Classic Stats Problem, New Twist: Romeo and Juliet meet for a date. Solve Without drawings.
$begingroup$
This question has been asked before here. It usually goes like:
Romeo and Juliet have a date at a given time, and each will arrive at the meeting place with a delay between 0 and 1 hour, with all pairs of delays being “equally likely," that is, according to a uniform probability law on the unit square. The first to arrive will wait for 15 minutes and will leave if the other has not arrived. What is the probability that they will meet?
The solution I always see is derived by drawing a picture such as in this excellent MIT video or in this Purdue PDF.
Derive the solution using only math (so no summing areas using drawings). Should be able to calculate the probability Romeo and Juliet would meet if both would only wait 10 or 5 minutes... or if one of the lovers could only wait for 5 minutes but the other could wait up to 30.
probability recreational-mathematics
asked Jan 12 at 20:04
zeluspzelusp
1056
$endgroup$
add a comment |
$begingroup$
This question has been asked before here. It usually goes like:
Romeo and Juliet have a date at a given time, and each will arrive at the meeting place with a delay between 0 and 1 hour, with all pairs of delays being “equally likely," that is, according to a uniform probability law on the unit square. The first to arrive will wait for 15 minutes and will leave if the other has not arrived. What is the probability that they will meet?
The solution I always see is derived by drawing a picture such as in this excellent MIT video or in this Purdue PDF.
Derive the solution using only math (so no summing areas using drawings). Should be able to calculate the probability Romeo and Juliet would meet if both would only wait 10 or 5 minutes... or if one of the lovers could only wait for 5 minutes but the other could wait up to 30.
probability recreational-mathematics
asked Jan 12 at 20:04
zeluspzelusp
1056
$endgroup$
add a comment |
$begingroup$
This question has been asked before here. It usually goes like:
Romeo and Juliet have a date at a given time, and each will arrive at the meeting place with a delay between 0 and 1 hour, with all pairs of delays being “equally likely," that is, according to a uniform probability law on the unit square. The first to arrive will wait for 15 minutes and will leave if the other has not arrived. What is the probability that they will meet?
The solution I always see is derived by drawing a picture such as in this excellent MIT video or in this Purdue PDF.
Derive the solution using only math (so no summing areas using drawings). Should be able to calculate the probability Romeo and Juliet would meet if both would only wait 10 or 5 minutes... or if one of the lovers could only wait for 5 minutes but the other could wait up to 30.
probability recreational-mathematics
asked Jan 12 at 20:04
zeluspzelusp
1056
$endgroup$
This question has been asked before here. It usually goes like:
Romeo and Juliet have a date at a given time, and each will arrive at the meeting place with a delay between 0 and 1 hour, with all pairs of delays being “equally likely," that is, according to a uniform probability law on the unit square. The first to arrive will wait for 15 minutes and will leave if the other has not arrived. What is the probability that they will meet?
The solution I always see is derived by drawing a picture such as in this excellent MIT video or in this Purdue PDF.
Derive the solution using only math (so no summing areas using drawings). Should be able to calculate the probability Romeo and Juliet would meet if both would only wait 10 or 5 minutes... or if one of the lovers could only wait for 5 minutes but the other could wait up to 30.
probability recreational-mathematics
probability recreational-mathematics
asked Jan 12 at 20:04
zeluspzelusp
1056
asked Jan 12 at 20:04
zeluspzelusp
1056
asked Jan 12 at 20:04
zeluspzelusp
1056
asked Jan 12 at 20:04
zeluspzelusp
1056
asked Jan 12 at 20:04
zeluspzelusp
1056
1056
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let it be that Romeo gives up after $r$ hours and Juliet after $j$ hours, where $0<r,j<1$.
Let $R$ and $J$ be iid random variables uniformly distributed over $[0,1]$ corresponding with the arrival time of Romeo and Julia respectively.
Based on the principle that $P(A)=mathbb Emathbf1_A$ for every event $A$ to be found is:$$P(Rleq Jleq R+r)+P(Jleq Rleq J+j)=$$$$int_0^1int_0^1[xleq yleq x+r]dydx+int_0^1int_0^1[yleq xleq y+j]dxdy$$where $$[text{condition on }x,y]$$ denotes the function $mathbb R^2tomathbb R$ that gives value $1$ if the condition is satisfied and gives value $0$ otherwise.
For the first term we find:
$begin{aligned}int_{0}^{1}int_{0}^{1}left[xleq yleq x+rright]dydx & =int_{0}^{1-r}int_{x}^{x+r}dydx+int_{1-r}^{1}int_{x}^{1}dydx\
& =int_{0}^{1-r}rdx+int_{1-r}^{1}1-xdx\
& =rleft(1-rright)+left[x-frac{1}{2}x^{2}right]_{1-r}^{1}\
& =r-frac{1}{2}r^{2}
end{aligned}
$
And similarly for the second $j-frac12j^2$ so the final answer is:$$r-frac12r^2+j-frac12j^2$$
answered Jan 13 at 15:18
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you so much for taking the time to answer this one :) my calculus is rusty so I tried verifying your solution using a graph. Here's what I came up with: Chart 1 and Chart 2. In the first chart I see that the answer is correct for equal wait times of 15 minutes (43.75%) but at an equal wait time of 30 minutes the probability of them meeting is around 80% when it ought to be 100%. And in the second chart I would expect that the probability of them meeting if Romeo waited the full hour would always be 100% but it's not.
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
... here's the python code I used to test your derivation per chance it's useful. Maybe I did something wrong?
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
Wait a sec, I think your solution could be right according to chart 1... I forgot that Juliet can arrive even after Romeo had waited 30 minutes. What's throwing me off now is that chart 2 shows a change in probability of them meeting despite the fact that the probability shouldn't change since Romeo is waiting the full one hour (Juliet's wait time should have no impact on their likely hood to meet since they would have 100% met at any given point within the hour). Is this just a strange corner case?
$endgroup$
– zelusp
Jan 14 at 6:20
$begingroup$
I just figured it out! Yes, if Romeo showed up first and waited a full hour, then there's a 100% chance that he'll meet Juliet even if she just hangs around for a second. However, Juliet can show up first. And in the extreme case that Romeo didn't show up until the end of the hour (even if he stays around for the following hour.. or all of eternity), then there's a 100% chance that he missed her... 100% of one half is only 50% of the whole. SO YOUR ANSWER'S GOLD!
$endgroup$
– zelusp
Jan 14 at 6:31
1
$begingroup$
You are welcome. Chart 1 corresponds with case $r=j$ so function $y=x-frac12x^2+x-frac12x^2=2x-x^2$. Chart 2 with case $r=1$ so with function $y=1-frac121^2+x-frac12x^2=frac12+x-frac12x^2$. Thank you for the python code.
$endgroup$
– drhab
Jan 14 at 14:55
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let it be that Romeo gives up after $r$ hours and Juliet after $j$ hours, where $0<r,j<1$.
Let $R$ and $J$ be iid random variables uniformly distributed over $[0,1]$ corresponding with the arrival time of Romeo and Julia respectively.
Based on the principle that $P(A)=mathbb Emathbf1_A$ for every event $A$ to be found is:$$P(Rleq Jleq R+r)+P(Jleq Rleq J+j)=$$$$int_0^1int_0^1[xleq yleq x+r]dydx+int_0^1int_0^1[yleq xleq y+j]dxdy$$where $$[text{condition on }x,y]$$ denotes the function $mathbb R^2tomathbb R$ that gives value $1$ if the condition is satisfied and gives value $0$ otherwise.
For the first term we find:
$begin{aligned}int_{0}^{1}int_{0}^{1}left[xleq yleq x+rright]dydx & =int_{0}^{1-r}int_{x}^{x+r}dydx+int_{1-r}^{1}int_{x}^{1}dydx\
& =int_{0}^{1-r}rdx+int_{1-r}^{1}1-xdx\
& =rleft(1-rright)+left[x-frac{1}{2}x^{2}right]_{1-r}^{1}\
& =r-frac{1}{2}r^{2}
end{aligned}
$
And similarly for the second $j-frac12j^2$ so the final answer is:$$r-frac12r^2+j-frac12j^2$$
answered Jan 13 at 15:18
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you so much for taking the time to answer this one :) my calculus is rusty so I tried verifying your solution using a graph. Here's what I came up with: Chart 1 and Chart 2. In the first chart I see that the answer is correct for equal wait times of 15 minutes (43.75%) but at an equal wait time of 30 minutes the probability of them meeting is around 80% when it ought to be 100%. And in the second chart I would expect that the probability of them meeting if Romeo waited the full hour would always be 100% but it's not.
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
... here's the python code I used to test your derivation per chance it's useful. Maybe I did something wrong?
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
Wait a sec, I think your solution could be right according to chart 1... I forgot that Juliet can arrive even after Romeo had waited 30 minutes. What's throwing me off now is that chart 2 shows a change in probability of them meeting despite the fact that the probability shouldn't change since Romeo is waiting the full one hour (Juliet's wait time should have no impact on their likely hood to meet since they would have 100% met at any given point within the hour). Is this just a strange corner case?
$endgroup$
– zelusp
Jan 14 at 6:20
$begingroup$
I just figured it out! Yes, if Romeo showed up first and waited a full hour, then there's a 100% chance that he'll meet Juliet even if she just hangs around for a second. However, Juliet can show up first. And in the extreme case that Romeo didn't show up until the end of the hour (even if he stays around for the following hour.. or all of eternity), then there's a 100% chance that he missed her... 100% of one half is only 50% of the whole. SO YOUR ANSWER'S GOLD!
$endgroup$
– zelusp
Jan 14 at 6:31
1
$begingroup$
You are welcome. Chart 1 corresponds with case $r=j$ so function $y=x-frac12x^2+x-frac12x^2=2x-x^2$. Chart 2 with case $r=1$ so with function $y=1-frac121^2+x-frac12x^2=frac12+x-frac12x^2$. Thank you for the python code.
$endgroup$
– drhab
Jan 14 at 14:55
|
show 1 more comment
$begingroup$
Let it be that Romeo gives up after $r$ hours and Juliet after $j$ hours, where $0<r,j<1$.
Let $R$ and $J$ be iid random variables uniformly distributed over $[0,1]$ corresponding with the arrival time of Romeo and Julia respectively.
Based on the principle that $P(A)=mathbb Emathbf1_A$ for every event $A$ to be found is:$$P(Rleq Jleq R+r)+P(Jleq Rleq J+j)=$$$$int_0^1int_0^1[xleq yleq x+r]dydx+int_0^1int_0^1[yleq xleq y+j]dxdy$$where $$[text{condition on }x,y]$$ denotes the function $mathbb R^2tomathbb R$ that gives value $1$ if the condition is satisfied and gives value $0$ otherwise.
For the first term we find:
$begin{aligned}int_{0}^{1}int_{0}^{1}left[xleq yleq x+rright]dydx & =int_{0}^{1-r}int_{x}^{x+r}dydx+int_{1-r}^{1}int_{x}^{1}dydx\
& =int_{0}^{1-r}rdx+int_{1-r}^{1}1-xdx\
& =rleft(1-rright)+left[x-frac{1}{2}x^{2}right]_{1-r}^{1}\
& =r-frac{1}{2}r^{2}
end{aligned}
$
And similarly for the second $j-frac12j^2$ so the final answer is:$$r-frac12r^2+j-frac12j^2$$
answered Jan 13 at 15:18
drhabdrhab
103k545136
$endgroup$
$begingroup$
Thank you so much for taking the time to answer this one :) my calculus is rusty so I tried verifying your solution using a graph. Here's what I came up with: Chart 1 and Chart 2. In the first chart I see that the answer is correct for equal wait times of 15 minutes (43.75%) but at an equal wait time of 30 minutes the probability of them meeting is around 80% when it ought to be 100%. And in the second chart I would expect that the probability of them meeting if Romeo waited the full hour would always be 100% but it's not.
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
... here's the python code I used to test your derivation per chance it's useful. Maybe I did something wrong?
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
Wait a sec, I think your solution could be right according to chart 1... I forgot that Juliet can arrive even after Romeo had waited 30 minutes. What's throwing me off now is that chart 2 shows a change in probability of them meeting despite the fact that the probability shouldn't change since Romeo is waiting the full one hour (Juliet's wait time should have no impact on their likely hood to meet since they would have 100% met at any given point within the hour). Is this just a strange corner case?
$endgroup$
– zelusp
Jan 14 at 6:20
$begingroup$
I just figured it out! Yes, if Romeo showed up first and waited a full hour, then there's a 100% chance that he'll meet Juliet even if she just hangs around for a second. However, Juliet can show up first. And in the extreme case that Romeo didn't show up until the end of the hour (even if he stays around for the following hour.. or all of eternity), then there's a 100% chance that he missed her... 100% of one half is only 50% of the whole. SO YOUR ANSWER'S GOLD!
$endgroup$
– zelusp
Jan 14 at 6:31
1
$begingroup$
You are welcome. Chart 1 corresponds with case $r=j$ so function $y=x-frac12x^2+x-frac12x^2=2x-x^2$. Chart 2 with case $r=1$ so with function $y=1-frac121^2+x-frac12x^2=frac12+x-frac12x^2$. Thank you for the python code.
$endgroup$
– drhab
Jan 14 at 14:55
|
show 1 more comment
$begingroup$
Let it be that Romeo gives up after $r$ hours and Juliet after $j$ hours, where $0<r,j<1$.
Let $R$ and $J$ be iid random variables uniformly distributed over $[0,1]$ corresponding with the arrival time of Romeo and Julia respectively.
Based on the principle that $P(A)=mathbb Emathbf1_A$ for every event $A$ to be found is:$$P(Rleq Jleq R+r)+P(Jleq Rleq J+j)=$$$$int_0^1int_0^1[xleq yleq x+r]dydx+int_0^1int_0^1[yleq xleq y+j]dxdy$$where $$[text{condition on }x,y]$$ denotes the function $mathbb R^2tomathbb R$ that gives value $1$ if the condition is satisfied and gives value $0$ otherwise.
For the first term we find:
$begin{aligned}int_{0}^{1}int_{0}^{1}left[xleq yleq x+rright]dydx & =int_{0}^{1-r}int_{x}^{x+r}dydx+int_{1-r}^{1}int_{x}^{1}dydx\
& =int_{0}^{1-r}rdx+int_{1-r}^{1}1-xdx\
& =rleft(1-rright)+left[x-frac{1}{2}x^{2}right]_{1-r}^{1}\
& =r-frac{1}{2}r^{2}
end{aligned}
$
And similarly for the second $j-frac12j^2$ so the final answer is:$$r-frac12r^2+j-frac12j^2$$
answered Jan 13 at 15:18
drhabdrhab
103k545136
$endgroup$
Let it be that Romeo gives up after $r$ hours and Juliet after $j$ hours, where $0<r,j<1$.
Let $R$ and $J$ be iid random variables uniformly distributed over $[0,1]$ corresponding with the arrival time of Romeo and Julia respectively.
Based on the principle that $P(A)=mathbb Emathbf1_A$ for every event $A$ to be found is:$$P(Rleq Jleq R+r)+P(Jleq Rleq J+j)=$$$$int_0^1int_0^1[xleq yleq x+r]dydx+int_0^1int_0^1[yleq xleq y+j]dxdy$$where $$[text{condition on }x,y]$$ denotes the function $mathbb R^2tomathbb R$ that gives value $1$ if the condition is satisfied and gives value $0$ otherwise.
For the first term we find:
$begin{aligned}int_{0}^{1}int_{0}^{1}left[xleq yleq x+rright]dydx & =int_{0}^{1-r}int_{x}^{x+r}dydx+int_{1-r}^{1}int_{x}^{1}dydx\
& =int_{0}^{1-r}rdx+int_{1-r}^{1}1-xdx\
& =rleft(1-rright)+left[x-frac{1}{2}x^{2}right]_{1-r}^{1}\
& =r-frac{1}{2}r^{2}
end{aligned}
$
And similarly for the second $j-frac12j^2$ so the final answer is:$$r-frac12r^2+j-frac12j^2$$
answered Jan 13 at 15:18
drhabdrhab
103k545136
answered Jan 13 at 15:18
drhabdrhab
103k545136
answered Jan 13 at 15:18
drhabdrhab
103k545136
answered Jan 13 at 15:18
drhabdrhab
103k545136
103k545136
$begingroup$
Thank you so much for taking the time to answer this one :) my calculus is rusty so I tried verifying your solution using a graph. Here's what I came up with: Chart 1 and Chart 2. In the first chart I see that the answer is correct for equal wait times of 15 minutes (43.75%) but at an equal wait time of 30 minutes the probability of them meeting is around 80% when it ought to be 100%. And in the second chart I would expect that the probability of them meeting if Romeo waited the full hour would always be 100% but it's not.
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
... here's the python code I used to test your derivation per chance it's useful. Maybe I did something wrong?
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
Wait a sec, I think your solution could be right according to chart 1... I forgot that Juliet can arrive even after Romeo had waited 30 minutes. What's throwing me off now is that chart 2 shows a change in probability of them meeting despite the fact that the probability shouldn't change since Romeo is waiting the full one hour (Juliet's wait time should have no impact on their likely hood to meet since they would have 100% met at any given point within the hour). Is this just a strange corner case?
$endgroup$
– zelusp
Jan 14 at 6:20
$begingroup$
I just figured it out! Yes, if Romeo showed up first and waited a full hour, then there's a 100% chance that he'll meet Juliet even if she just hangs around for a second. However, Juliet can show up first. And in the extreme case that Romeo didn't show up until the end of the hour (even if he stays around for the following hour.. or all of eternity), then there's a 100% chance that he missed her... 100% of one half is only 50% of the whole. SO YOUR ANSWER'S GOLD!
$endgroup$
– zelusp
Jan 14 at 6:31
1
$begingroup$
You are welcome. Chart 1 corresponds with case $r=j$ so function $y=x-frac12x^2+x-frac12x^2=2x-x^2$. Chart 2 with case $r=1$ so with function $y=1-frac121^2+x-frac12x^2=frac12+x-frac12x^2$. Thank you for the python code.
$endgroup$
– drhab
Jan 14 at 14:55
|
show 1 more comment
$begingroup$
Thank you so much for taking the time to answer this one :) my calculus is rusty so I tried verifying your solution using a graph. Here's what I came up with: Chart 1 and Chart 2. In the first chart I see that the answer is correct for equal wait times of 15 minutes (43.75%) but at an equal wait time of 30 minutes the probability of them meeting is around 80% when it ought to be 100%. And in the second chart I would expect that the probability of them meeting if Romeo waited the full hour would always be 100% but it's not.
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
... here's the python code I used to test your derivation per chance it's useful. Maybe I did something wrong?
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
Wait a sec, I think your solution could be right according to chart 1... I forgot that Juliet can arrive even after Romeo had waited 30 minutes. What's throwing me off now is that chart 2 shows a change in probability of them meeting despite the fact that the probability shouldn't change since Romeo is waiting the full one hour (Juliet's wait time should have no impact on their likely hood to meet since they would have 100% met at any given point within the hour). Is this just a strange corner case?
$endgroup$
– zelusp
Jan 14 at 6:20
$begingroup$
I just figured it out! Yes, if Romeo showed up first and waited a full hour, then there's a 100% chance that he'll meet Juliet even if she just hangs around for a second. However, Juliet can show up first. And in the extreme case that Romeo didn't show up until the end of the hour (even if he stays around for the following hour.. or all of eternity), then there's a 100% chance that he missed her... 100% of one half is only 50% of the whole. SO YOUR ANSWER'S GOLD!
$endgroup$
– zelusp
Jan 14 at 6:31
1
$begingroup$
You are welcome. Chart 1 corresponds with case $r=j$ so function $y=x-frac12x^2+x-frac12x^2=2x-x^2$. Chart 2 with case $r=1$ so with function $y=1-frac121^2+x-frac12x^2=frac12+x-frac12x^2$. Thank you for the python code.
$endgroup$
– drhab
Jan 14 at 14:55
$begingroup$
Thank you so much for taking the time to answer this one :) my calculus is rusty so I tried verifying your solution using a graph. Here's what I came up with: Chart 1 and Chart 2. In the first chart I see that the answer is correct for equal wait times of 15 minutes (43.75%) but at an equal wait time of 30 minutes the probability of them meeting is around 80% when it ought to be 100%. And in the second chart I would expect that the probability of them meeting if Romeo waited the full hour would always be 100% but it's not.
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
Thank you so much for taking the time to answer this one :) my calculus is rusty so I tried verifying your solution using a graph. Here's what I came up with: Chart 1 and Chart 2. In the first chart I see that the answer is correct for equal wait times of 15 minutes (43.75%) but at an equal wait time of 30 minutes the probability of them meeting is around 80% when it ought to be 100%. And in the second chart I would expect that the probability of them meeting if Romeo waited the full hour would always be 100% but it's not.
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
... here's the python code I used to test your derivation per chance it's useful. Maybe I did something wrong?
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
... here's the python code I used to test your derivation per chance it's useful. Maybe I did something wrong?
$endgroup$
– zelusp
Jan 14 at 2:19
$begingroup$
Wait a sec, I think your solution could be right according to chart 1... I forgot that Juliet can arrive even after Romeo had waited 30 minutes. What's throwing me off now is that chart 2 shows a change in probability of them meeting despite the fact that the probability shouldn't change since Romeo is waiting the full one hour (Juliet's wait time should have no impact on their likely hood to meet since they would have 100% met at any given point within the hour). Is this just a strange corner case?
$endgroup$
– zelusp
Jan 14 at 6:20
$begingroup$
Wait a sec, I think your solution could be right according to chart 1... I forgot that Juliet can arrive even after Romeo had waited 30 minutes. What's throwing me off now is that chart 2 shows a change in probability of them meeting despite the fact that the probability shouldn't change since Romeo is waiting the full one hour (Juliet's wait time should have no impact on their likely hood to meet since they would have 100% met at any given point within the hour). Is this just a strange corner case?
$endgroup$
– zelusp
Jan 14 at 6:20
$begingroup$
I just figured it out! Yes, if Romeo showed up first and waited a full hour, then there's a 100% chance that he'll meet Juliet even if she just hangs around for a second. However, Juliet can show up first. And in the extreme case that Romeo didn't show up until the end of the hour (even if he stays around for the following hour.. or all of eternity), then there's a 100% chance that he missed her... 100% of one half is only 50% of the whole. SO YOUR ANSWER'S GOLD!
$endgroup$
– zelusp
Jan 14 at 6:31
$begingroup$
I just figured it out! Yes, if Romeo showed up first and waited a full hour, then there's a 100% chance that he'll meet Juliet even if she just hangs around for a second. However, Juliet can show up first. And in the extreme case that Romeo didn't show up until the end of the hour (even if he stays around for the following hour.. or all of eternity), then there's a 100% chance that he missed her... 100% of one half is only 50% of the whole. SO YOUR ANSWER'S GOLD!
$endgroup$
– zelusp
Jan 14 at 6:31
1
1
$begingroup$
You are welcome. Chart 1 corresponds with case $r=j$ so function $y=x-frac12x^2+x-frac12x^2=2x-x^2$. Chart 2 with case $r=1$ so with function $y=1-frac121^2+x-frac12x^2=frac12+x-frac12x^2$. Thank you for the python code.
$endgroup$
– drhab
Jan 14 at 14:55
$begingroup$
You are welcome. Chart 1 corresponds with case $r=j$ so function $y=x-frac12x^2+x-frac12x^2=2x-x^2$. Chart 2 with case $r=1$ so with function $y=1-frac121^2+x-frac12x^2=frac12+x-frac12x^2$. Thank you for the python code.
$endgroup$
– drhab
Jan 14 at 14:55
|
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