Mixing
Simplify $(-1)^{n-1}frac{800}{(pi^2){(2n-1)^2}}$ [closed]
Im having problems simplifying this equation, I know that the answer should be $$frac{800}{(pi^2)}cdot frac{(-1)^{n+1}}{(2n-1)^2}$$ however I'm not entirely sure why.. If someone could explain it would be a great help
algebra-precalculus
edited Nov 22 '18 at 12:03
Barry Cipra
59.3k653125
asked Nov 22 '18 at 11:32
S.BullS.Bull
61
closed as off-topic by amWhy, Glorfindel, Jyrki Lahtonen, Servaes, user126154 Nov 22 '18 at 13:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Glorfindel, user126154
If this question can be reworded to fit the rules in the help center, please edit the question.
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Im having problems simplifying this equation, I know that the answer should be $$frac{800}{(pi^2)}cdot frac{(-1)^{n+1}}{(2n-1)^2}$$ however I'm not entirely sure why.. If someone could explain it would be a great help
algebra-precalculus
edited Nov 22 '18 at 12:03
Barry Cipra
59.3k653125
asked Nov 22 '18 at 11:32
S.BullS.Bull
61
closed as off-topic by amWhy, Glorfindel, Jyrki Lahtonen, Servaes, user126154 Nov 22 '18 at 13:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Glorfindel, user126154
If this question can be reworded to fit the rules in the help center, please edit the question.
1
I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
– Matti P.
Nov 22 '18 at 11:42
3
Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
– Matti P.
Nov 22 '18 at 11:42
1
Where is the equation?
– William Elliot
Nov 22 '18 at 11:56
1
@MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
– Barry Cipra
Nov 22 '18 at 11:57
1
I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
– Barry Cipra
Nov 22 '18 at 12:05
|
show 1 more comment
Im having problems simplifying this equation, I know that the answer should be $$frac{800}{(pi^2)}cdot frac{(-1)^{n+1}}{(2n-1)^2}$$ however I'm not entirely sure why.. If someone could explain it would be a great help
algebra-precalculus
edited Nov 22 '18 at 12:03
Barry Cipra
59.3k653125
asked Nov 22 '18 at 11:32
S.BullS.Bull
61
Im having problems simplifying this equation, I know that the answer should be $$frac{800}{(pi^2)}cdot frac{(-1)^{n+1}}{(2n-1)^2}$$ however I'm not entirely sure why.. If someone could explain it would be a great help
algebra-precalculus
algebra-precalculus
edited Nov 22 '18 at 12:03
Barry Cipra
59.3k653125
asked Nov 22 '18 at 11:32
S.BullS.Bull
61
edited Nov 22 '18 at 12:03
Barry Cipra
59.3k653125
asked Nov 22 '18 at 11:32
S.BullS.Bull
61
edited Nov 22 '18 at 12:03
Barry Cipra
59.3k653125
edited Nov 22 '18 at 12:03
Barry Cipra
59.3k653125
edited Nov 22 '18 at 12:03
Barry Cipra
59.3k653125
59.3k653125
asked Nov 22 '18 at 11:32
S.BullS.Bull
61
asked Nov 22 '18 at 11:32
S.BullS.Bull
61
asked Nov 22 '18 at 11:32
S.BullS.Bull
61
61
closed as off-topic by amWhy, Glorfindel, Jyrki Lahtonen, Servaes, user126154 Nov 22 '18 at 13:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Glorfindel, user126154
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Glorfindel, Jyrki Lahtonen, Servaes, user126154 Nov 22 '18 at 13:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Glorfindel, user126154
If this question can be reworded to fit the rules in the help center, please edit the question.
1
I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
– Matti P.
Nov 22 '18 at 11:42
3
Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
– Matti P.
Nov 22 '18 at 11:42
1
Where is the equation?
– William Elliot
Nov 22 '18 at 11:56
1
@MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
– Barry Cipra
Nov 22 '18 at 11:57
1
I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
– Barry Cipra
Nov 22 '18 at 12:05
|
show 1 more comment
1
I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
– Matti P.
Nov 22 '18 at 11:42
3
Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
– Matti P.
Nov 22 '18 at 11:42
1
Where is the equation?
– William Elliot
Nov 22 '18 at 11:56
1
@MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
– Barry Cipra
Nov 22 '18 at 11:57
1
I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
– Barry Cipra
Nov 22 '18 at 12:05
1
1
I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
– Matti P.
Nov 22 '18 at 11:42
I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
– Matti P.
Nov 22 '18 at 11:42
3
3
Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
– Matti P.
Nov 22 '18 at 11:42
Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
– Matti P.
Nov 22 '18 at 11:42
1
1
Where is the equation?
– William Elliot
Nov 22 '18 at 11:56
Where is the equation?
– William Elliot
Nov 22 '18 at 11:56
1
1
@MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
– Barry Cipra
Nov 22 '18 at 11:57
@MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
– Barry Cipra
Nov 22 '18 at 11:57
1
1
I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
– Barry Cipra
Nov 22 '18 at 12:05
I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
– Barry Cipra
Nov 22 '18 at 12:05
|
show 1 more comment
1 Answer
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active
oldest
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It seems that there are two questions here:
The mathematical question 'why is $(-1)^{n-1}$ equal to $(-1)^{n+1}$?'
and the aesthetic question 'why is the expression with all constants (numbers) in one fraction on the left and all stuff that depends on $n$ together in another fraction on the right considered simpler than the original form?'
The mathematical question has been answered in the comments. As for the aesthetic question: of course this is subjective, but the form with constants separated from the rest helps see the structure. For instance in some computations you can help yourself by 'summarizing' the entire term $frac{800}{pi^2}$ as $C$ (or some other letter of your choice) so that it won't distract and initimidate you as much when you continue to do stuff with the rest of the equation. Another example of a context where the simplification would be useful: if this were one term in an infinite sum
$$sum_{n=1}^infty frac{800}{pi^2} frac{(-1)^{n+1}}{(2n - 1)^2}$$ as suggested in the comments, then a next simplifying step could be to rewrite the entire sum as:
$$frac{800}{pi^2} sum_{n=1}^infty frac{(-1)^{n+1}}{(2n - 1)^2}$$
And this further simplification would be harder to see if you had not did the previous simplification (of changing the expression in the title to the one in the post).
answered Nov 22 '18 at 12:37
VincentVincent
3,07911228
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It seems that there are two questions here:
The mathematical question 'why is $(-1)^{n-1}$ equal to $(-1)^{n+1}$?'
and the aesthetic question 'why is the expression with all constants (numbers) in one fraction on the left and all stuff that depends on $n$ together in another fraction on the right considered simpler than the original form?'
The mathematical question has been answered in the comments. As for the aesthetic question: of course this is subjective, but the form with constants separated from the rest helps see the structure. For instance in some computations you can help yourself by 'summarizing' the entire term $frac{800}{pi^2}$ as $C$ (or some other letter of your choice) so that it won't distract and initimidate you as much when you continue to do stuff with the rest of the equation. Another example of a context where the simplification would be useful: if this were one term in an infinite sum
$$sum_{n=1}^infty frac{800}{pi^2} frac{(-1)^{n+1}}{(2n - 1)^2}$$ as suggested in the comments, then a next simplifying step could be to rewrite the entire sum as:
$$frac{800}{pi^2} sum_{n=1}^infty frac{(-1)^{n+1}}{(2n - 1)^2}$$
And this further simplification would be harder to see if you had not did the previous simplification (of changing the expression in the title to the one in the post).
answered Nov 22 '18 at 12:37
VincentVincent
3,07911228
add a comment |
It seems that there are two questions here:
The mathematical question 'why is $(-1)^{n-1}$ equal to $(-1)^{n+1}$?'
and the aesthetic question 'why is the expression with all constants (numbers) in one fraction on the left and all stuff that depends on $n$ together in another fraction on the right considered simpler than the original form?'
The mathematical question has been answered in the comments. As for the aesthetic question: of course this is subjective, but the form with constants separated from the rest helps see the structure. For instance in some computations you can help yourself by 'summarizing' the entire term $frac{800}{pi^2}$ as $C$ (or some other letter of your choice) so that it won't distract and initimidate you as much when you continue to do stuff with the rest of the equation. Another example of a context where the simplification would be useful: if this were one term in an infinite sum
$$sum_{n=1}^infty frac{800}{pi^2} frac{(-1)^{n+1}}{(2n - 1)^2}$$ as suggested in the comments, then a next simplifying step could be to rewrite the entire sum as:
$$frac{800}{pi^2} sum_{n=1}^infty frac{(-1)^{n+1}}{(2n - 1)^2}$$
And this further simplification would be harder to see if you had not did the previous simplification (of changing the expression in the title to the one in the post).
answered Nov 22 '18 at 12:37
VincentVincent
3,07911228
add a comment |
It seems that there are two questions here:
The mathematical question 'why is $(-1)^{n-1}$ equal to $(-1)^{n+1}$?'
and the aesthetic question 'why is the expression with all constants (numbers) in one fraction on the left and all stuff that depends on $n$ together in another fraction on the right considered simpler than the original form?'
The mathematical question has been answered in the comments. As for the aesthetic question: of course this is subjective, but the form with constants separated from the rest helps see the structure. For instance in some computations you can help yourself by 'summarizing' the entire term $frac{800}{pi^2}$ as $C$ (or some other letter of your choice) so that it won't distract and initimidate you as much when you continue to do stuff with the rest of the equation. Another example of a context where the simplification would be useful: if this were one term in an infinite sum
$$sum_{n=1}^infty frac{800}{pi^2} frac{(-1)^{n+1}}{(2n - 1)^2}$$ as suggested in the comments, then a next simplifying step could be to rewrite the entire sum as:
$$frac{800}{pi^2} sum_{n=1}^infty frac{(-1)^{n+1}}{(2n - 1)^2}$$
And this further simplification would be harder to see if you had not did the previous simplification (of changing the expression in the title to the one in the post).
answered Nov 22 '18 at 12:37
VincentVincent
3,07911228
It seems that there are two questions here:
The mathematical question 'why is $(-1)^{n-1}$ equal to $(-1)^{n+1}$?'
and the aesthetic question 'why is the expression with all constants (numbers) in one fraction on the left and all stuff that depends on $n$ together in another fraction on the right considered simpler than the original form?'
The mathematical question has been answered in the comments. As for the aesthetic question: of course this is subjective, but the form with constants separated from the rest helps see the structure. For instance in some computations you can help yourself by 'summarizing' the entire term $frac{800}{pi^2}$ as $C$ (or some other letter of your choice) so that it won't distract and initimidate you as much when you continue to do stuff with the rest of the equation. Another example of a context where the simplification would be useful: if this were one term in an infinite sum
$$sum_{n=1}^infty frac{800}{pi^2} frac{(-1)^{n+1}}{(2n - 1)^2}$$ as suggested in the comments, then a next simplifying step could be to rewrite the entire sum as:
$$frac{800}{pi^2} sum_{n=1}^infty frac{(-1)^{n+1}}{(2n - 1)^2}$$
And this further simplification would be harder to see if you had not did the previous simplification (of changing the expression in the title to the one in the post).
answered Nov 22 '18 at 12:37
VincentVincent
3,07911228
answered Nov 22 '18 at 12:37
VincentVincent
3,07911228
answered Nov 22 '18 at 12:37
VincentVincent
3,07911228
answered Nov 22 '18 at 12:37
VincentVincent
3,07911228
3,07911228
add a comment |
add a comment |
1
I don't get it. You mention an expression, and then you say "I know that the answer should be .. " ... and then the same expression. Did you accidentally write the same expression twice or did I not understand the question?
– Matti P.
Nov 22 '18 at 11:42
3
Also, if the problem is just simplifying the given expression, I would say that it's already in its simplest form.
– Matti P.
Nov 22 '18 at 11:42
1
Where is the equation?
– William Elliot
Nov 22 '18 at 11:56
1
@MattiP., someone edited the title to make it identical with the expression in the text. However, the only differences in the original were the order of the terms and the power of $-1$, which was $n-1$ in the title instead of $n+1$. (If those differences are what the OP finds puzzling, the commutative law for multiplication accounts for the reordering of the terms, and $$(-1)^{n+1}=(-1)^{(n-1)+2}=(-1)^2(-1)^{n-1}=1(-1)^{n-1}=(-1)^{n-1}$$ accounts for the change from $n-1$ to $n+1$.)
– Barry Cipra
Nov 22 '18 at 11:57
1
I decided to edit the title back to its original appearance (but in built-up fraction form). It's possible that "simplify" here simply means pulling all the constant stuff out front and collecting the terms with $n$'s.
– Barry Cipra
Nov 22 '18 at 12:05