Average of (harmonic) average speeds?

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Suppose I have travel speeds $x_i$ at times $z_i$ ($i = 1, ldots, n$) from point $A$ to point $B$.
Furthermore, assume that I have travel speeds $y_i$ at times $z_i$ ($i = 1, ldots, n$) from point $B$ to point $A$.
Now, I know that the average speed at time $z_i$ ($i = 1, ldots, n$) is given by the harmonic mean
$$frac2x_iy_ix_i + y_i.$$
Here is my question:
What formula do we use to compute for the average of the (harmonic) average speeds, say, per quarter?
I am confused as to whether we still need to use the harmonic average formula (since we are still dealing with equal travel distances [from $A$ to $B$ or from $B$ to $A$]), or if we could already use the (ordinary) arithmetic average formula.
algebra-precalculus average
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up vote
1
down vote
favorite
Suppose I have travel speeds $x_i$ at times $z_i$ ($i = 1, ldots, n$) from point $A$ to point $B$.
Furthermore, assume that I have travel speeds $y_i$ at times $z_i$ ($i = 1, ldots, n$) from point $B$ to point $A$.
Now, I know that the average speed at time $z_i$ ($i = 1, ldots, n$) is given by the harmonic mean
$$frac2x_iy_ix_i + y_i.$$
Here is my question:
What formula do we use to compute for the average of the (harmonic) average speeds, say, per quarter?
I am confused as to whether we still need to use the harmonic average formula (since we are still dealing with equal travel distances [from $A$ to $B$ or from $B$ to $A$]), or if we could already use the (ordinary) arithmetic average formula.
algebra-precalculus average
What do you mean by average speed at time $i$ ? Consider the following - 1) An object can be travelling only from A to B or B to A ,but not BOTH, so average of what? 2) Time is continuous, not discreet, so why only integral times ?
â DevashishKaushik
Aug 21 at 9:58
@DevashishKaushik, by average speed at time $i$ I meant the average of the speeds $x_i$ and $y_i$, which I am sure you will agree to be the harmonic average of $x_i$ and $y_i$. And no, the times are not necessarily integral, I meant the $i$'s to be indices.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:40
I have edited my question in response to your comment, @DevashishKaushik. I hope the question makes more sense now.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:43
+1 Yes, it's quite clear what you are asking, as evidenced by the presence of 2 answers. :D
â DevashishKaushik
Aug 21 at 12:13
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose I have travel speeds $x_i$ at times $z_i$ ($i = 1, ldots, n$) from point $A$ to point $B$.
Furthermore, assume that I have travel speeds $y_i$ at times $z_i$ ($i = 1, ldots, n$) from point $B$ to point $A$.
Now, I know that the average speed at time $z_i$ ($i = 1, ldots, n$) is given by the harmonic mean
$$frac2x_iy_ix_i + y_i.$$
Here is my question:
What formula do we use to compute for the average of the (harmonic) average speeds, say, per quarter?
I am confused as to whether we still need to use the harmonic average formula (since we are still dealing with equal travel distances [from $A$ to $B$ or from $B$ to $A$]), or if we could already use the (ordinary) arithmetic average formula.
algebra-precalculus average
Suppose I have travel speeds $x_i$ at times $z_i$ ($i = 1, ldots, n$) from point $A$ to point $B$.
Furthermore, assume that I have travel speeds $y_i$ at times $z_i$ ($i = 1, ldots, n$) from point $B$ to point $A$.
Now, I know that the average speed at time $z_i$ ($i = 1, ldots, n$) is given by the harmonic mean
$$frac2x_iy_ix_i + y_i.$$
Here is my question:
What formula do we use to compute for the average of the (harmonic) average speeds, say, per quarter?
I am confused as to whether we still need to use the harmonic average formula (since we are still dealing with equal travel distances [from $A$ to $B$ or from $B$ to $A$]), or if we could already use the (ordinary) arithmetic average formula.
algebra-precalculus average
edited Aug 21 at 10:42
asked Aug 21 at 8:23
Jose Arnaldo Bebita Dris
5,07531941
5,07531941
What do you mean by average speed at time $i$ ? Consider the following - 1) An object can be travelling only from A to B or B to A ,but not BOTH, so average of what? 2) Time is continuous, not discreet, so why only integral times ?
â DevashishKaushik
Aug 21 at 9:58
@DevashishKaushik, by average speed at time $i$ I meant the average of the speeds $x_i$ and $y_i$, which I am sure you will agree to be the harmonic average of $x_i$ and $y_i$. And no, the times are not necessarily integral, I meant the $i$'s to be indices.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:40
I have edited my question in response to your comment, @DevashishKaushik. I hope the question makes more sense now.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:43
+1 Yes, it's quite clear what you are asking, as evidenced by the presence of 2 answers. :D
â DevashishKaushik
Aug 21 at 12:13
add a comment |Â
What do you mean by average speed at time $i$ ? Consider the following - 1) An object can be travelling only from A to B or B to A ,but not BOTH, so average of what? 2) Time is continuous, not discreet, so why only integral times ?
â DevashishKaushik
Aug 21 at 9:58
@DevashishKaushik, by average speed at time $i$ I meant the average of the speeds $x_i$ and $y_i$, which I am sure you will agree to be the harmonic average of $x_i$ and $y_i$. And no, the times are not necessarily integral, I meant the $i$'s to be indices.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:40
I have edited my question in response to your comment, @DevashishKaushik. I hope the question makes more sense now.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:43
+1 Yes, it's quite clear what you are asking, as evidenced by the presence of 2 answers. :D
â DevashishKaushik
Aug 21 at 12:13
What do you mean by average speed at time $i$ ? Consider the following - 1) An object can be travelling only from A to B or B to A ,but not BOTH, so average of what? 2) Time is continuous, not discreet, so why only integral times ?
â DevashishKaushik
Aug 21 at 9:58
What do you mean by average speed at time $i$ ? Consider the following - 1) An object can be travelling only from A to B or B to A ,but not BOTH, so average of what? 2) Time is continuous, not discreet, so why only integral times ?
â DevashishKaushik
Aug 21 at 9:58
@DevashishKaushik, by average speed at time $i$ I meant the average of the speeds $x_i$ and $y_i$, which I am sure you will agree to be the harmonic average of $x_i$ and $y_i$. And no, the times are not necessarily integral, I meant the $i$'s to be indices.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:40
@DevashishKaushik, by average speed at time $i$ I meant the average of the speeds $x_i$ and $y_i$, which I am sure you will agree to be the harmonic average of $x_i$ and $y_i$. And no, the times are not necessarily integral, I meant the $i$'s to be indices.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:40
I have edited my question in response to your comment, @DevashishKaushik. I hope the question makes more sense now.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:43
I have edited my question in response to your comment, @DevashishKaushik. I hope the question makes more sense now.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:43
+1 Yes, it's quite clear what you are asking, as evidenced by the presence of 2 answers. :D
â DevashishKaushik
Aug 21 at 12:13
+1 Yes, it's quite clear what you are asking, as evidenced by the presence of 2 answers. :D
â DevashishKaushik
Aug 21 at 12:13
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Calling this the âaverage speedâ is ambiguous (and this ambiguity may have contributed to your confusion). If (as in this case) two speeds are measured over the same distance $s$, in the sense that the distance was covered in times $t_1$ and $t_2$ and the corresponding speeds $v_1=frac st_1$ and $v_2=frac st_2$ are calculated, then the harmonic mean is the speed that would have been calculated if the two measurements had been combined into one:
$$
frac2st_1+t_2=frac2sfrac sv_1+frac sv_2=frac2v_1v_2v_1+v_2;.
$$
Now the question is what you want to calculate for the quarter. You could just take the (arithmetic) average of the speeds if that's what you happen to be interested in. However, if you want to answer the same question again for the entire quarter that was answered by the harmonic mean for two measurements, namely, what speed would have been measured if all the measurements in the quarter had formed a single combined measurement, then you need to take the generalized harmonic mean of all the speeds measured:
$$
fracnssum_i=1^nt_i=fracnssum_i=1^nfrac sv_i=left(frac1nsum_i=1^nv_i^-1right)^-1;.
$$
You'll get the same result whether you first combine the speeds pairwise for each time $z_i$ and then combine the resulting speeds, or whether you combine all the $x_i$ and $y_i$ together in one go.
Basically, what you're doing is just averaging inverse speeds instead of speeds, so as long as you take the reciprocal before and after each averaging operation, you can treat this like a normal average.
(Note the terminological connection to the harmonic numbers and the harmonic series, which also involve reciprocals.)
Thanks, @joriki! This answer articulates to a certain extent what I had in mind.
â Jose Arnaldo Bebita Dris
Aug 21 at 11:18
add a comment |Â
up vote
3
down vote
I'll just add to the nice answer from @joriki.
Every "average" has a certain use. You don't just use an average without having a reason behind it (not just statistics but physics). For example, if you wanted to find the velocity that you can use in the equation for kinetic energy instead of summing over all objects (for example, calculating the "average" velocity of molecules at some temperature), you would calculate the quadratic average (root mean square) because you would get the definition of average from $(sum m_i)v_avg^2=sum m_iv_i^2$. Another example is the center of mass. In linear gravity (on the surface of the earth, not too far), the center of mass is the linear average of position over all objects. But when you are dealing with Newtonian gravity (big picture, planets and stuff), it's the harmonic average that determines the point around which the objects orbit.
So - there is no one "right" average - each has a different meaning, and is used differently.
That being said, back to your question. If these velocities are fundamental individual results (measurements with a radar or something) and you are doing statistics on these measurements... then use the arithmetic average, because that's what you do in statistical analysis when you are looking for the most likely value of some quantity, when a Gaussian distribution is assumed (symmetric to deviations up or down). But I don't think this applies to your situation.
Averaging inverse velocities is just like averaging times... which is usual in sports. And in sports, even doing statistics, you will do statistics with times, not velocities (and thus compute average times). So... as far as I understand, if talking about velocities is simply a convenient conversion, but the time is the true thing that was measured, then I'd say that just use the harmonic average for everything, because talking about velocities is just a "mask" over the true quantities in the system and velocities themselves were not really measured or detected directly. If I was doing this... I'd use harmonic average, or just stop pretending that we are talking about velocities (using times instead).
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Calling this the âaverage speedâ is ambiguous (and this ambiguity may have contributed to your confusion). If (as in this case) two speeds are measured over the same distance $s$, in the sense that the distance was covered in times $t_1$ and $t_2$ and the corresponding speeds $v_1=frac st_1$ and $v_2=frac st_2$ are calculated, then the harmonic mean is the speed that would have been calculated if the two measurements had been combined into one:
$$
frac2st_1+t_2=frac2sfrac sv_1+frac sv_2=frac2v_1v_2v_1+v_2;.
$$
Now the question is what you want to calculate for the quarter. You could just take the (arithmetic) average of the speeds if that's what you happen to be interested in. However, if you want to answer the same question again for the entire quarter that was answered by the harmonic mean for two measurements, namely, what speed would have been measured if all the measurements in the quarter had formed a single combined measurement, then you need to take the generalized harmonic mean of all the speeds measured:
$$
fracnssum_i=1^nt_i=fracnssum_i=1^nfrac sv_i=left(frac1nsum_i=1^nv_i^-1right)^-1;.
$$
You'll get the same result whether you first combine the speeds pairwise for each time $z_i$ and then combine the resulting speeds, or whether you combine all the $x_i$ and $y_i$ together in one go.
Basically, what you're doing is just averaging inverse speeds instead of speeds, so as long as you take the reciprocal before and after each averaging operation, you can treat this like a normal average.
(Note the terminological connection to the harmonic numbers and the harmonic series, which also involve reciprocals.)
Thanks, @joriki! This answer articulates to a certain extent what I had in mind.
â Jose Arnaldo Bebita Dris
Aug 21 at 11:18
add a comment |Â
up vote
2
down vote
accepted
Calling this the âaverage speedâ is ambiguous (and this ambiguity may have contributed to your confusion). If (as in this case) two speeds are measured over the same distance $s$, in the sense that the distance was covered in times $t_1$ and $t_2$ and the corresponding speeds $v_1=frac st_1$ and $v_2=frac st_2$ are calculated, then the harmonic mean is the speed that would have been calculated if the two measurements had been combined into one:
$$
frac2st_1+t_2=frac2sfrac sv_1+frac sv_2=frac2v_1v_2v_1+v_2;.
$$
Now the question is what you want to calculate for the quarter. You could just take the (arithmetic) average of the speeds if that's what you happen to be interested in. However, if you want to answer the same question again for the entire quarter that was answered by the harmonic mean for two measurements, namely, what speed would have been measured if all the measurements in the quarter had formed a single combined measurement, then you need to take the generalized harmonic mean of all the speeds measured:
$$
fracnssum_i=1^nt_i=fracnssum_i=1^nfrac sv_i=left(frac1nsum_i=1^nv_i^-1right)^-1;.
$$
You'll get the same result whether you first combine the speeds pairwise for each time $z_i$ and then combine the resulting speeds, or whether you combine all the $x_i$ and $y_i$ together in one go.
Basically, what you're doing is just averaging inverse speeds instead of speeds, so as long as you take the reciprocal before and after each averaging operation, you can treat this like a normal average.
(Note the terminological connection to the harmonic numbers and the harmonic series, which also involve reciprocals.)
Thanks, @joriki! This answer articulates to a certain extent what I had in mind.
â Jose Arnaldo Bebita Dris
Aug 21 at 11:18
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Calling this the âaverage speedâ is ambiguous (and this ambiguity may have contributed to your confusion). If (as in this case) two speeds are measured over the same distance $s$, in the sense that the distance was covered in times $t_1$ and $t_2$ and the corresponding speeds $v_1=frac st_1$ and $v_2=frac st_2$ are calculated, then the harmonic mean is the speed that would have been calculated if the two measurements had been combined into one:
$$
frac2st_1+t_2=frac2sfrac sv_1+frac sv_2=frac2v_1v_2v_1+v_2;.
$$
Now the question is what you want to calculate for the quarter. You could just take the (arithmetic) average of the speeds if that's what you happen to be interested in. However, if you want to answer the same question again for the entire quarter that was answered by the harmonic mean for two measurements, namely, what speed would have been measured if all the measurements in the quarter had formed a single combined measurement, then you need to take the generalized harmonic mean of all the speeds measured:
$$
fracnssum_i=1^nt_i=fracnssum_i=1^nfrac sv_i=left(frac1nsum_i=1^nv_i^-1right)^-1;.
$$
You'll get the same result whether you first combine the speeds pairwise for each time $z_i$ and then combine the resulting speeds, or whether you combine all the $x_i$ and $y_i$ together in one go.
Basically, what you're doing is just averaging inverse speeds instead of speeds, so as long as you take the reciprocal before and after each averaging operation, you can treat this like a normal average.
(Note the terminological connection to the harmonic numbers and the harmonic series, which also involve reciprocals.)
Calling this the âaverage speedâ is ambiguous (and this ambiguity may have contributed to your confusion). If (as in this case) two speeds are measured over the same distance $s$, in the sense that the distance was covered in times $t_1$ and $t_2$ and the corresponding speeds $v_1=frac st_1$ and $v_2=frac st_2$ are calculated, then the harmonic mean is the speed that would have been calculated if the two measurements had been combined into one:
$$
frac2st_1+t_2=frac2sfrac sv_1+frac sv_2=frac2v_1v_2v_1+v_2;.
$$
Now the question is what you want to calculate for the quarter. You could just take the (arithmetic) average of the speeds if that's what you happen to be interested in. However, if you want to answer the same question again for the entire quarter that was answered by the harmonic mean for two measurements, namely, what speed would have been measured if all the measurements in the quarter had formed a single combined measurement, then you need to take the generalized harmonic mean of all the speeds measured:
$$
fracnssum_i=1^nt_i=fracnssum_i=1^nfrac sv_i=left(frac1nsum_i=1^nv_i^-1right)^-1;.
$$
You'll get the same result whether you first combine the speeds pairwise for each time $z_i$ and then combine the resulting speeds, or whether you combine all the $x_i$ and $y_i$ together in one go.
Basically, what you're doing is just averaging inverse speeds instead of speeds, so as long as you take the reciprocal before and after each averaging operation, you can treat this like a normal average.
(Note the terminological connection to the harmonic numbers and the harmonic series, which also involve reciprocals.)
edited Aug 21 at 11:06
answered Aug 21 at 11:00
joriki
166k10180331
166k10180331
Thanks, @joriki! This answer articulates to a certain extent what I had in mind.
â Jose Arnaldo Bebita Dris
Aug 21 at 11:18
add a comment |Â
Thanks, @joriki! This answer articulates to a certain extent what I had in mind.
â Jose Arnaldo Bebita Dris
Aug 21 at 11:18
Thanks, @joriki! This answer articulates to a certain extent what I had in mind.
â Jose Arnaldo Bebita Dris
Aug 21 at 11:18
Thanks, @joriki! This answer articulates to a certain extent what I had in mind.
â Jose Arnaldo Bebita Dris
Aug 21 at 11:18
add a comment |Â
up vote
3
down vote
I'll just add to the nice answer from @joriki.
Every "average" has a certain use. You don't just use an average without having a reason behind it (not just statistics but physics). For example, if you wanted to find the velocity that you can use in the equation for kinetic energy instead of summing over all objects (for example, calculating the "average" velocity of molecules at some temperature), you would calculate the quadratic average (root mean square) because you would get the definition of average from $(sum m_i)v_avg^2=sum m_iv_i^2$. Another example is the center of mass. In linear gravity (on the surface of the earth, not too far), the center of mass is the linear average of position over all objects. But when you are dealing with Newtonian gravity (big picture, planets and stuff), it's the harmonic average that determines the point around which the objects orbit.
So - there is no one "right" average - each has a different meaning, and is used differently.
That being said, back to your question. If these velocities are fundamental individual results (measurements with a radar or something) and you are doing statistics on these measurements... then use the arithmetic average, because that's what you do in statistical analysis when you are looking for the most likely value of some quantity, when a Gaussian distribution is assumed (symmetric to deviations up or down). But I don't think this applies to your situation.
Averaging inverse velocities is just like averaging times... which is usual in sports. And in sports, even doing statistics, you will do statistics with times, not velocities (and thus compute average times). So... as far as I understand, if talking about velocities is simply a convenient conversion, but the time is the true thing that was measured, then I'd say that just use the harmonic average for everything, because talking about velocities is just a "mask" over the true quantities in the system and velocities themselves were not really measured or detected directly. If I was doing this... I'd use harmonic average, or just stop pretending that we are talking about velocities (using times instead).
add a comment |Â
up vote
3
down vote
I'll just add to the nice answer from @joriki.
Every "average" has a certain use. You don't just use an average without having a reason behind it (not just statistics but physics). For example, if you wanted to find the velocity that you can use in the equation for kinetic energy instead of summing over all objects (for example, calculating the "average" velocity of molecules at some temperature), you would calculate the quadratic average (root mean square) because you would get the definition of average from $(sum m_i)v_avg^2=sum m_iv_i^2$. Another example is the center of mass. In linear gravity (on the surface of the earth, not too far), the center of mass is the linear average of position over all objects. But when you are dealing with Newtonian gravity (big picture, planets and stuff), it's the harmonic average that determines the point around which the objects orbit.
So - there is no one "right" average - each has a different meaning, and is used differently.
That being said, back to your question. If these velocities are fundamental individual results (measurements with a radar or something) and you are doing statistics on these measurements... then use the arithmetic average, because that's what you do in statistical analysis when you are looking for the most likely value of some quantity, when a Gaussian distribution is assumed (symmetric to deviations up or down). But I don't think this applies to your situation.
Averaging inverse velocities is just like averaging times... which is usual in sports. And in sports, even doing statistics, you will do statistics with times, not velocities (and thus compute average times). So... as far as I understand, if talking about velocities is simply a convenient conversion, but the time is the true thing that was measured, then I'd say that just use the harmonic average for everything, because talking about velocities is just a "mask" over the true quantities in the system and velocities themselves were not really measured or detected directly. If I was doing this... I'd use harmonic average, or just stop pretending that we are talking about velocities (using times instead).
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I'll just add to the nice answer from @joriki.
Every "average" has a certain use. You don't just use an average without having a reason behind it (not just statistics but physics). For example, if you wanted to find the velocity that you can use in the equation for kinetic energy instead of summing over all objects (for example, calculating the "average" velocity of molecules at some temperature), you would calculate the quadratic average (root mean square) because you would get the definition of average from $(sum m_i)v_avg^2=sum m_iv_i^2$. Another example is the center of mass. In linear gravity (on the surface of the earth, not too far), the center of mass is the linear average of position over all objects. But when you are dealing with Newtonian gravity (big picture, planets and stuff), it's the harmonic average that determines the point around which the objects orbit.
So - there is no one "right" average - each has a different meaning, and is used differently.
That being said, back to your question. If these velocities are fundamental individual results (measurements with a radar or something) and you are doing statistics on these measurements... then use the arithmetic average, because that's what you do in statistical analysis when you are looking for the most likely value of some quantity, when a Gaussian distribution is assumed (symmetric to deviations up or down). But I don't think this applies to your situation.
Averaging inverse velocities is just like averaging times... which is usual in sports. And in sports, even doing statistics, you will do statistics with times, not velocities (and thus compute average times). So... as far as I understand, if talking about velocities is simply a convenient conversion, but the time is the true thing that was measured, then I'd say that just use the harmonic average for everything, because talking about velocities is just a "mask" over the true quantities in the system and velocities themselves were not really measured or detected directly. If I was doing this... I'd use harmonic average, or just stop pretending that we are talking about velocities (using times instead).
I'll just add to the nice answer from @joriki.
Every "average" has a certain use. You don't just use an average without having a reason behind it (not just statistics but physics). For example, if you wanted to find the velocity that you can use in the equation for kinetic energy instead of summing over all objects (for example, calculating the "average" velocity of molecules at some temperature), you would calculate the quadratic average (root mean square) because you would get the definition of average from $(sum m_i)v_avg^2=sum m_iv_i^2$. Another example is the center of mass. In linear gravity (on the surface of the earth, not too far), the center of mass is the linear average of position over all objects. But when you are dealing with Newtonian gravity (big picture, planets and stuff), it's the harmonic average that determines the point around which the objects orbit.
So - there is no one "right" average - each has a different meaning, and is used differently.
That being said, back to your question. If these velocities are fundamental individual results (measurements with a radar or something) and you are doing statistics on these measurements... then use the arithmetic average, because that's what you do in statistical analysis when you are looking for the most likely value of some quantity, when a Gaussian distribution is assumed (symmetric to deviations up or down). But I don't think this applies to your situation.
Averaging inverse velocities is just like averaging times... which is usual in sports. And in sports, even doing statistics, you will do statistics with times, not velocities (and thus compute average times). So... as far as I understand, if talking about velocities is simply a convenient conversion, but the time is the true thing that was measured, then I'd say that just use the harmonic average for everything, because talking about velocities is just a "mask" over the true quantities in the system and velocities themselves were not really measured or detected directly. If I was doing this... I'd use harmonic average, or just stop pretending that we are talking about velocities (using times instead).
answered Aug 21 at 11:41
orion
12k11832
12k11832
add a comment |Â
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What do you mean by average speed at time $i$ ? Consider the following - 1) An object can be travelling only from A to B or B to A ,but not BOTH, so average of what? 2) Time is continuous, not discreet, so why only integral times ?
â DevashishKaushik
Aug 21 at 9:58
@DevashishKaushik, by average speed at time $i$ I meant the average of the speeds $x_i$ and $y_i$, which I am sure you will agree to be the harmonic average of $x_i$ and $y_i$. And no, the times are not necessarily integral, I meant the $i$'s to be indices.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:40
I have edited my question in response to your comment, @DevashishKaushik. I hope the question makes more sense now.
â Jose Arnaldo Bebita Dris
Aug 21 at 10:43
+1 Yes, it's quite clear what you are asking, as evidenced by the presence of 2 answers. :D
â DevashishKaushik
Aug 21 at 12:13