Mixing
Converting a Product to a Sum
$begingroup$
How can I convert $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]$$ to a sum? I have been trying to solve this product by inductive reasoning but I figured it was too complex... Is there an agebraic/deductive way of converting this expression to a sum?
algebra-precalculus products
edited Jan 30 at 22:14
Kevin Long
3,58121431
asked Jan 30 at 22:03
A. LavieA. Lavie
274
$endgroup$
add a comment |
$begingroup$
How can I convert $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]$$ to a sum? I have been trying to solve this product by inductive reasoning but I figured it was too complex... Is there an agebraic/deductive way of converting this expression to a sum?
algebra-precalculus products
edited Jan 30 at 22:14
Kevin Long
3,58121431
asked Jan 30 at 22:03
A. LavieA. Lavie
274
$endgroup$
$begingroup$
If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 22:12
$begingroup$
@AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
$endgroup$
– Yves Daoust
Jan 30 at 22:15
add a comment |
$begingroup$
How can I convert $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]$$ to a sum? I have been trying to solve this product by inductive reasoning but I figured it was too complex... Is there an agebraic/deductive way of converting this expression to a sum?
algebra-precalculus products
edited Jan 30 at 22:14
Kevin Long
3,58121431
asked Jan 30 at 22:03
A. LavieA. Lavie
274
$endgroup$
How can I convert $$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]$$ to a sum? I have been trying to solve this product by inductive reasoning but I figured it was too complex... Is there an agebraic/deductive way of converting this expression to a sum?
algebra-precalculus products
algebra-precalculus products
edited Jan 30 at 22:14
Kevin Long
3,58121431
asked Jan 30 at 22:03
A. LavieA. Lavie
274
edited Jan 30 at 22:14
Kevin Long
3,58121431
asked Jan 30 at 22:03
A. LavieA. Lavie
274
edited Jan 30 at 22:14
Kevin Long
3,58121431
edited Jan 30 at 22:14
Kevin Long
3,58121431
edited Jan 30 at 22:14
Kevin Long
3,58121431
3,58121431
asked Jan 30 at 22:03
A. LavieA. Lavie
274
asked Jan 30 at 22:03
A. LavieA. Lavie
274
asked Jan 30 at 22:03
A. LavieA. Lavie
274
274
$begingroup$
If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 22:12
$begingroup$
@AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
$endgroup$
– Yves Daoust
Jan 30 at 22:15
add a comment |
$begingroup$
If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 22:12
$begingroup$
@AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
$endgroup$
– Yves Daoust
Jan 30 at 22:15
$begingroup$
If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 22:12
$begingroup$
If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 22:12
$begingroup$
@AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
$endgroup$
– Yves Daoust
Jan 30 at 22:15
$begingroup$
@AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
$endgroup$
– Yves Daoust
Jan 30 at 22:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{x^i-1}{x-1}=frac{displaystyleprod_{i=1}^k(x^i-1)}{(x-1)^k}.$$
Now the expression
$$prod_{i=1}^k(x^i-1)$$
yields irregular polynomials.
$$x-1$$
$$x^3-x^2-x+1$$
$$x^6-x^5-x^4+x^2+x-1$$
$$x^{10}-x^9-x^8+2x^5-x^2-x+1$$
$$x^{15}-x^{14}-x^{13}+x^{10}+x^9+x^8-x^7-x^6-x^5+x^2+x-1$$
$$cdots$$
answered Jan 30 at 22:22
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
Yes... is there a general expression (a sum) which I can write for each value of k?
$endgroup$
– A. Lavie
Jan 30 at 22:47
$begingroup$
$prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
$endgroup$
– Thies Heidecke
Jan 30 at 22:57
$begingroup$
@A.Lavie: there isn't.
$endgroup$
– Yves Daoust
Jan 31 at 8:03
add a comment |
$begingroup$
The sums are just geometric series which you can sum, getting
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{1-x^i}{1-x}$$
The denominator is just $(1-x)^k$. You can expand the numerator as a sum getting
$$1-sum_{i=1}^kx^i+sum_{i=1}^{k-1}sum_{j=i+1}^kx^{i+j}-sum_{i=1}^{k-2}sum_{j=i+1}^{k-1}sum_{m=j+1}^kx^{i+j+m}+ldots$$
but I don't think that is helping.
answered Jan 30 at 22:17
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{x^i-1}{x-1}=frac{displaystyleprod_{i=1}^k(x^i-1)}{(x-1)^k}.$$
Now the expression
$$prod_{i=1}^k(x^i-1)$$
yields irregular polynomials.
$$x-1$$
$$x^3-x^2-x+1$$
$$x^6-x^5-x^4+x^2+x-1$$
$$x^{10}-x^9-x^8+2x^5-x^2-x+1$$
$$x^{15}-x^{14}-x^{13}+x^{10}+x^9+x^8-x^7-x^6-x^5+x^2+x-1$$
$$cdots$$
answered Jan 30 at 22:22
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
Yes... is there a general expression (a sum) which I can write for each value of k?
$endgroup$
– A. Lavie
Jan 30 at 22:47
$begingroup$
$prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
$endgroup$
– Thies Heidecke
Jan 30 at 22:57
$begingroup$
@A.Lavie: there isn't.
$endgroup$
– Yves Daoust
Jan 31 at 8:03
add a comment |
$begingroup$
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{x^i-1}{x-1}=frac{displaystyleprod_{i=1}^k(x^i-1)}{(x-1)^k}.$$
Now the expression
$$prod_{i=1}^k(x^i-1)$$
yields irregular polynomials.
$$x-1$$
$$x^3-x^2-x+1$$
$$x^6-x^5-x^4+x^2+x-1$$
$$x^{10}-x^9-x^8+2x^5-x^2-x+1$$
$$x^{15}-x^{14}-x^{13}+x^{10}+x^9+x^8-x^7-x^6-x^5+x^2+x-1$$
$$cdots$$
answered Jan 30 at 22:22
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
Yes... is there a general expression (a sum) which I can write for each value of k?
$endgroup$
– A. Lavie
Jan 30 at 22:47
$begingroup$
$prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
$endgroup$
– Thies Heidecke
Jan 30 at 22:57
$begingroup$
@A.Lavie: there isn't.
$endgroup$
– Yves Daoust
Jan 31 at 8:03
add a comment |
$begingroup$
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{x^i-1}{x-1}=frac{displaystyleprod_{i=1}^k(x^i-1)}{(x-1)^k}.$$
Now the expression
$$prod_{i=1}^k(x^i-1)$$
yields irregular polynomials.
$$x-1$$
$$x^3-x^2-x+1$$
$$x^6-x^5-x^4+x^2+x-1$$
$$x^{10}-x^9-x^8+2x^5-x^2-x+1$$
$$x^{15}-x^{14}-x^{13}+x^{10}+x^9+x^8-x^7-x^6-x^5+x^2+x-1$$
$$cdots$$
answered Jan 30 at 22:22
Yves DaoustYves Daoust
132k676230
$endgroup$
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{x^i-1}{x-1}=frac{displaystyleprod_{i=1}^k(x^i-1)}{(x-1)^k}.$$
Now the expression
$$prod_{i=1}^k(x^i-1)$$
yields irregular polynomials.
$$x-1$$
$$x^3-x^2-x+1$$
$$x^6-x^5-x^4+x^2+x-1$$
$$x^{10}-x^9-x^8+2x^5-x^2-x+1$$
$$x^{15}-x^{14}-x^{13}+x^{10}+x^9+x^8-x^7-x^6-x^5+x^2+x-1$$
$$cdots$$
answered Jan 30 at 22:22
Yves DaoustYves Daoust
132k676230
edited Jan 30 at 22:27
answered Jan 30 at 22:22
Yves DaoustYves Daoust
132k676230
answered Jan 30 at 22:22
Yves DaoustYves Daoust
132k676230
answered Jan 30 at 22:22
Yves DaoustYves Daoust
132k676230
132k676230
$begingroup$
Yes... is there a general expression (a sum) which I can write for each value of k?
$endgroup$
– A. Lavie
Jan 30 at 22:47
$begingroup$
$prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
$endgroup$
– Thies Heidecke
Jan 30 at 22:57
$begingroup$
@A.Lavie: there isn't.
$endgroup$
– Yves Daoust
Jan 31 at 8:03
add a comment |
$begingroup$
Yes... is there a general expression (a sum) which I can write for each value of k?
$endgroup$
– A. Lavie
Jan 30 at 22:47
$begingroup$
$prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
$endgroup$
– Thies Heidecke
Jan 30 at 22:57
$begingroup$
@A.Lavie: there isn't.
$endgroup$
– Yves Daoust
Jan 31 at 8:03
$begingroup$
Yes... is there a general expression (a sum) which I can write for each value of k?
$endgroup$
– A. Lavie
Jan 30 at 22:47
$begingroup$
Yes... is there a general expression (a sum) which I can write for each value of k?
$endgroup$
– A. Lavie
Jan 30 at 22:47
$begingroup$
$prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
$endgroup$
– Thies Heidecke
Jan 30 at 22:57
$begingroup$
$prod_{i=1}^k (x^i-1)$ can be expressed via a Pochhammer q symbol $(-1)^k(x;x)_k$.
$endgroup$
– Thies Heidecke
Jan 30 at 22:57
$begingroup$
@A.Lavie: there isn't.
$endgroup$
– Yves Daoust
Jan 31 at 8:03
$begingroup$
@A.Lavie: there isn't.
$endgroup$
– Yves Daoust
Jan 31 at 8:03
add a comment |
$begingroup$
The sums are just geometric series which you can sum, getting
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{1-x^i}{1-x}$$
The denominator is just $(1-x)^k$. You can expand the numerator as a sum getting
$$1-sum_{i=1}^kx^i+sum_{i=1}^{k-1}sum_{j=i+1}^kx^{i+j}-sum_{i=1}^{k-2}sum_{j=i+1}^{k-1}sum_{m=j+1}^kx^{i+j+m}+ldots$$
but I don't think that is helping.
answered Jan 30 at 22:17
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
The sums are just geometric series which you can sum, getting
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{1-x^i}{1-x}$$
The denominator is just $(1-x)^k$. You can expand the numerator as a sum getting
$$1-sum_{i=1}^kx^i+sum_{i=1}^{k-1}sum_{j=i+1}^kx^{i+j}-sum_{i=1}^{k-2}sum_{j=i+1}^{k-1}sum_{m=j+1}^kx^{i+j+m}+ldots$$
but I don't think that is helping.
answered Jan 30 at 22:17
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
The sums are just geometric series which you can sum, getting
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{1-x^i}{1-x}$$
The denominator is just $(1-x)^k$. You can expand the numerator as a sum getting
$$1-sum_{i=1}^kx^i+sum_{i=1}^{k-1}sum_{j=i+1}^kx^{i+j}-sum_{i=1}^{k-2}sum_{j=i+1}^{k-1}sum_{m=j+1}^kx^{i+j+m}+ldots$$
but I don't think that is helping.
answered Jan 30 at 22:17
Ross MillikanRoss Millikan
301k24200375
$endgroup$
The sums are just geometric series which you can sum, getting
$$prod_{i=1}^kleft[sum_{j=0}^{i-1}x^jright]=prod_{i=1}^kfrac{1-x^i}{1-x}$$
The denominator is just $(1-x)^k$. You can expand the numerator as a sum getting
$$1-sum_{i=1}^kx^i+sum_{i=1}^{k-1}sum_{j=i+1}^kx^{i+j}-sum_{i=1}^{k-2}sum_{j=i+1}^{k-1}sum_{m=j+1}^kx^{i+j+m}+ldots$$
but I don't think that is helping.
answered Jan 30 at 22:17
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 22:17
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 22:17
Ross MillikanRoss Millikan
301k24200375
answered Jan 30 at 22:17
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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$begingroup$
If the logarithm is a legal operation in your case (i.e. the numbers are positive) it will be a good option.
$endgroup$
– AlgebraicallyClosed
Jan 30 at 22:12
$begingroup$
@AlgebraicallyClosed: keeping the signs apart is no big deal. But you will get logarithms of sums, which is nothing convenient.
$endgroup$
– Yves Daoust
Jan 30 at 22:15