What is the significance of the number 2.87682 popping up in my calculations?
$begingroup$
My graph for cube size 4
I've been working on a proof to determine the surface area of a cube based on how many times you cut it vertically in half (cut the width in half) and then cut the resulting prisms in half a given number of times. I think that the equation is exponential, based on the graph that I generated in a java program that I wrote, but when I try to find the exponent based on the "y = Pe^(rt)" equation, the r value is consistent between x values, but not between y values.
For example, if I have a cube with side length 3, and cut it and its prisms in half 0 times, the resulting surface area of all remaining objects will be 54. If I cut it one time the total surface area becomes 72. The resulting r value for this when using regression is around 0.287682. However if I use regression for the next set of points (1, 72) and (2, 108), the value jumps to 0.405465.
However, if I have a cube with side length 4, and cut it and its prisms in half 0 times, the resulting surface area of all remaining objects will be 96. If I cut it one time the total surface area becomes 128. The r value is STILL 0.287682. And when I use the next set of points (1, 128) and (2, 192), the value is STILL 0.405465.
Why does this value change? Are .287682 and .405465 special in some way, or is that just the way that exponential functions work? My real question is, if this function isn't exponential, then how can it be written?
exponential-function area
asked May 31 '18 at 3:50
Matthew AverillMatthew Averill
253
$endgroup$
add a comment |
$begingroup$
My graph for cube size 4
I've been working on a proof to determine the surface area of a cube based on how many times you cut it vertically in half (cut the width in half) and then cut the resulting prisms in half a given number of times. I think that the equation is exponential, based on the graph that I generated in a java program that I wrote, but when I try to find the exponent based on the "y = Pe^(rt)" equation, the r value is consistent between x values, but not between y values.
For example, if I have a cube with side length 3, and cut it and its prisms in half 0 times, the resulting surface area of all remaining objects will be 54. If I cut it one time the total surface area becomes 72. The resulting r value for this when using regression is around 0.287682. However if I use regression for the next set of points (1, 72) and (2, 108), the value jumps to 0.405465.
However, if I have a cube with side length 4, and cut it and its prisms in half 0 times, the resulting surface area of all remaining objects will be 96. If I cut it one time the total surface area becomes 128. The r value is STILL 0.287682. And when I use the next set of points (1, 128) and (2, 192), the value is STILL 0.405465.
Why does this value change? Are .287682 and .405465 special in some way, or is that just the way that exponential functions work? My real question is, if this function isn't exponential, then how can it be written?
exponential-function area
asked May 31 '18 at 3:50
Matthew AverillMatthew Averill
253
$endgroup$
$begingroup$
This suggests that the points you are trying to plot are not on an exponential curve. And indeed, the ratio of successive y values isn't the same. Are you asking why the r values you get don't depend on the side length? The reason is because they depend only on the successive ratios which don't depend on the side length.
$endgroup$
– Ian
May 31 '18 at 3:52
$begingroup$
A way to see this graphically is to plot log(y) vs x, you will find that the graph is in fact still convex, not linear.
$endgroup$
– Ian
May 31 '18 at 3:58
$begingroup$
I'm sorry, my initial question wasn't accurate. I'm interested in finding out how to write this relationship as an equation, as I haven't found a way to yet.
$endgroup$
– Matthew Averill
May 31 '18 at 3:59
1
$begingroup$
You begin with 6A where A is a given side's area, then you add 2A then 4A then ... then $2^nA$. The overall expression can be evaluated using the finite geometric sum formula.
$endgroup$
– Ian
May 31 '18 at 4:21
add a comment |
$begingroup$
My graph for cube size 4
I've been working on a proof to determine the surface area of a cube based on how many times you cut it vertically in half (cut the width in half) and then cut the resulting prisms in half a given number of times. I think that the equation is exponential, based on the graph that I generated in a java program that I wrote, but when I try to find the exponent based on the "y = Pe^(rt)" equation, the r value is consistent between x values, but not between y values.
For example, if I have a cube with side length 3, and cut it and its prisms in half 0 times, the resulting surface area of all remaining objects will be 54. If I cut it one time the total surface area becomes 72. The resulting r value for this when using regression is around 0.287682. However if I use regression for the next set of points (1, 72) and (2, 108), the value jumps to 0.405465.
However, if I have a cube with side length 4, and cut it and its prisms in half 0 times, the resulting surface area of all remaining objects will be 96. If I cut it one time the total surface area becomes 128. The r value is STILL 0.287682. And when I use the next set of points (1, 128) and (2, 192), the value is STILL 0.405465.
Why does this value change? Are .287682 and .405465 special in some way, or is that just the way that exponential functions work? My real question is, if this function isn't exponential, then how can it be written?
exponential-function area
asked May 31 '18 at 3:50
Matthew AverillMatthew Averill
253
$endgroup$
My graph for cube size 4
I've been working on a proof to determine the surface area of a cube based on how many times you cut it vertically in half (cut the width in half) and then cut the resulting prisms in half a given number of times. I think that the equation is exponential, based on the graph that I generated in a java program that I wrote, but when I try to find the exponent based on the "y = Pe^(rt)" equation, the r value is consistent between x values, but not between y values.
For example, if I have a cube with side length 3, and cut it and its prisms in half 0 times, the resulting surface area of all remaining objects will be 54. If I cut it one time the total surface area becomes 72. The resulting r value for this when using regression is around 0.287682. However if I use regression for the next set of points (1, 72) and (2, 108), the value jumps to 0.405465.
However, if I have a cube with side length 4, and cut it and its prisms in half 0 times, the resulting surface area of all remaining objects will be 96. If I cut it one time the total surface area becomes 128. The r value is STILL 0.287682. And when I use the next set of points (1, 128) and (2, 192), the value is STILL 0.405465.
Why does this value change? Are .287682 and .405465 special in some way, or is that just the way that exponential functions work? My real question is, if this function isn't exponential, then how can it be written?
exponential-function area
exponential-function area
asked May 31 '18 at 3:50
Matthew AverillMatthew Averill
253
asked May 31 '18 at 3:50
Matthew AverillMatthew Averill
253
edited Dec 16 '18 at 23:58
Matthew Averill
asked May 31 '18 at 3:50
Matthew AverillMatthew Averill
253
asked May 31 '18 at 3:50
Matthew AverillMatthew Averill
253
asked May 31 '18 at 3:50
Matthew AverillMatthew Averill
253
253
$begingroup$
This suggests that the points you are trying to plot are not on an exponential curve. And indeed, the ratio of successive y values isn't the same. Are you asking why the r values you get don't depend on the side length? The reason is because they depend only on the successive ratios which don't depend on the side length.
$endgroup$
– Ian
May 31 '18 at 3:52
$begingroup$
A way to see this graphically is to plot log(y) vs x, you will find that the graph is in fact still convex, not linear.
$endgroup$
– Ian
May 31 '18 at 3:58
$begingroup$
I'm sorry, my initial question wasn't accurate. I'm interested in finding out how to write this relationship as an equation, as I haven't found a way to yet.
$endgroup$
– Matthew Averill
May 31 '18 at 3:59
1
$begingroup$
You begin with 6A where A is a given side's area, then you add 2A then 4A then ... then $2^nA$. The overall expression can be evaluated using the finite geometric sum formula.
$endgroup$
– Ian
May 31 '18 at 4:21
add a comment |
$begingroup$
This suggests that the points you are trying to plot are not on an exponential curve. And indeed, the ratio of successive y values isn't the same. Are you asking why the r values you get don't depend on the side length? The reason is because they depend only on the successive ratios which don't depend on the side length.
$endgroup$
– Ian
May 31 '18 at 3:52
$begingroup$
A way to see this graphically is to plot log(y) vs x, you will find that the graph is in fact still convex, not linear.
$endgroup$
– Ian
May 31 '18 at 3:58
$begingroup$
I'm sorry, my initial question wasn't accurate. I'm interested in finding out how to write this relationship as an equation, as I haven't found a way to yet.
$endgroup$
– Matthew Averill
May 31 '18 at 3:59
1
$begingroup$
You begin with 6A where A is a given side's area, then you add 2A then 4A then ... then $2^nA$. The overall expression can be evaluated using the finite geometric sum formula.
$endgroup$
– Ian
May 31 '18 at 4:21
$begingroup$
This suggests that the points you are trying to plot are not on an exponential curve. And indeed, the ratio of successive y values isn't the same. Are you asking why the r values you get don't depend on the side length? The reason is because they depend only on the successive ratios which don't depend on the side length.
$endgroup$
– Ian
May 31 '18 at 3:52
$begingroup$
This suggests that the points you are trying to plot are not on an exponential curve. And indeed, the ratio of successive y values isn't the same. Are you asking why the r values you get don't depend on the side length? The reason is because they depend only on the successive ratios which don't depend on the side length.
$endgroup$
– Ian
May 31 '18 at 3:52
$begingroup$
A way to see this graphically is to plot log(y) vs x, you will find that the graph is in fact still convex, not linear.
$endgroup$
– Ian
May 31 '18 at 3:58
$begingroup$
A way to see this graphically is to plot log(y) vs x, you will find that the graph is in fact still convex, not linear.
$endgroup$
– Ian
May 31 '18 at 3:58
$begingroup$
I'm sorry, my initial question wasn't accurate. I'm interested in finding out how to write this relationship as an equation, as I haven't found a way to yet.
$endgroup$
– Matthew Averill
May 31 '18 at 3:59
$begingroup$
I'm sorry, my initial question wasn't accurate. I'm interested in finding out how to write this relationship as an equation, as I haven't found a way to yet.
$endgroup$
– Matthew Averill
May 31 '18 at 3:59
1
1
$begingroup$
You begin with 6A where A is a given side's area, then you add 2A then 4A then ... then $2^nA$. The overall expression can be evaluated using the finite geometric sum formula.
$endgroup$
– Ian
May 31 '18 at 4:21
$begingroup$
You begin with 6A where A is a given side's area, then you add 2A then 4A then ... then $2^nA$. The overall expression can be evaluated using the finite geometric sum formula.
$endgroup$
– Ian
May 31 '18 at 4:21
add a comment |
1 Answer
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$begingroup$
As your graph demonstrates, the total area after $n$ cuts is $6A+sum_{k=1}^n 2^k A = (4 + 2^{n+1})A$, where $A$ is the area of one face. Here I have used the formula for the finite geometric sum
$$sum_{k=1}^n r^k = frac{r-r^{n+1}}{1-r}.$$
This is not exponential until you subtract off the $4A$, which is why the successive ratios increase with $n$. On the other hand, the successive ratios also do not depend explicitly on $A$, which is why you got the same ratios when you took two different choices of $A$. It turns out that the exponential factor $r$ in the usual exponential regression (which is to say linear regression applied to $ln(y) = rt+ln(P)$) depends only on these successive ratios and not on the absolute numbers. The absolute size of the numbers is entirely absorbed into the prefactor $P$.
answered Jun 2 '18 at 4:36
IanIan
68.7k25390
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add a comment |
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$begingroup$
As your graph demonstrates, the total area after $n$ cuts is $6A+sum_{k=1}^n 2^k A = (4 + 2^{n+1})A$, where $A$ is the area of one face. Here I have used the formula for the finite geometric sum
$$sum_{k=1}^n r^k = frac{r-r^{n+1}}{1-r}.$$
This is not exponential until you subtract off the $4A$, which is why the successive ratios increase with $n$. On the other hand, the successive ratios also do not depend explicitly on $A$, which is why you got the same ratios when you took two different choices of $A$. It turns out that the exponential factor $r$ in the usual exponential regression (which is to say linear regression applied to $ln(y) = rt+ln(P)$) depends only on these successive ratios and not on the absolute numbers. The absolute size of the numbers is entirely absorbed into the prefactor $P$.
answered Jun 2 '18 at 4:36
IanIan
68.7k25390
$endgroup$
add a comment |
$begingroup$
As your graph demonstrates, the total area after $n$ cuts is $6A+sum_{k=1}^n 2^k A = (4 + 2^{n+1})A$, where $A$ is the area of one face. Here I have used the formula for the finite geometric sum
$$sum_{k=1}^n r^k = frac{r-r^{n+1}}{1-r}.$$
This is not exponential until you subtract off the $4A$, which is why the successive ratios increase with $n$. On the other hand, the successive ratios also do not depend explicitly on $A$, which is why you got the same ratios when you took two different choices of $A$. It turns out that the exponential factor $r$ in the usual exponential regression (which is to say linear regression applied to $ln(y) = rt+ln(P)$) depends only on these successive ratios and not on the absolute numbers. The absolute size of the numbers is entirely absorbed into the prefactor $P$.
answered Jun 2 '18 at 4:36
IanIan
68.7k25390
$endgroup$
add a comment |
$begingroup$
As your graph demonstrates, the total area after $n$ cuts is $6A+sum_{k=1}^n 2^k A = (4 + 2^{n+1})A$, where $A$ is the area of one face. Here I have used the formula for the finite geometric sum
$$sum_{k=1}^n r^k = frac{r-r^{n+1}}{1-r}.$$
This is not exponential until you subtract off the $4A$, which is why the successive ratios increase with $n$. On the other hand, the successive ratios also do not depend explicitly on $A$, which is why you got the same ratios when you took two different choices of $A$. It turns out that the exponential factor $r$ in the usual exponential regression (which is to say linear regression applied to $ln(y) = rt+ln(P)$) depends only on these successive ratios and not on the absolute numbers. The absolute size of the numbers is entirely absorbed into the prefactor $P$.
answered Jun 2 '18 at 4:36
IanIan
68.7k25390
$endgroup$
As your graph demonstrates, the total area after $n$ cuts is $6A+sum_{k=1}^n 2^k A = (4 + 2^{n+1})A$, where $A$ is the area of one face. Here I have used the formula for the finite geometric sum
$$sum_{k=1}^n r^k = frac{r-r^{n+1}}{1-r}.$$
This is not exponential until you subtract off the $4A$, which is why the successive ratios increase with $n$. On the other hand, the successive ratios also do not depend explicitly on $A$, which is why you got the same ratios when you took two different choices of $A$. It turns out that the exponential factor $r$ in the usual exponential regression (which is to say linear regression applied to $ln(y) = rt+ln(P)$) depends only on these successive ratios and not on the absolute numbers. The absolute size of the numbers is entirely absorbed into the prefactor $P$.
answered Jun 2 '18 at 4:36
IanIan
68.7k25390
answered Jun 2 '18 at 4:36
IanIan
68.7k25390
answered Jun 2 '18 at 4:36
IanIan
68.7k25390
answered Jun 2 '18 at 4:36
IanIan
68.7k25390
68.7k25390
add a comment |
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$begingroup$
This suggests that the points you are trying to plot are not on an exponential curve. And indeed, the ratio of successive y values isn't the same. Are you asking why the r values you get don't depend on the side length? The reason is because they depend only on the successive ratios which don't depend on the side length.
$endgroup$
– Ian
May 31 '18 at 3:52
$begingroup$
A way to see this graphically is to plot log(y) vs x, you will find that the graph is in fact still convex, not linear.
$endgroup$
– Ian
May 31 '18 at 3:58
$begingroup$
I'm sorry, my initial question wasn't accurate. I'm interested in finding out how to write this relationship as an equation, as I haven't found a way to yet.
$endgroup$
– Matthew Averill
May 31 '18 at 3:59
1
$begingroup$
You begin with 6A where A is a given side's area, then you add 2A then 4A then ... then $2^nA$. The overall expression can be evaluated using the finite geometric sum formula.
$endgroup$
– Ian
May 31 '18 at 4:21