Mixing
Simplifying $frac1{1+x}+frac2{1+x^2}+frac4{1+x^4}+frac8{1+x^8}+frac{16}{x^{16}-1}$
$begingroup$
We need to simplify $$dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{1+x^8}+dfrac{16}{x^{16}-1}$$
The last denominator can be factored and we can get all the other denominators as factors of $x^{16}-1$. I tried handling the expressions in pairs,starting from the right.I also tried to take a common factor of two out of the numerators to help simplify,but that has yielded nothing.I then tried multiplying all the fractions to get $x^{16}-1$ in the denominator but that worsens things(I think so anyway).
So after doing the above things(and much more),I feel like I am running out of ideas.A really small hint will be appreciated.
algebra-precalculus
edited Jan 29 at 20:06
Martin Sleziak
44.9k10122277
asked May 2 '14 at 5:45
rah4927rah4927
1,6711339
$endgroup$
add a comment |
$begingroup$
We need to simplify $$dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{1+x^8}+dfrac{16}{x^{16}-1}$$
The last denominator can be factored and we can get all the other denominators as factors of $x^{16}-1$. I tried handling the expressions in pairs,starting from the right.I also tried to take a common factor of two out of the numerators to help simplify,but that has yielded nothing.I then tried multiplying all the fractions to get $x^{16}-1$ in the denominator but that worsens things(I think so anyway).
So after doing the above things(and much more),I feel like I am running out of ideas.A really small hint will be appreciated.
algebra-precalculus
edited Jan 29 at 20:06
Martin Sleziak
44.9k10122277
asked May 2 '14 at 5:45
rah4927rah4927
1,6711339
$endgroup$
2
$begingroup$
If you split that last factor into two fractions with denominators $(x^8pm 1)$
$endgroup$
– Test123
May 2 '14 at 5:50
add a comment |
$begingroup$
We need to simplify $$dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{1+x^8}+dfrac{16}{x^{16}-1}$$
The last denominator can be factored and we can get all the other denominators as factors of $x^{16}-1$. I tried handling the expressions in pairs,starting from the right.I also tried to take a common factor of two out of the numerators to help simplify,but that has yielded nothing.I then tried multiplying all the fractions to get $x^{16}-1$ in the denominator but that worsens things(I think so anyway).
So after doing the above things(and much more),I feel like I am running out of ideas.A really small hint will be appreciated.
algebra-precalculus
edited Jan 29 at 20:06
Martin Sleziak
44.9k10122277
asked May 2 '14 at 5:45
rah4927rah4927
1,6711339
$endgroup$
We need to simplify $$dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{1+x^8}+dfrac{16}{x^{16}-1}$$
The last denominator can be factored and we can get all the other denominators as factors of $x^{16}-1$. I tried handling the expressions in pairs,starting from the right.I also tried to take a common factor of two out of the numerators to help simplify,but that has yielded nothing.I then tried multiplying all the fractions to get $x^{16}-1$ in the denominator but that worsens things(I think so anyway).
So after doing the above things(and much more),I feel like I am running out of ideas.A really small hint will be appreciated.
algebra-precalculus
algebra-precalculus
edited Jan 29 at 20:06
Martin Sleziak
44.9k10122277
asked May 2 '14 at 5:45
rah4927rah4927
1,6711339
edited Jan 29 at 20:06
Martin Sleziak
44.9k10122277
asked May 2 '14 at 5:45
rah4927rah4927
1,6711339
edited Jan 29 at 20:06
Martin Sleziak
44.9k10122277
edited Jan 29 at 20:06
Martin Sleziak
44.9k10122277
edited Jan 29 at 20:06
Martin Sleziak
44.9k10122277
44.9k10122277
asked May 2 '14 at 5:45
rah4927rah4927
1,6711339
asked May 2 '14 at 5:45
rah4927rah4927
1,6711339
asked May 2 '14 at 5:45
rah4927rah4927
1,6711339
1,6711339
2
$begingroup$
If you split that last factor into two fractions with denominators $(x^8pm 1)$
$endgroup$
– Test123
May 2 '14 at 5:50
add a comment |
2
$begingroup$
If you split that last factor into two fractions with denominators $(x^8pm 1)$
$endgroup$
– Test123
May 2 '14 at 5:50
2
2
$begingroup$
If you split that last factor into two fractions with denominators $(x^8pm 1)$
$endgroup$
– Test123
May 2 '14 at 5:50
$begingroup$
If you split that last factor into two fractions with denominators $(x^8pm 1)$
$endgroup$
– Test123
May 2 '14 at 5:50
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$frac{16}{x^{16}-1}=frac{8}{x^8-1}+frac{-8}{x^8+1}$$
So the $4^{th}$ term of the original sum and the $2^{nd}$ part of the decomposition above are canceled. You are left with:
$$
dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{x^{8}-1}
$$
Continue similarly. In the end you will get $frac{1}{x-1}$.
answered May 2 '14 at 5:56
Test123Test123
2,792828
$endgroup$
$begingroup$
Ah,I should have thought of PFD in resolving this problem.Thanks a lot for your help.
$endgroup$
– rah4927
May 2 '14 at 5:58
$begingroup$
@rah4927 This is what I suggested originally in my comment.
$endgroup$
– Test123
May 2 '14 at 5:59
$begingroup$
I thought you were asking me to factor the denominator(sorry I didn't read your comment properly).
$endgroup$
– rah4927
May 2 '14 at 6:00
$begingroup$
@Test123. I think that you will get $frac{1}{x-1}$
$endgroup$
– Claude Leibovici
May 2 '14 at 6:50
2
$begingroup$
If you knew the number of typo's I can make !
$endgroup$
– Claude Leibovici
May 2 '14 at 6:53
|
show 2 more comments
$begingroup$
More generally,
$$ sum_{n=0}^N dfrac{2^n}{1+x^{2^n}} = dfrac{2^{N+1}}{1-x^{2^{N+1}}} - dfrac{1}{1-x}$$
as can be proven by induction.
answered May 2 '14 at 6:16
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
Similarly, $$sum_{n=0}^N dfrac{3^n (2 + x^{3^n})}{1 + x^{3^n} + x^{2 cdot 3^n}} = dfrac{3^{N+1}}{1 - x^{3^{N+1}}} - dfrac{1}{1 - x}$$
$endgroup$
– Robert Israel
May 2 '14 at 15:14
add a comment |
$begingroup$
Hint: Add $dfrac{1}{1-x}$ to the given expression and see the sum telescope.
answered May 13 '14 at 11:21
rah4927rah4927
1,6711339
$endgroup$
add a comment |
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3 Answers
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$begingroup$
$$frac{16}{x^{16}-1}=frac{8}{x^8-1}+frac{-8}{x^8+1}$$
So the $4^{th}$ term of the original sum and the $2^{nd}$ part of the decomposition above are canceled. You are left with:
$$
dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{x^{8}-1}
$$
Continue similarly. In the end you will get $frac{1}{x-1}$.
answered May 2 '14 at 5:56
Test123Test123
2,792828
$endgroup$
$begingroup$
Ah,I should have thought of PFD in resolving this problem.Thanks a lot for your help.
$endgroup$
– rah4927
May 2 '14 at 5:58
$begingroup$
@rah4927 This is what I suggested originally in my comment.
$endgroup$
– Test123
May 2 '14 at 5:59
$begingroup$
I thought you were asking me to factor the denominator(sorry I didn't read your comment properly).
$endgroup$
– rah4927
May 2 '14 at 6:00
$begingroup$
@Test123. I think that you will get $frac{1}{x-1}$
$endgroup$
– Claude Leibovici
May 2 '14 at 6:50
2
$begingroup$
If you knew the number of typo's I can make !
$endgroup$
– Claude Leibovici
May 2 '14 at 6:53
|
show 2 more comments
$begingroup$
$$frac{16}{x^{16}-1}=frac{8}{x^8-1}+frac{-8}{x^8+1}$$
So the $4^{th}$ term of the original sum and the $2^{nd}$ part of the decomposition above are canceled. You are left with:
$$
dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{x^{8}-1}
$$
Continue similarly. In the end you will get $frac{1}{x-1}$.
answered May 2 '14 at 5:56
Test123Test123
2,792828
$endgroup$
$begingroup$
Ah,I should have thought of PFD in resolving this problem.Thanks a lot for your help.
$endgroup$
– rah4927
May 2 '14 at 5:58
$begingroup$
@rah4927 This is what I suggested originally in my comment.
$endgroup$
– Test123
May 2 '14 at 5:59
$begingroup$
I thought you were asking me to factor the denominator(sorry I didn't read your comment properly).
$endgroup$
– rah4927
May 2 '14 at 6:00
$begingroup$
@Test123. I think that you will get $frac{1}{x-1}$
$endgroup$
– Claude Leibovici
May 2 '14 at 6:50
2
$begingroup$
If you knew the number of typo's I can make !
$endgroup$
– Claude Leibovici
May 2 '14 at 6:53
|
show 2 more comments
$begingroup$
$$frac{16}{x^{16}-1}=frac{8}{x^8-1}+frac{-8}{x^8+1}$$
So the $4^{th}$ term of the original sum and the $2^{nd}$ part of the decomposition above are canceled. You are left with:
$$
dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{x^{8}-1}
$$
Continue similarly. In the end you will get $frac{1}{x-1}$.
answered May 2 '14 at 5:56
Test123Test123
2,792828
$endgroup$
$$frac{16}{x^{16}-1}=frac{8}{x^8-1}+frac{-8}{x^8+1}$$
So the $4^{th}$ term of the original sum and the $2^{nd}$ part of the decomposition above are canceled. You are left with:
$$
dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{x^{8}-1}
$$
Continue similarly. In the end you will get $frac{1}{x-1}$.
answered May 2 '14 at 5:56
Test123Test123
2,792828
edited Feb 14 '17 at 11:17
answered May 2 '14 at 5:56
Test123Test123
2,792828
answered May 2 '14 at 5:56
Test123Test123
2,792828
answered May 2 '14 at 5:56
Test123Test123
2,792828
2,792828
$begingroup$
Ah,I should have thought of PFD in resolving this problem.Thanks a lot for your help.
$endgroup$
– rah4927
May 2 '14 at 5:58
$begingroup$
@rah4927 This is what I suggested originally in my comment.
$endgroup$
– Test123
May 2 '14 at 5:59
$begingroup$
I thought you were asking me to factor the denominator(sorry I didn't read your comment properly).
$endgroup$
– rah4927
May 2 '14 at 6:00
$begingroup$
@Test123. I think that you will get $frac{1}{x-1}$
$endgroup$
– Claude Leibovici
May 2 '14 at 6:50
2
$begingroup$
If you knew the number of typo's I can make !
$endgroup$
– Claude Leibovici
May 2 '14 at 6:53
|
show 2 more comments
$begingroup$
Ah,I should have thought of PFD in resolving this problem.Thanks a lot for your help.
$endgroup$
– rah4927
May 2 '14 at 5:58
$begingroup$
@rah4927 This is what I suggested originally in my comment.
$endgroup$
– Test123
May 2 '14 at 5:59
$begingroup$
I thought you were asking me to factor the denominator(sorry I didn't read your comment properly).
$endgroup$
– rah4927
May 2 '14 at 6:00
$begingroup$
@Test123. I think that you will get $frac{1}{x-1}$
$endgroup$
– Claude Leibovici
May 2 '14 at 6:50
2
$begingroup$
If you knew the number of typo's I can make !
$endgroup$
– Claude Leibovici
May 2 '14 at 6:53
$begingroup$
Ah,I should have thought of PFD in resolving this problem.Thanks a lot for your help.
$endgroup$
– rah4927
May 2 '14 at 5:58
$begingroup$
Ah,I should have thought of PFD in resolving this problem.Thanks a lot for your help.
$endgroup$
– rah4927
May 2 '14 at 5:58
$begingroup$
@rah4927 This is what I suggested originally in my comment.
$endgroup$
– Test123
May 2 '14 at 5:59
$begingroup$
@rah4927 This is what I suggested originally in my comment.
$endgroup$
– Test123
May 2 '14 at 5:59
$begingroup$
I thought you were asking me to factor the denominator(sorry I didn't read your comment properly).
$endgroup$
– rah4927
May 2 '14 at 6:00
$begingroup$
I thought you were asking me to factor the denominator(sorry I didn't read your comment properly).
$endgroup$
– rah4927
May 2 '14 at 6:00
$begingroup$
@Test123. I think that you will get $frac{1}{x-1}$
$endgroup$
– Claude Leibovici
May 2 '14 at 6:50
$begingroup$
@Test123. I think that you will get $frac{1}{x-1}$
$endgroup$
– Claude Leibovici
May 2 '14 at 6:50
2
2
$begingroup$
If you knew the number of typo's I can make !
$endgroup$
– Claude Leibovici
May 2 '14 at 6:53
$begingroup$
If you knew the number of typo's I can make !
$endgroup$
– Claude Leibovici
May 2 '14 at 6:53
|
show 2 more comments
$begingroup$
More generally,
$$ sum_{n=0}^N dfrac{2^n}{1+x^{2^n}} = dfrac{2^{N+1}}{1-x^{2^{N+1}}} - dfrac{1}{1-x}$$
as can be proven by induction.
answered May 2 '14 at 6:16
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
Similarly, $$sum_{n=0}^N dfrac{3^n (2 + x^{3^n})}{1 + x^{3^n} + x^{2 cdot 3^n}} = dfrac{3^{N+1}}{1 - x^{3^{N+1}}} - dfrac{1}{1 - x}$$
$endgroup$
– Robert Israel
May 2 '14 at 15:14
add a comment |
$begingroup$
More generally,
$$ sum_{n=0}^N dfrac{2^n}{1+x^{2^n}} = dfrac{2^{N+1}}{1-x^{2^{N+1}}} - dfrac{1}{1-x}$$
as can be proven by induction.
answered May 2 '14 at 6:16
Robert IsraelRobert Israel
330k23219473
$endgroup$
$begingroup$
Similarly, $$sum_{n=0}^N dfrac{3^n (2 + x^{3^n})}{1 + x^{3^n} + x^{2 cdot 3^n}} = dfrac{3^{N+1}}{1 - x^{3^{N+1}}} - dfrac{1}{1 - x}$$
$endgroup$
– Robert Israel
May 2 '14 at 15:14
add a comment |
$begingroup$
More generally,
$$ sum_{n=0}^N dfrac{2^n}{1+x^{2^n}} = dfrac{2^{N+1}}{1-x^{2^{N+1}}} - dfrac{1}{1-x}$$
as can be proven by induction.
answered May 2 '14 at 6:16
Robert IsraelRobert Israel
330k23219473
$endgroup$
More generally,
$$ sum_{n=0}^N dfrac{2^n}{1+x^{2^n}} = dfrac{2^{N+1}}{1-x^{2^{N+1}}} - dfrac{1}{1-x}$$
as can be proven by induction.
answered May 2 '14 at 6:16
Robert IsraelRobert Israel
330k23219473
edited May 2 '14 at 15:15
answered May 2 '14 at 6:16
Robert IsraelRobert Israel
330k23219473
answered May 2 '14 at 6:16
Robert IsraelRobert Israel
330k23219473
answered May 2 '14 at 6:16
Robert IsraelRobert Israel
330k23219473
330k23219473
$begingroup$
Similarly, $$sum_{n=0}^N dfrac{3^n (2 + x^{3^n})}{1 + x^{3^n} + x^{2 cdot 3^n}} = dfrac{3^{N+1}}{1 - x^{3^{N+1}}} - dfrac{1}{1 - x}$$
$endgroup$
– Robert Israel
May 2 '14 at 15:14
add a comment |
$begingroup$
Similarly, $$sum_{n=0}^N dfrac{3^n (2 + x^{3^n})}{1 + x^{3^n} + x^{2 cdot 3^n}} = dfrac{3^{N+1}}{1 - x^{3^{N+1}}} - dfrac{1}{1 - x}$$
$endgroup$
– Robert Israel
May 2 '14 at 15:14
$begingroup$
Similarly, $$sum_{n=0}^N dfrac{3^n (2 + x^{3^n})}{1 + x^{3^n} + x^{2 cdot 3^n}} = dfrac{3^{N+1}}{1 - x^{3^{N+1}}} - dfrac{1}{1 - x}$$
$endgroup$
– Robert Israel
May 2 '14 at 15:14
$begingroup$
Similarly, $$sum_{n=0}^N dfrac{3^n (2 + x^{3^n})}{1 + x^{3^n} + x^{2 cdot 3^n}} = dfrac{3^{N+1}}{1 - x^{3^{N+1}}} - dfrac{1}{1 - x}$$
$endgroup$
– Robert Israel
May 2 '14 at 15:14
add a comment |
$begingroup$
Hint: Add $dfrac{1}{1-x}$ to the given expression and see the sum telescope.
answered May 13 '14 at 11:21
rah4927rah4927
1,6711339
$endgroup$
add a comment |
$begingroup$
Hint: Add $dfrac{1}{1-x}$ to the given expression and see the sum telescope.
answered May 13 '14 at 11:21
rah4927rah4927
1,6711339
$endgroup$
add a comment |
$begingroup$
Hint: Add $dfrac{1}{1-x}$ to the given expression and see the sum telescope.
answered May 13 '14 at 11:21
rah4927rah4927
1,6711339
$endgroup$
Hint: Add $dfrac{1}{1-x}$ to the given expression and see the sum telescope.
answered May 13 '14 at 11:21
rah4927rah4927
1,6711339
answered May 13 '14 at 11:21
rah4927rah4927
1,6711339
answered May 13 '14 at 11:21
rah4927rah4927
1,6711339
answered May 13 '14 at 11:21
rah4927rah4927
1,6711339
1,6711339
add a comment |
add a comment |
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2
$begingroup$
If you split that last factor into two fractions with denominators $(x^8pm 1)$
$endgroup$
– Test123
May 2 '14 at 5:50