how to find slope of discrete point?
$begingroup$
I am wondering if it is possible to find the slope at each point in the following dataset,
% X Y
%===================
0.7761 0.5715
0.794 0.5729
0.8117 0.5744
0.8292 0.5762
0.8465 0.5782
0.8637 0.5804
0.8807 0.5828
0.8977 0.5853
0.9144 0.5879
0.9311 0.5907
0.9477 0.5937
0.9641 0.5968
0.9805 0.6
0.9967 0.6033
1.0129 0.6067
I understand that the slope can be obtained using the difference of the two neighboring points by
$$m = frac{y_2-y_1}{x_2-x_1}$$
and, the angle that each point made with the $x$-axis is essentially the $atan$ of $m$
$$theta = tan^{-1}(m) $$
But, is it possible to calculate the slope without using the above formula? without trying to curve-fit the points.
vector-analysis angle slope
asked Jan 8 at 13:51
BeeTiauBeeTiau
758
$endgroup$
add a comment |
$begingroup$
I am wondering if it is possible to find the slope at each point in the following dataset,
% X Y
%===================
0.7761 0.5715
0.794 0.5729
0.8117 0.5744
0.8292 0.5762
0.8465 0.5782
0.8637 0.5804
0.8807 0.5828
0.8977 0.5853
0.9144 0.5879
0.9311 0.5907
0.9477 0.5937
0.9641 0.5968
0.9805 0.6
0.9967 0.6033
1.0129 0.6067
I understand that the slope can be obtained using the difference of the two neighboring points by
$$m = frac{y_2-y_1}{x_2-x_1}$$
and, the angle that each point made with the $x$-axis is essentially the $atan$ of $m$
$$theta = tan^{-1}(m) $$
But, is it possible to calculate the slope without using the above formula? without trying to curve-fit the points.
vector-analysis angle slope
asked Jan 8 at 13:51
BeeTiauBeeTiau
758
$endgroup$
$begingroup$
The formula that you're using for the slope $m$ assumes that the function $y=f(x)$ is linear between the points. There are other methods, too. It's possible to assume a quadratic relation, etc ... But why do you ask if it's possible to calculate without using the formula? What's wrong with using the formula?
$endgroup$
– Matti P.
Jan 8 at 13:54
$begingroup$
There's nothing wrong with the formula. The reason is that the method is heavily dependent on the sampling rate of the points. I am curious if there is any other method or approach that can be used.
$endgroup$
– BeeTiau
Jan 8 at 13:56
1
$begingroup$
@BeeTiau: I'd take Matti P's suggestion, and assume a reasonable polynomial shape (why polynomial? Because they're super-easy to differentiate), differentiate that, and get the derivative value at the desired points. This method has the advantage of smoothing out your data a bit (fitting a curve is a summation process, and hence smoothing, whereas raw differentiation like you did makes graphs more jagged and noisy). As for sampling rate, fitting a reasonable polynomial (you might look up cubic splines) will mitigate that.
$endgroup$
– Adrian Keister
Jan 8 at 14:06
add a comment |
$begingroup$
I am wondering if it is possible to find the slope at each point in the following dataset,
% X Y
%===================
0.7761 0.5715
0.794 0.5729
0.8117 0.5744
0.8292 0.5762
0.8465 0.5782
0.8637 0.5804
0.8807 0.5828
0.8977 0.5853
0.9144 0.5879
0.9311 0.5907
0.9477 0.5937
0.9641 0.5968
0.9805 0.6
0.9967 0.6033
1.0129 0.6067
I understand that the slope can be obtained using the difference of the two neighboring points by
$$m = frac{y_2-y_1}{x_2-x_1}$$
and, the angle that each point made with the $x$-axis is essentially the $atan$ of $m$
$$theta = tan^{-1}(m) $$
But, is it possible to calculate the slope without using the above formula? without trying to curve-fit the points.
vector-analysis angle slope
asked Jan 8 at 13:51
BeeTiauBeeTiau
758
$endgroup$
I am wondering if it is possible to find the slope at each point in the following dataset,
% X Y
%===================
0.7761 0.5715
0.794 0.5729
0.8117 0.5744
0.8292 0.5762
0.8465 0.5782
0.8637 0.5804
0.8807 0.5828
0.8977 0.5853
0.9144 0.5879
0.9311 0.5907
0.9477 0.5937
0.9641 0.5968
0.9805 0.6
0.9967 0.6033
1.0129 0.6067
I understand that the slope can be obtained using the difference of the two neighboring points by
$$m = frac{y_2-y_1}{x_2-x_1}$$
and, the angle that each point made with the $x$-axis is essentially the $atan$ of $m$
$$theta = tan^{-1}(m) $$
But, is it possible to calculate the slope without using the above formula? without trying to curve-fit the points.
vector-analysis angle slope
vector-analysis angle slope
asked Jan 8 at 13:51
BeeTiauBeeTiau
758
asked Jan 8 at 13:51
BeeTiauBeeTiau
758
asked Jan 8 at 13:51
BeeTiauBeeTiau
758
asked Jan 8 at 13:51
BeeTiauBeeTiau
758
asked Jan 8 at 13:51
BeeTiauBeeTiau
758
758
$begingroup$
The formula that you're using for the slope $m$ assumes that the function $y=f(x)$ is linear between the points. There are other methods, too. It's possible to assume a quadratic relation, etc ... But why do you ask if it's possible to calculate without using the formula? What's wrong with using the formula?
$endgroup$
– Matti P.
Jan 8 at 13:54
$begingroup$
There's nothing wrong with the formula. The reason is that the method is heavily dependent on the sampling rate of the points. I am curious if there is any other method or approach that can be used.
$endgroup$
– BeeTiau
Jan 8 at 13:56
1
$begingroup$
@BeeTiau: I'd take Matti P's suggestion, and assume a reasonable polynomial shape (why polynomial? Because they're super-easy to differentiate), differentiate that, and get the derivative value at the desired points. This method has the advantage of smoothing out your data a bit (fitting a curve is a summation process, and hence smoothing, whereas raw differentiation like you did makes graphs more jagged and noisy). As for sampling rate, fitting a reasonable polynomial (you might look up cubic splines) will mitigate that.
$endgroup$
– Adrian Keister
Jan 8 at 14:06
add a comment |
$begingroup$
The formula that you're using for the slope $m$ assumes that the function $y=f(x)$ is linear between the points. There are other methods, too. It's possible to assume a quadratic relation, etc ... But why do you ask if it's possible to calculate without using the formula? What's wrong with using the formula?
$endgroup$
– Matti P.
Jan 8 at 13:54
$begingroup$
There's nothing wrong with the formula. The reason is that the method is heavily dependent on the sampling rate of the points. I am curious if there is any other method or approach that can be used.
$endgroup$
– BeeTiau
Jan 8 at 13:56
1
$begingroup$
@BeeTiau: I'd take Matti P's suggestion, and assume a reasonable polynomial shape (why polynomial? Because they're super-easy to differentiate), differentiate that, and get the derivative value at the desired points. This method has the advantage of smoothing out your data a bit (fitting a curve is a summation process, and hence smoothing, whereas raw differentiation like you did makes graphs more jagged and noisy). As for sampling rate, fitting a reasonable polynomial (you might look up cubic splines) will mitigate that.
$endgroup$
– Adrian Keister
Jan 8 at 14:06
$begingroup$
The formula that you're using for the slope $m$ assumes that the function $y=f(x)$ is linear between the points. There are other methods, too. It's possible to assume a quadratic relation, etc ... But why do you ask if it's possible to calculate without using the formula? What's wrong with using the formula?
$endgroup$
– Matti P.
Jan 8 at 13:54
$begingroup$
The formula that you're using for the slope $m$ assumes that the function $y=f(x)$ is linear between the points. There are other methods, too. It's possible to assume a quadratic relation, etc ... But why do you ask if it's possible to calculate without using the formula? What's wrong with using the formula?
$endgroup$
– Matti P.
Jan 8 at 13:54
$begingroup$
There's nothing wrong with the formula. The reason is that the method is heavily dependent on the sampling rate of the points. I am curious if there is any other method or approach that can be used.
$endgroup$
– BeeTiau
Jan 8 at 13:56
$begingroup$
There's nothing wrong with the formula. The reason is that the method is heavily dependent on the sampling rate of the points. I am curious if there is any other method or approach that can be used.
$endgroup$
– BeeTiau
Jan 8 at 13:56
1
1
$begingroup$
@BeeTiau: I'd take Matti P's suggestion, and assume a reasonable polynomial shape (why polynomial? Because they're super-easy to differentiate), differentiate that, and get the derivative value at the desired points. This method has the advantage of smoothing out your data a bit (fitting a curve is a summation process, and hence smoothing, whereas raw differentiation like you did makes graphs more jagged and noisy). As for sampling rate, fitting a reasonable polynomial (you might look up cubic splines) will mitigate that.
$endgroup$
– Adrian Keister
Jan 8 at 14:06
$begingroup$
@BeeTiau: I'd take Matti P's suggestion, and assume a reasonable polynomial shape (why polynomial? Because they're super-easy to differentiate), differentiate that, and get the derivative value at the desired points. This method has the advantage of smoothing out your data a bit (fitting a curve is a summation process, and hence smoothing, whereas raw differentiation like you did makes graphs more jagged and noisy). As for sampling rate, fitting a reasonable polynomial (you might look up cubic splines) will mitigate that.
$endgroup$
– Adrian Keister
Jan 8 at 14:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In my experience,
when trying to estimate the slope at a point,
it is better to use the slope of the line
between the preceding and following point.
This is analogous to the fact that
$f'(x)$
is more accurately estimated by
$(f(x+h)-f(x-h))/(2h)$
(error of order $h^2$)
than by
$(f(x+h)-f(x))/(h)$
(error of order $h$).
answered Jan 8 at 15:26
marty cohenmarty cohen
74.9k549130
$endgroup$
$begingroup$
Thanks. Meaning that there is no way we can estimate the slope without knowing the functions (or trend?) of the data?
$endgroup$
– BeeTiau
Jan 8 at 15:45
$begingroup$
All you need are a series of (x, y) values.
$endgroup$
– marty cohen
Jan 8 at 20:06
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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oldest
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$begingroup$
In my experience,
when trying to estimate the slope at a point,
it is better to use the slope of the line
between the preceding and following point.
This is analogous to the fact that
$f'(x)$
is more accurately estimated by
$(f(x+h)-f(x-h))/(2h)$
(error of order $h^2$)
than by
$(f(x+h)-f(x))/(h)$
(error of order $h$).
answered Jan 8 at 15:26
marty cohenmarty cohen
74.9k549130
$endgroup$
$begingroup$
Thanks. Meaning that there is no way we can estimate the slope without knowing the functions (or trend?) of the data?
$endgroup$
– BeeTiau
Jan 8 at 15:45
$begingroup$
All you need are a series of (x, y) values.
$endgroup$
– marty cohen
Jan 8 at 20:06
add a comment |
$begingroup$
In my experience,
when trying to estimate the slope at a point,
it is better to use the slope of the line
between the preceding and following point.
This is analogous to the fact that
$f'(x)$
is more accurately estimated by
$(f(x+h)-f(x-h))/(2h)$
(error of order $h^2$)
than by
$(f(x+h)-f(x))/(h)$
(error of order $h$).
answered Jan 8 at 15:26
marty cohenmarty cohen
74.9k549130
$endgroup$
$begingroup$
Thanks. Meaning that there is no way we can estimate the slope without knowing the functions (or trend?) of the data?
$endgroup$
– BeeTiau
Jan 8 at 15:45
$begingroup$
All you need are a series of (x, y) values.
$endgroup$
– marty cohen
Jan 8 at 20:06
add a comment |
$begingroup$
In my experience,
when trying to estimate the slope at a point,
it is better to use the slope of the line
between the preceding and following point.
This is analogous to the fact that
$f'(x)$
is more accurately estimated by
$(f(x+h)-f(x-h))/(2h)$
(error of order $h^2$)
than by
$(f(x+h)-f(x))/(h)$
(error of order $h$).
answered Jan 8 at 15:26
marty cohenmarty cohen
74.9k549130
$endgroup$
In my experience,
when trying to estimate the slope at a point,
it is better to use the slope of the line
between the preceding and following point.
This is analogous to the fact that
$f'(x)$
is more accurately estimated by
$(f(x+h)-f(x-h))/(2h)$
(error of order $h^2$)
than by
$(f(x+h)-f(x))/(h)$
(error of order $h$).
answered Jan 8 at 15:26
marty cohenmarty cohen
74.9k549130
answered Jan 8 at 15:26
marty cohenmarty cohen
74.9k549130
answered Jan 8 at 15:26
marty cohenmarty cohen
74.9k549130
answered Jan 8 at 15:26
marty cohenmarty cohen
74.9k549130
74.9k549130
$begingroup$
Thanks. Meaning that there is no way we can estimate the slope without knowing the functions (or trend?) of the data?
$endgroup$
– BeeTiau
Jan 8 at 15:45
$begingroup$
All you need are a series of (x, y) values.
$endgroup$
– marty cohen
Jan 8 at 20:06
add a comment |
$begingroup$
Thanks. Meaning that there is no way we can estimate the slope without knowing the functions (or trend?) of the data?
$endgroup$
– BeeTiau
Jan 8 at 15:45
$begingroup$
All you need are a series of (x, y) values.
$endgroup$
– marty cohen
Jan 8 at 20:06
$begingroup$
Thanks. Meaning that there is no way we can estimate the slope without knowing the functions (or trend?) of the data?
$endgroup$
– BeeTiau
Jan 8 at 15:45
$begingroup$
Thanks. Meaning that there is no way we can estimate the slope without knowing the functions (or trend?) of the data?
$endgroup$
– BeeTiau
Jan 8 at 15:45
$begingroup$
All you need are a series of (x, y) values.
$endgroup$
– marty cohen
Jan 8 at 20:06
$begingroup$
All you need are a series of (x, y) values.
$endgroup$
– marty cohen
Jan 8 at 20:06
add a comment |
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$begingroup$
The formula that you're using for the slope $m$ assumes that the function $y=f(x)$ is linear between the points. There are other methods, too. It's possible to assume a quadratic relation, etc ... But why do you ask if it's possible to calculate without using the formula? What's wrong with using the formula?
$endgroup$
– Matti P.
Jan 8 at 13:54
$begingroup$
There's nothing wrong with the formula. The reason is that the method is heavily dependent on the sampling rate of the points. I am curious if there is any other method or approach that can be used.
$endgroup$
– BeeTiau
Jan 8 at 13:56
1
$begingroup$
@BeeTiau: I'd take Matti P's suggestion, and assume a reasonable polynomial shape (why polynomial? Because they're super-easy to differentiate), differentiate that, and get the derivative value at the desired points. This method has the advantage of smoothing out your data a bit (fitting a curve is a summation process, and hence smoothing, whereas raw differentiation like you did makes graphs more jagged and noisy). As for sampling rate, fitting a reasonable polynomial (you might look up cubic splines) will mitigate that.
$endgroup$
– Adrian Keister
Jan 8 at 14:06