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Impulse train: why an indeterminate result?
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I have an impulse train given by
$$frac1R+1+fracsum_k=1^R cos(frac2k pi xR+1)R+1$$
It seems obvious to me that, for $x=0$, the function returns $1$. This is because $cos (0)=1$, and we therefore end up with $frac1R+1+fracRR+1=fracR+1R+1=1$.
However, my math software (Mathematica) gives an indeterminate result at $x=0$. Usually this means there is a division by $0$ somewhere. But I can't see any reason for this function to produce an indeterminate result. (This is about the maths of the situation, not the way the software works.)
Can anyone explain?
sequences-and-series trigonometry
asked Jul 22 at 21:38
Richard Burke-Ward
1428
 |Â
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up vote
2
down vote
favorite
I have an impulse train given by
$$frac1R+1+fracsum_k=1^R cos(frac2k pi xR+1)R+1$$
It seems obvious to me that, for $x=0$, the function returns $1$. This is because $cos (0)=1$, and we therefore end up with $frac1R+1+fracRR+1=fracR+1R+1=1$.
However, my math software (Mathematica) gives an indeterminate result at $x=0$. Usually this means there is a division by $0$ somewhere. But I can't see any reason for this function to produce an indeterminate result. (This is about the maths of the situation, not the way the software works.)
Can anyone explain?
sequences-and-series trigonometry
asked Jul 22 at 21:38
Richard Burke-Ward
1428
1
It is indeed the case that, when $x = 0$, the expression evaluates to $1$ (assuming that $R neq -1$). With that being said, the only real question is why your software did what it did.
– Omnomnomnom
Jul 22 at 21:44
You're right about the math; I'm not sure why Mathematica would stumble here. Maybe it's nervous about $R$? I would try setting R to be some specific value and checking what it says then.
– rwbogl
Jul 22 at 21:45
OK, thanks. I'll go over to the software site and ask there. (Assigning a value to $R$ makes no difference, BTW.)
– Richard Burke-Ward
Jul 22 at 21:49
SymPy has a similar problem. Defining just the sum, then substituting $x = 0$ yields an unevaluated sum $sum_k = 1^R 1$, which is very strange! I'm a little disappointed in both CAS's.
– rwbogl
Jul 22 at 21:54
Could you show the result obtained for $sum_k=1^R cos(frac2k pi xR+1)$ ? Whatever it could be, could you try a Taylor expansion of it around $x=0$. Thanks.
– Claude Leibovici
Jul 23 at 5:22
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have an impulse train given by
$$frac1R+1+fracsum_k=1^R cos(frac2k pi xR+1)R+1$$
It seems obvious to me that, for $x=0$, the function returns $1$. This is because $cos (0)=1$, and we therefore end up with $frac1R+1+fracRR+1=fracR+1R+1=1$.
However, my math software (Mathematica) gives an indeterminate result at $x=0$. Usually this means there is a division by $0$ somewhere. But I can't see any reason for this function to produce an indeterminate result. (This is about the maths of the situation, not the way the software works.)
Can anyone explain?
sequences-and-series trigonometry
asked Jul 22 at 21:38
Richard Burke-Ward
1428
I have an impulse train given by
$$frac1R+1+fracsum_k=1^R cos(frac2k pi xR+1)R+1$$
It seems obvious to me that, for $x=0$, the function returns $1$. This is because $cos (0)=1$, and we therefore end up with $frac1R+1+fracRR+1=fracR+1R+1=1$.
However, my math software (Mathematica) gives an indeterminate result at $x=0$. Usually this means there is a division by $0$ somewhere. But I can't see any reason for this function to produce an indeterminate result. (This is about the maths of the situation, not the way the software works.)
Can anyone explain?
sequences-and-series trigonometry
asked Jul 22 at 21:38
Richard Burke-Ward
1428
asked Jul 22 at 21:38
Richard Burke-Ward
1428
asked Jul 22 at 21:38
Richard Burke-Ward
1428
asked Jul 22 at 21:38
Richard Burke-Ward
1428
1428
1
It is indeed the case that, when $x = 0$, the expression evaluates to $1$ (assuming that $R neq -1$). With that being said, the only real question is why your software did what it did.
– Omnomnomnom
Jul 22 at 21:44
You're right about the math; I'm not sure why Mathematica would stumble here. Maybe it's nervous about $R$? I would try setting R to be some specific value and checking what it says then.
– rwbogl
Jul 22 at 21:45
OK, thanks. I'll go over to the software site and ask there. (Assigning a value to $R$ makes no difference, BTW.)
– Richard Burke-Ward
Jul 22 at 21:49
SymPy has a similar problem. Defining just the sum, then substituting $x = 0$ yields an unevaluated sum $sum_k = 1^R 1$, which is very strange! I'm a little disappointed in both CAS's.
– rwbogl
Jul 22 at 21:54
Could you show the result obtained for $sum_k=1^R cos(frac2k pi xR+1)$ ? Whatever it could be, could you try a Taylor expansion of it around $x=0$. Thanks.
– Claude Leibovici
Jul 23 at 5:22
 |Â
show 1 more comment
1
It is indeed the case that, when $x = 0$, the expression evaluates to $1$ (assuming that $R neq -1$). With that being said, the only real question is why your software did what it did.
– Omnomnomnom
Jul 22 at 21:44
You're right about the math; I'm not sure why Mathematica would stumble here. Maybe it's nervous about $R$? I would try setting R to be some specific value and checking what it says then.
– rwbogl
Jul 22 at 21:45
OK, thanks. I'll go over to the software site and ask there. (Assigning a value to $R$ makes no difference, BTW.)
– Richard Burke-Ward
Jul 22 at 21:49
SymPy has a similar problem. Defining just the sum, then substituting $x = 0$ yields an unevaluated sum $sum_k = 1^R 1$, which is very strange! I'm a little disappointed in both CAS's.
– rwbogl
Jul 22 at 21:54
Could you show the result obtained for $sum_k=1^R cos(frac2k pi xR+1)$ ? Whatever it could be, could you try a Taylor expansion of it around $x=0$. Thanks.
– Claude Leibovici
Jul 23 at 5:22
1
1
It is indeed the case that, when $x = 0$, the expression evaluates to $1$ (assuming that $R neq -1$). With that being said, the only real question is why your software did what it did.
– Omnomnomnom
Jul 22 at 21:44
It is indeed the case that, when $x = 0$, the expression evaluates to $1$ (assuming that $R neq -1$). With that being said, the only real question is why your software did what it did.
– Omnomnomnom
Jul 22 at 21:44
You're right about the math; I'm not sure why Mathematica would stumble here. Maybe it's nervous about $R$? I would try setting R to be some specific value and checking what it says then.
– rwbogl
Jul 22 at 21:45
You're right about the math; I'm not sure why Mathematica would stumble here. Maybe it's nervous about $R$? I would try setting R to be some specific value and checking what it says then.
– rwbogl
Jul 22 at 21:45
OK, thanks. I'll go over to the software site and ask there. (Assigning a value to $R$ makes no difference, BTW.)
– Richard Burke-Ward
Jul 22 at 21:49
OK, thanks. I'll go over to the software site and ask there. (Assigning a value to $R$ makes no difference, BTW.)
– Richard Burke-Ward
Jul 22 at 21:49
SymPy has a similar problem. Defining just the sum, then substituting $x = 0$ yields an unevaluated sum $sum_k = 1^R 1$, which is very strange! I'm a little disappointed in both CAS's.
– rwbogl
Jul 22 at 21:54
SymPy has a similar problem. Defining just the sum, then substituting $x = 0$ yields an unevaluated sum $sum_k = 1^R 1$, which is very strange! I'm a little disappointed in both CAS's.
– rwbogl
Jul 22 at 21:54
Could you show the result obtained for $sum_k=1^R cos(frac2k pi xR+1)$ ? Whatever it could be, could you try a Taylor expansion of it around $x=0$. Thanks.
– Claude Leibovici
Jul 23 at 5:22
Could you show the result obtained for $sum_k=1^R cos(frac2k pi xR+1)$ ? Whatever it could be, could you try a Taylor expansion of it around $x=0$. Thanks.
– Claude Leibovici
Jul 23 at 5:22
 |Â
show 1 more comment
1 Answer
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I set the problem in Wolfram Development Platform and the limit is properly given.
The only thing I noticed is that the simplification of the result is not as good as it could be since, using another CAS,
$$sum_k=1^R cos left(frac2 kpi xR+1right)=frac12 left(sin left(fracpi (2 R+1) xR+1right) csc left(fracpi
xR+1right)-1right)$$ making
$$frac1R+1+fracsum_k=1^R cos(frac2k pi xR+1)R+1=fracsin left(fracpi (2 R+1) xR+1right) csc left(fracpi
xR+1right)+12 (R+1)$$ Probably, the indetermination comes from the limit of $csc(t)$ when $tto 0$ while, using Taylor expansion
$$sin left(fracpi (2 R+1) xR+1right)=frac( pi (2 R+1) ) R+1x-frac( pi (2R+1) )^3 6
(R+1)^3x^3+Oleft(x^5right)$$
$$csc left(fracpi xR+1right)=fracR+1pi x+fracpi x6 (R+1)+frac7 pi ^3 x^3360
(R+1)^3+Oleft(x^5right)$$
$$sin left(fracpi (2 R+1) xR+1right),csc left(fracpi xR+1right)=(2 R+1)-frac2 pi ^2 R (2 R+1) 3 (R+1)x^2+Oleft(x^4right)$$
answered Jul 23 at 5:09
Claude Leibovici
111k1055126
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I set the problem in Wolfram Development Platform and the limit is properly given.
The only thing I noticed is that the simplification of the result is not as good as it could be since, using another CAS,
$$sum_k=1^R cos left(frac2 kpi xR+1right)=frac12 left(sin left(fracpi (2 R+1) xR+1right) csc left(fracpi
xR+1right)-1right)$$ making
$$frac1R+1+fracsum_k=1^R cos(frac2k pi xR+1)R+1=fracsin left(fracpi (2 R+1) xR+1right) csc left(fracpi
xR+1right)+12 (R+1)$$ Probably, the indetermination comes from the limit of $csc(t)$ when $tto 0$ while, using Taylor expansion
$$sin left(fracpi (2 R+1) xR+1right)=frac( pi (2 R+1) ) R+1x-frac( pi (2R+1) )^3 6
(R+1)^3x^3+Oleft(x^5right)$$
$$csc left(fracpi xR+1right)=fracR+1pi x+fracpi x6 (R+1)+frac7 pi ^3 x^3360
(R+1)^3+Oleft(x^5right)$$
$$sin left(fracpi (2 R+1) xR+1right),csc left(fracpi xR+1right)=(2 R+1)-frac2 pi ^2 R (2 R+1) 3 (R+1)x^2+Oleft(x^4right)$$
answered Jul 23 at 5:09
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
accepted
I set the problem in Wolfram Development Platform and the limit is properly given.
The only thing I noticed is that the simplification of the result is not as good as it could be since, using another CAS,
$$sum_k=1^R cos left(frac2 kpi xR+1right)=frac12 left(sin left(fracpi (2 R+1) xR+1right) csc left(fracpi
xR+1right)-1right)$$ making
$$frac1R+1+fracsum_k=1^R cos(frac2k pi xR+1)R+1=fracsin left(fracpi (2 R+1) xR+1right) csc left(fracpi
xR+1right)+12 (R+1)$$ Probably, the indetermination comes from the limit of $csc(t)$ when $tto 0$ while, using Taylor expansion
$$sin left(fracpi (2 R+1) xR+1right)=frac( pi (2 R+1) ) R+1x-frac( pi (2R+1) )^3 6
(R+1)^3x^3+Oleft(x^5right)$$
$$csc left(fracpi xR+1right)=fracR+1pi x+fracpi x6 (R+1)+frac7 pi ^3 x^3360
(R+1)^3+Oleft(x^5right)$$
$$sin left(fracpi (2 R+1) xR+1right),csc left(fracpi xR+1right)=(2 R+1)-frac2 pi ^2 R (2 R+1) 3 (R+1)x^2+Oleft(x^4right)$$
answered Jul 23 at 5:09
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I set the problem in Wolfram Development Platform and the limit is properly given.
The only thing I noticed is that the simplification of the result is not as good as it could be since, using another CAS,
$$sum_k=1^R cos left(frac2 kpi xR+1right)=frac12 left(sin left(fracpi (2 R+1) xR+1right) csc left(fracpi
xR+1right)-1right)$$ making
$$frac1R+1+fracsum_k=1^R cos(frac2k pi xR+1)R+1=fracsin left(fracpi (2 R+1) xR+1right) csc left(fracpi
xR+1right)+12 (R+1)$$ Probably, the indetermination comes from the limit of $csc(t)$ when $tto 0$ while, using Taylor expansion
$$sin left(fracpi (2 R+1) xR+1right)=frac( pi (2 R+1) ) R+1x-frac( pi (2R+1) )^3 6
(R+1)^3x^3+Oleft(x^5right)$$
$$csc left(fracpi xR+1right)=fracR+1pi x+fracpi x6 (R+1)+frac7 pi ^3 x^3360
(R+1)^3+Oleft(x^5right)$$
$$sin left(fracpi (2 R+1) xR+1right),csc left(fracpi xR+1right)=(2 R+1)-frac2 pi ^2 R (2 R+1) 3 (R+1)x^2+Oleft(x^4right)$$
answered Jul 23 at 5:09
Claude Leibovici
111k1055126
I set the problem in Wolfram Development Platform and the limit is properly given.
The only thing I noticed is that the simplification of the result is not as good as it could be since, using another CAS,
$$sum_k=1^R cos left(frac2 kpi xR+1right)=frac12 left(sin left(fracpi (2 R+1) xR+1right) csc left(fracpi
xR+1right)-1right)$$ making
$$frac1R+1+fracsum_k=1^R cos(frac2k pi xR+1)R+1=fracsin left(fracpi (2 R+1) xR+1right) csc left(fracpi
xR+1right)+12 (R+1)$$ Probably, the indetermination comes from the limit of $csc(t)$ when $tto 0$ while, using Taylor expansion
$$sin left(fracpi (2 R+1) xR+1right)=frac( pi (2 R+1) ) R+1x-frac( pi (2R+1) )^3 6
(R+1)^3x^3+Oleft(x^5right)$$
$$csc left(fracpi xR+1right)=fracR+1pi x+fracpi x6 (R+1)+frac7 pi ^3 x^3360
(R+1)^3+Oleft(x^5right)$$
$$sin left(fracpi (2 R+1) xR+1right),csc left(fracpi xR+1right)=(2 R+1)-frac2 pi ^2 R (2 R+1) 3 (R+1)x^2+Oleft(x^4right)$$
answered Jul 23 at 5:09
Claude Leibovici
111k1055126
answered Jul 23 at 5:09
Claude Leibovici
111k1055126
answered Jul 23 at 5:09
Claude Leibovici
111k1055126
answered Jul 23 at 5:09
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
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1
It is indeed the case that, when $x = 0$, the expression evaluates to $1$ (assuming that $R neq -1$). With that being said, the only real question is why your software did what it did.
– Omnomnomnom
Jul 22 at 21:44
You're right about the math; I'm not sure why Mathematica would stumble here. Maybe it's nervous about $R$? I would try setting R to be some specific value and checking what it says then.
– rwbogl
Jul 22 at 21:45
OK, thanks. I'll go over to the software site and ask there. (Assigning a value to $R$ makes no difference, BTW.)
– Richard Burke-Ward
Jul 22 at 21:49
SymPy has a similar problem. Defining just the sum, then substituting $x = 0$ yields an unevaluated sum $sum_k = 1^R 1$, which is very strange! I'm a little disappointed in both CAS's.
– rwbogl
Jul 22 at 21:54
Could you show the result obtained for $sum_k=1^R cos(frac2k pi xR+1)$ ? Whatever it could be, could you try a Taylor expansion of it around $x=0$. Thanks.
– Claude Leibovici
Jul 23 at 5:22