17:46
Naved THE SheikhNaved THE Sheikh
186
asked Jan 22 at 17:46
Naved THE SheikhNaved THE Sheikh
186
asked Jan 22 at 17:46
Naved THE SheikhNaved THE Sheikh
186
186
1
$begingroup$
The ansatz that you are using only works when your rational function has a denominator that is a $textbf{linear term}$ squared. The correct ansatz here is $frac{x}{(x^{2} + 1)^{2}} = frac{A}{x + i} + frac{B}{x-i} + frac{Cx + D}{(x+i)^{2}} + frac{Ex + F}{(x - i)^{2}}$. I'm not sure if you wanted your partial fractions to have complex coefficients, but this will give you the complete partial fraction expansion of $frac{x}{(x^{2} + 1)^{2}}$
$endgroup$
– Adam Higgins
Jan 22 at 17:55
1
$begingroup$
This is right. You start off with a fraction in the form you require, and when you do the calculations they show that this is the right decomposition. You didn't get "no solution", just an unexpected solution.
$endgroup$
– Mark Bennet
Jan 22 at 17:56
1
$begingroup$
@AdamHiggins You can decompose into partial fractions in the way suggested in the question. For example if integrating a rational function over the reals the factorisation of the denominator into linear and quadratic factors with real coefficients does make sense.
$endgroup$
– Mark Bennet
Jan 22 at 17:58
1
$begingroup$
@MarkBennet I'm not sure I understand your point
$endgroup$
– Adam Higgins
Jan 22 at 18:00
1
$begingroup$
@AdamHiggins You say this only works with linear factors. It can be done perfectly well with irreducible quadratic factors.
$endgroup$
– Mark Bennet
Jan 22 at 18:01
|
show 4 more comments
1
$begingroup$
The ansatz that you are using only works when your rational function has a denominator that is a $textbf{linear term}$ squared. The correct ansatz here is $frac{x}{(x^{2} + 1)^{2}} = frac{A}{x + i} + frac{B}{x-i} + frac{Cx + D}{(x+i)^{2}} + frac{Ex + F}{(x - i)^{2}}$. I'm not sure if you wanted your partial fractions to have complex coefficients, but this will give you the complete partial fraction expansion of $frac{x}{(x^{2} + 1)^{2}}$
$endgroup$
– Adam Higgins
Jan 22 at 17:55
1
$begingroup$
This is right. You start off with a fraction in the form you require, and when you do the calculations they show that this is the right decomposition. You didn't get "no solution", just an unexpected solution.
$endgroup$
– Mark Bennet
Jan 22 at 17:56
1
$begingroup$
@AdamHiggins You can decompose into partial fractions in the way suggested in the question. For example if integrating a rational function over the reals the factorisation of the denominator into linear and quadratic factors with real coefficients does make sense.
$endgroup$
– Mark Bennet
Jan 22 at 17:58
1
$begingroup$
@MarkBennet I'm not sure I understand your point
$endgroup$
– Adam Higgins
Jan 22 at 18:00
1
$begingroup$
@AdamHiggins You say this only works with linear factors. It can be done perfectly well with irreducible quadratic factors.
$endgroup$
– Mark Bennet
Jan 22 at 18:01
1
1
$begingroup$
The ansatz that you are using only works when your rational function has a denominator that is a $textbf{linear term}$ squared. The correct ansatz here is $frac{x}{(x^{2} + 1)^{2}} = frac{A}{x + i} + frac{B}{x-i} + frac{Cx + D}{(x+i)^{2}} + frac{Ex + F}{(x - i)^{2}}$. I'm not sure if you wanted your partial fractions to have complex coefficients, but this will give you the complete partial fraction expansion of $frac{x}{(x^{2} + 1)^{2}}$
$endgroup$
– Adam Higgins
Jan 22 at 17:55
$begingroup$
The ansatz that you are using only works when your rational function has a denominator that is a $textbf{linear term}$ squared. The correct ansatz here is $frac{x}{(x^{2} + 1)^{2}} = frac{A}{x + i} + frac{B}{x-i} + frac{Cx + D}{(x+i)^{2}} + frac{Ex + F}{(x - i)^{2}}$. I'm not sure if you wanted your partial fractions to have complex coefficients, but this will give you the complete partial fraction expansion of $frac{x}{(x^{2} + 1)^{2}}$
$endgroup$
– Adam Higgins
Jan 22 at 17:55
1
1
$begingroup$
This is right. You start off with a fraction in the form you require, and when you do the calculations they show that this is the right decomposition. You didn't get "no solution", just an unexpected solution.
$endgroup$
– Mark Bennet
Jan 22 at 17:56
$begingroup$
This is right. You start off with a fraction in the form you require, and when you do the calculations they show that this is the right decomposition. You didn't get "no solution", just an unexpected solution.
$endgroup$
– Mark Bennet
Jan 22 at 17:56
1
1
$begingroup$
@AdamHiggins You can decompose into partial fractions in the way suggested in the question. For example if integrating a rational function over the reals the factorisation of the denominator into linear and quadratic factors with real coefficients does make sense.
$endgroup$
– Mark Bennet
Jan 22 at 17:58
$begingroup$
@AdamHiggins You can decompose into partial fractions in the way suggested in the question. For example if integrating a rational function over the reals the factorisation of the denominator into linear and quadratic factors with real coefficients does make sense.
$endgroup$
– Mark Bennet
Jan 22 at 17:58
1
1
$begingroup$
@MarkBennet I'm not sure I understand your point
$endgroup$
– Adam Higgins
Jan 22 at 18:00
$begingroup$
@MarkBennet I'm not sure I understand your point
$endgroup$
– Adam Higgins
Jan 22 at 18:00
1
1
$begingroup$
@AdamHiggins You say this only works with linear factors. It can be done perfectly well with irreducible quadratic factors.
$endgroup$
– Mark Bennet
Jan 22 at 18:01
$begingroup$
@AdamHiggins You say this only works with linear factors. It can be done perfectly well with irreducible quadratic factors.
$endgroup$
– Mark Bennet
Jan 22 at 18:01
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Well, of course the solution of the form $frac{Ax+B}{x^2+1}+frac{Cx+D}{(x^2+1)^2}$ is $frac{x}{(x^2+1)^2}$; that's what you started with. If you want another partial fractions expression, use complex numbers. If the point of the exercise is to integrate the function, you'd be better off substituting $y=x^2+1$.
answered Jan 22 at 17:57
J.G.J.G.
29.5k22946
$endgroup$
$begingroup$
The cope of breaking $$frac{x}{((x)^2+1)^2}$$ into Partial Fractions is to evaluate the value of it's Inverse Laplace Transform. Will it help, if I use complex numbers here?
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:13
$begingroup$
@NavedTHESheikh Definitely.
$endgroup$
– J.G.
Jan 22 at 18:18
add a comment |
$begingroup$
Hint: Use that $$x^2+1=(x-i)(x+i)$$ where $$i^2=-1$$
answered Jan 22 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.4k42866
$endgroup$
$begingroup$
Thank you so much. I am yet trying it out. Will tell you if it works. 😉😊
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:14
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
Well, of course the solution of the form $frac{Ax+B}{x^2+1}+frac{Cx+D}{(x^2+1)^2}$ is $frac{x}{(x^2+1)^2}$; that's what you started with. If you want another partial fractions expression, use complex numbers. If the point of the exercise is to integrate the function, you'd be better off substituting $y=x^2+1$.
answered Jan 22 at 17:57
J.G.J.G.
29.5k22946
$endgroup$
$begingroup$
The cope of breaking $$frac{x}{((x)^2+1)^2}$$ into Partial Fractions is to evaluate the value of it's Inverse Laplace Transform. Will it help, if I use complex numbers here?
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:13
$begingroup$
@NavedTHESheikh Definitely.
$endgroup$
– J.G.
Jan 22 at 18:18
add a comment |
$begingroup$
Well, of course the solution of the form $frac{Ax+B}{x^2+1}+frac{Cx+D}{(x^2+1)^2}$ is $frac{x}{(x^2+1)^2}$; that's what you started with. If you want another partial fractions expression, use complex numbers. If the point of the exercise is to integrate the function, you'd be better off substituting $y=x^2+1$.
answered Jan 22 at 17:57
J.G.J.G.
29.5k22946
$endgroup$
$begingroup$
The cope of breaking $$frac{x}{((x)^2+1)^2}$$ into Partial Fractions is to evaluate the value of it's Inverse Laplace Transform. Will it help, if I use complex numbers here?
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:13
$begingroup$
@NavedTHESheikh Definitely.
$endgroup$
– J.G.
Jan 22 at 18:18
add a comment |
$begingroup$
Well, of course the solution of the form $frac{Ax+B}{x^2+1}+frac{Cx+D}{(x^2+1)^2}$ is $frac{x}{(x^2+1)^2}$; that's what you started with. If you want another partial fractions expression, use complex numbers. If the point of the exercise is to integrate the function, you'd be better off substituting $y=x^2+1$.
answered Jan 22 at 17:57
J.G.J.G.
29.5k22946
$endgroup$
Well, of course the solution of the form $frac{Ax+B}{x^2+1}+frac{Cx+D}{(x^2+1)^2}$ is $frac{x}{(x^2+1)^2}$; that's what you started with. If you want another partial fractions expression, use complex numbers. If the point of the exercise is to integrate the function, you'd be better off substituting $y=x^2+1$.
answered Jan 22 at 17:57
J.G.J.G.
29.5k22946
answered Jan 22 at 17:57
J.G.J.G.
29.5k22946
answered Jan 22 at 17:57
J.G.J.G.
29.5k22946
answered Jan 22 at 17:57
J.G.J.G.
29.5k22946
29.5k22946
$begingroup$
The cope of breaking $$frac{x}{((x)^2+1)^2}$$ into Partial Fractions is to evaluate the value of it's Inverse Laplace Transform. Will it help, if I use complex numbers here?
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:13
$begingroup$
@NavedTHESheikh Definitely.
$endgroup$
– J.G.
Jan 22 at 18:18
add a comment |
$begingroup$
The cope of breaking $$frac{x}{((x)^2+1)^2}$$ into Partial Fractions is to evaluate the value of it's Inverse Laplace Transform. Will it help, if I use complex numbers here?
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:13
$begingroup$
@NavedTHESheikh Definitely.
$endgroup$
– J.G.
Jan 22 at 18:18
$begingroup$
The cope of breaking $$frac{x}{((x)^2+1)^2}$$ into Partial Fractions is to evaluate the value of it's Inverse Laplace Transform. Will it help, if I use complex numbers here?
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:13
$begingroup$
The cope of breaking $$frac{x}{((x)^2+1)^2}$$ into Partial Fractions is to evaluate the value of it's Inverse Laplace Transform. Will it help, if I use complex numbers here?
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:13
$begingroup$
@NavedTHESheikh Definitely.
$endgroup$
– J.G.
Jan 22 at 18:18
$begingroup$
@NavedTHESheikh Definitely.
$endgroup$
– J.G.
Jan 22 at 18:18
add a comment |
$begingroup$
Hint: Use that $$x^2+1=(x-i)(x+i)$$ where $$i^2=-1$$
answered Jan 22 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.4k42866
$endgroup$
$begingroup$
Thank you so much. I am yet trying it out. Will tell you if it works. 😉😊
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:14
add a comment |
$begingroup$
Hint: Use that $$x^2+1=(x-i)(x+i)$$ where $$i^2=-1$$
answered Jan 22 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.4k42866
$endgroup$
$begingroup$
Thank you so much. I am yet trying it out. Will tell you if it works. 😉😊
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:14
add a comment |
$begingroup$
Hint: Use that $$x^2+1=(x-i)(x+i)$$ where $$i^2=-1$$
answered Jan 22 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.4k42866
$endgroup$
Hint: Use that $$x^2+1=(x-i)(x+i)$$ where $$i^2=-1$$
answered Jan 22 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.4k42866
answered Jan 22 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.4k42866
answered Jan 22 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.4k42866
answered Jan 22 at 17:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.4k42866
77.4k42866
$begingroup$
Thank you so much. I am yet trying it out. Will tell you if it works. 😉😊
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:14
add a comment |
$begingroup$
Thank you so much. I am yet trying it out. Will tell you if it works. 😉😊
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:14
$begingroup$
Thank you so much. I am yet trying it out. Will tell you if it works. 😉😊
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:14
$begingroup$
Thank you so much. I am yet trying it out. Will tell you if it works. 😉😊
$endgroup$
– Naved THE Sheikh
Jan 22 at 18:14
add a comment |
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$begingroup$
The ansatz that you are using only works when your rational function has a denominator that is a $textbf{linear term}$ squared. The correct ansatz here is $frac{x}{(x^{2} + 1)^{2}} = frac{A}{x + i} + frac{B}{x-i} + frac{Cx + D}{(x+i)^{2}} + frac{Ex + F}{(x - i)^{2}}$. I'm not sure if you wanted your partial fractions to have complex coefficients, but this will give you the complete partial fraction expansion of $frac{x}{(x^{2} + 1)^{2}}$
$endgroup$
– Adam Higgins
Jan 22 at 17:55
1
$begingroup$
This is right. You start off with a fraction in the form you require, and when you do the calculations they show that this is the right decomposition. You didn't get "no solution", just an unexpected solution.
$endgroup$
– Mark Bennet
Jan 22 at 17:56
1
$begingroup$
@AdamHiggins You can decompose into partial fractions in the way suggested in the question. For example if integrating a rational function over the reals the factorisation of the denominator into linear and quadratic factors with real coefficients does make sense.
$endgroup$
– Mark Bennet
Jan 22 at 17:58
1
$begingroup$
@MarkBennet I'm not sure I understand your point
$endgroup$
– Adam Higgins
Jan 22 at 18:00
1
$begingroup$
@AdamHiggins You say this only works with linear factors. It can be done perfectly well with irreducible quadratic factors.
$endgroup$
– Mark Bennet
Jan 22 at 18:01