Mixing
Powers of Irreducible Polynomials in Partial Fractions
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Fractions are normally decomposed by finding the numerator that goes with each individual term of the denominator of the original fraction. However, when a denominator term is repeated, the solution algorithm involves finding numerators for ascending powers of that term. So there appears to be a difference in how we treat repeated and non-repeated terms.
I thought it might be insightful to work out a couple of examples with almost-repeated terms. We can find, either trivially or via partial fraction decomposition, that $$fracx + 5x^2 = frac1x + frac5x^2$$ Compare this to the decomposition of a similar fraction $$fracx + 5(x + varepsilon)(x) = fracvarepsilon - 5varepsilon(x + varepsilon) + frac5varepsilon x$$ where $varepsilon$ is an arbitrarily small constant. To me, it's not particularly clear that the right-hand side (RHS) of each equation relates to that of the other. We can't set $varepsilon$ to $0$ in the second RHS, but even using limits to let $varepsilon$ approach $0$ doesn't yield the first RHS, as I would have guessed. Furthermore, the solution process does not harmonize the degree disparity between the top and bottom RHS; the top has the form $fracdegree 0degree 1 + fracdegree 0degree 2$, while the bottom has the form $fracdegree 0degree 1 + fracdegree 0degree 1$.
A less trivial example, comparing $$fracx^2 + 1(x^2)(x + 3) = -frac19x + frac13x^2 + frac109(x + 3)$$ to $$fracx^2 + 1(x + varepsilon)(x)(x + 3) = frac13varepsilon x + fracvarepsilon^2 + 1varepsilon(varepsilon - 3)(x + varepsilon) + frac103(3 - varepsilon)(x + 3)$$ reveals a similar absence of clarity. Confusingly, one gets the urge to let some of the $varepsilon$ approach $0$ while letting others approach $3$, but even arbitrarily allowing this much freedom won't allow us to reach the top RHS in this pair.
Why is this analysis not providing more insight into the repeated term rule for partial fraction decomposition, and will some other analysis in a similar vein work better?
calculus limits intuition fractions partial-fractions
asked Jul 25 at 19:17
user10478
1628
add a comment |Â
up vote
4
down vote
favorite
Fractions are normally decomposed by finding the numerator that goes with each individual term of the denominator of the original fraction. However, when a denominator term is repeated, the solution algorithm involves finding numerators for ascending powers of that term. So there appears to be a difference in how we treat repeated and non-repeated terms.
I thought it might be insightful to work out a couple of examples with almost-repeated terms. We can find, either trivially or via partial fraction decomposition, that $$fracx + 5x^2 = frac1x + frac5x^2$$ Compare this to the decomposition of a similar fraction $$fracx + 5(x + varepsilon)(x) = fracvarepsilon - 5varepsilon(x + varepsilon) + frac5varepsilon x$$ where $varepsilon$ is an arbitrarily small constant. To me, it's not particularly clear that the right-hand side (RHS) of each equation relates to that of the other. We can't set $varepsilon$ to $0$ in the second RHS, but even using limits to let $varepsilon$ approach $0$ doesn't yield the first RHS, as I would have guessed. Furthermore, the solution process does not harmonize the degree disparity between the top and bottom RHS; the top has the form $fracdegree 0degree 1 + fracdegree 0degree 2$, while the bottom has the form $fracdegree 0degree 1 + fracdegree 0degree 1$.
A less trivial example, comparing $$fracx^2 + 1(x^2)(x + 3) = -frac19x + frac13x^2 + frac109(x + 3)$$ to $$fracx^2 + 1(x + varepsilon)(x)(x + 3) = frac13varepsilon x + fracvarepsilon^2 + 1varepsilon(varepsilon - 3)(x + varepsilon) + frac103(3 - varepsilon)(x + 3)$$ reveals a similar absence of clarity. Confusingly, one gets the urge to let some of the $varepsilon$ approach $0$ while letting others approach $3$, but even arbitrarily allowing this much freedom won't allow us to reach the top RHS in this pair.
Why is this analysis not providing more insight into the repeated term rule for partial fraction decomposition, and will some other analysis in a similar vein work better?
calculus limits intuition fractions partial-fractions
asked Jul 25 at 19:17
user10478
1628
Just because $a = b+c$ and $a = l+r$ does not mean that $l to b$ and $r to c$ when some parameter is varied.
– Math Lover
Jul 25 at 19:38
add a comment |Â
up vote
4
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up vote
4
down vote
favorite
Fractions are normally decomposed by finding the numerator that goes with each individual term of the denominator of the original fraction. However, when a denominator term is repeated, the solution algorithm involves finding numerators for ascending powers of that term. So there appears to be a difference in how we treat repeated and non-repeated terms.
I thought it might be insightful to work out a couple of examples with almost-repeated terms. We can find, either trivially or via partial fraction decomposition, that $$fracx + 5x^2 = frac1x + frac5x^2$$ Compare this to the decomposition of a similar fraction $$fracx + 5(x + varepsilon)(x) = fracvarepsilon - 5varepsilon(x + varepsilon) + frac5varepsilon x$$ where $varepsilon$ is an arbitrarily small constant. To me, it's not particularly clear that the right-hand side (RHS) of each equation relates to that of the other. We can't set $varepsilon$ to $0$ in the second RHS, but even using limits to let $varepsilon$ approach $0$ doesn't yield the first RHS, as I would have guessed. Furthermore, the solution process does not harmonize the degree disparity between the top and bottom RHS; the top has the form $fracdegree 0degree 1 + fracdegree 0degree 2$, while the bottom has the form $fracdegree 0degree 1 + fracdegree 0degree 1$.
A less trivial example, comparing $$fracx^2 + 1(x^2)(x + 3) = -frac19x + frac13x^2 + frac109(x + 3)$$ to $$fracx^2 + 1(x + varepsilon)(x)(x + 3) = frac13varepsilon x + fracvarepsilon^2 + 1varepsilon(varepsilon - 3)(x + varepsilon) + frac103(3 - varepsilon)(x + 3)$$ reveals a similar absence of clarity. Confusingly, one gets the urge to let some of the $varepsilon$ approach $0$ while letting others approach $3$, but even arbitrarily allowing this much freedom won't allow us to reach the top RHS in this pair.
Why is this analysis not providing more insight into the repeated term rule for partial fraction decomposition, and will some other analysis in a similar vein work better?
calculus limits intuition fractions partial-fractions
asked Jul 25 at 19:17
user10478
1628
Fractions are normally decomposed by finding the numerator that goes with each individual term of the denominator of the original fraction. However, when a denominator term is repeated, the solution algorithm involves finding numerators for ascending powers of that term. So there appears to be a difference in how we treat repeated and non-repeated terms.
I thought it might be insightful to work out a couple of examples with almost-repeated terms. We can find, either trivially or via partial fraction decomposition, that $$fracx + 5x^2 = frac1x + frac5x^2$$ Compare this to the decomposition of a similar fraction $$fracx + 5(x + varepsilon)(x) = fracvarepsilon - 5varepsilon(x + varepsilon) + frac5varepsilon x$$ where $varepsilon$ is an arbitrarily small constant. To me, it's not particularly clear that the right-hand side (RHS) of each equation relates to that of the other. We can't set $varepsilon$ to $0$ in the second RHS, but even using limits to let $varepsilon$ approach $0$ doesn't yield the first RHS, as I would have guessed. Furthermore, the solution process does not harmonize the degree disparity between the top and bottom RHS; the top has the form $fracdegree 0degree 1 + fracdegree 0degree 2$, while the bottom has the form $fracdegree 0degree 1 + fracdegree 0degree 1$.
A less trivial example, comparing $$fracx^2 + 1(x^2)(x + 3) = -frac19x + frac13x^2 + frac109(x + 3)$$ to $$fracx^2 + 1(x + varepsilon)(x)(x + 3) = frac13varepsilon x + fracvarepsilon^2 + 1varepsilon(varepsilon - 3)(x + varepsilon) + frac103(3 - varepsilon)(x + 3)$$ reveals a similar absence of clarity. Confusingly, one gets the urge to let some of the $varepsilon$ approach $0$ while letting others approach $3$, but even arbitrarily allowing this much freedom won't allow us to reach the top RHS in this pair.
Why is this analysis not providing more insight into the repeated term rule for partial fraction decomposition, and will some other analysis in a similar vein work better?
calculus limits intuition fractions partial-fractions
asked Jul 25 at 19:17
user10478
1628
asked Jul 25 at 19:17
user10478
1628
asked Jul 25 at 19:17
user10478
1628
asked Jul 25 at 19:17
user10478
1628
1628
Just because $a = b+c$ and $a = l+r$ does not mean that $l to b$ and $r to c$ when some parameter is varied.
– Math Lover
Jul 25 at 19:38
add a comment |Â
Just because $a = b+c$ and $a = l+r$ does not mean that $l to b$ and $r to c$ when some parameter is varied.
– Math Lover
Jul 25 at 19:38
Just because $a = b+c$ and $a = l+r$ does not mean that $l to b$ and $r to c$ when some parameter is varied.
– Math Lover
Jul 25 at 19:38
Just because $a = b+c$ and $a = l+r$ does not mean that $l to b$ and $r to c$ when some parameter is varied.
– Math Lover
Jul 25 at 19:38
add a comment |Â
2 Answers
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up vote
0
down vote
In your first example, Maclaurin expansion with respect to the variable $epsilon$ gives (after a short calculation)
$$
fracepsilon-5epsilon (x+epsilon) = -frac5epsilon x + left(frac1x + frac5x^2 right) + O(epsilon).
$$
Cancel the $5/epsilon x$ terms and then let $epsilon to 0$.
answered Jul 25 at 19:46
Hans Lundmark
32.7k563109
add a comment |Â
up vote
0
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Note that a rational function $R(x) = fracf(x)q(x)$ can be expressed as a summation of multiple rational functions in different ways. The standard way is to write the expression such that the denominators are the powers of irreducible polynomials. But there are multiple ways to decompose a given expression. For example,
$$fracx+5x(x+e) = fracxx(x+e)+frac5x(x+e) = frac1x+e+frac5x(x+e). tag1$$
Obviously, $$lim_e to 0 frac1x+e+frac5x(x+e) = frac1x+ frac5x^2.tag2$$
But $(1)$ might not be preferred for carrying out integration of $fracx+5x(x+e)$.
answered Jul 25 at 19:49
Math Lover
12.3k21232
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
In your first example, Maclaurin expansion with respect to the variable $epsilon$ gives (after a short calculation)
$$
fracepsilon-5epsilon (x+epsilon) = -frac5epsilon x + left(frac1x + frac5x^2 right) + O(epsilon).
$$
Cancel the $5/epsilon x$ terms and then let $epsilon to 0$.
answered Jul 25 at 19:46
Hans Lundmark
32.7k563109
add a comment |Â
up vote
0
down vote
In your first example, Maclaurin expansion with respect to the variable $epsilon$ gives (after a short calculation)
$$
fracepsilon-5epsilon (x+epsilon) = -frac5epsilon x + left(frac1x + frac5x^2 right) + O(epsilon).
$$
Cancel the $5/epsilon x$ terms and then let $epsilon to 0$.
answered Jul 25 at 19:46
Hans Lundmark
32.7k563109
add a comment |Â
up vote
0
down vote
up vote
0
down vote
In your first example, Maclaurin expansion with respect to the variable $epsilon$ gives (after a short calculation)
$$
fracepsilon-5epsilon (x+epsilon) = -frac5epsilon x + left(frac1x + frac5x^2 right) + O(epsilon).
$$
Cancel the $5/epsilon x$ terms and then let $epsilon to 0$.
answered Jul 25 at 19:46
Hans Lundmark
32.7k563109
In your first example, Maclaurin expansion with respect to the variable $epsilon$ gives (after a short calculation)
$$
fracepsilon-5epsilon (x+epsilon) = -frac5epsilon x + left(frac1x + frac5x^2 right) + O(epsilon).
$$
Cancel the $5/epsilon x$ terms and then let $epsilon to 0$.
answered Jul 25 at 19:46
Hans Lundmark
32.7k563109
answered Jul 25 at 19:46
Hans Lundmark
32.7k563109
answered Jul 25 at 19:46
Hans Lundmark
32.7k563109
answered Jul 25 at 19:46
Hans Lundmark
32.7k563109
32.7k563109
add a comment |Â
add a comment |Â
up vote
0
down vote
Note that a rational function $R(x) = fracf(x)q(x)$ can be expressed as a summation of multiple rational functions in different ways. The standard way is to write the expression such that the denominators are the powers of irreducible polynomials. But there are multiple ways to decompose a given expression. For example,
$$fracx+5x(x+e) = fracxx(x+e)+frac5x(x+e) = frac1x+e+frac5x(x+e). tag1$$
Obviously, $$lim_e to 0 frac1x+e+frac5x(x+e) = frac1x+ frac5x^2.tag2$$
But $(1)$ might not be preferred for carrying out integration of $fracx+5x(x+e)$.
answered Jul 25 at 19:49
Math Lover
12.3k21232
add a comment |Â
up vote
0
down vote
Note that a rational function $R(x) = fracf(x)q(x)$ can be expressed as a summation of multiple rational functions in different ways. The standard way is to write the expression such that the denominators are the powers of irreducible polynomials. But there are multiple ways to decompose a given expression. For example,
$$fracx+5x(x+e) = fracxx(x+e)+frac5x(x+e) = frac1x+e+frac5x(x+e). tag1$$
Obviously, $$lim_e to 0 frac1x+e+frac5x(x+e) = frac1x+ frac5x^2.tag2$$
But $(1)$ might not be preferred for carrying out integration of $fracx+5x(x+e)$.
answered Jul 25 at 19:49
Math Lover
12.3k21232
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that a rational function $R(x) = fracf(x)q(x)$ can be expressed as a summation of multiple rational functions in different ways. The standard way is to write the expression such that the denominators are the powers of irreducible polynomials. But there are multiple ways to decompose a given expression. For example,
$$fracx+5x(x+e) = fracxx(x+e)+frac5x(x+e) = frac1x+e+frac5x(x+e). tag1$$
Obviously, $$lim_e to 0 frac1x+e+frac5x(x+e) = frac1x+ frac5x^2.tag2$$
But $(1)$ might not be preferred for carrying out integration of $fracx+5x(x+e)$.
answered Jul 25 at 19:49
Math Lover
12.3k21232
Note that a rational function $R(x) = fracf(x)q(x)$ can be expressed as a summation of multiple rational functions in different ways. The standard way is to write the expression such that the denominators are the powers of irreducible polynomials. But there are multiple ways to decompose a given expression. For example,
$$fracx+5x(x+e) = fracxx(x+e)+frac5x(x+e) = frac1x+e+frac5x(x+e). tag1$$
Obviously, $$lim_e to 0 frac1x+e+frac5x(x+e) = frac1x+ frac5x^2.tag2$$
But $(1)$ might not be preferred for carrying out integration of $fracx+5x(x+e)$.
answered Jul 25 at 19:49
Math Lover
12.3k21232
edited Jul 25 at 20:17
answered Jul 25 at 19:49
Math Lover
12.3k21232
answered Jul 25 at 19:49
Math Lover
12.3k21232
answered Jul 25 at 19:49
Math Lover
12.3k21232
12.3k21232
add a comment |Â
add a comment |Â
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Just because $a = b+c$ and $a = l+r$ does not mean that $l to b$ and $r to c$ when some parameter is varied.
– Math Lover
Jul 25 at 19:38