I will post it here when I have a volunteer.
Thanks.
Here it is anyway…
I have a complex math problem that needs solving in my research project that I don’t have the time to solve.
It will need calculus and differentiation skills.
Let me know if you are interested.
For:
{f(w, b) = \frac{1}{m}\Large\sum_{i = 1}^{m}\left[\frac{\lvert w^{2}x^{(i)}-w(y^{(i)}-b)\rvert}{cos(tan^{-1}(\frac{y_o^{(i)}}{x_o^{(i)}}))}\right]}
And:
\LARGE{\frac{y_o^{(i)}}{x_o^{(i)}} = \frac{2w^{2}b -wx^{(i)}+ y^{(i)} + b}{w^2x^{(i)} - w(y^{(i)} - b)}}
Solve the simultaneous equations:
\Large{\frac{\partial{f(w, b)}}{\partial w } = 0} \text{ and } \Large{\frac{\partial{f(w, b)}}{\partial b } = 0}
For the variables w and b.
x^{(i)} \text{ and } y^{(i)} are each values from one of two different sets of m real numbers from i = 1 \text{ to } i = m
Hi! I took a look at your problem and here’s how I solved it:
The objective function is:
f(w, b) = \frac{1}{m} \sum_{i=1}^{m} \left[ \frac{w^2 x^{(i)} - w(y^{(i)} - b)}{\cos\left(\tan^{-1}\left(\frac{y^{(i)}}{x_0^{(i)}}\right)\right)} \right]
Let:
C^{(i)} = \cos\left(\tan^{-1}\left(\frac{y^{(i)}}{x_0^{(i)}}\right)\right)
Because, C^{(i)} is constant with respect to w and b.
Partial derivative w.r.t. w is:
\frac{\partial f}{\partial w} = \frac{1}{m} \sum_{i=1}^{m} \frac{1}{C^{(i)}} \left(2w x^{(i)} - (y^{(i)} - b)\right)
Also, partial derivative w.r.t. b is:
\frac{\partial f}{\partial b} = \frac{1}{m} \sum_{i=1}^{m} \frac{w}{C^{(i)}} = \frac{w}{m} \sum_{i=1}^{m} \frac{1}{C^{(i)}}
Now I set the gradients to zero. From:
\frac{\partial f}{\partial b} = 0 \Rightarrow w = 0
Substitute w = 0 into \frac{\partial f}{\partial w}:
\sum_{i=1}^{m} \frac{(y^{(i)} - b)}{C^{(i)}} = 0
b = \frac{\sum_{i=1}^{m} \frac{y^{(i)}}{C^{(i)}}}{\sum_{i=1}^{m} \frac{1}{C^{(i)}}}
The final solution is:
w^* = 0
b^* = \dfrac{\sum \frac{y^{(i)}}{C^{(i)}}}{\sum \frac{1}{C^{(i)}}}
You have a mistake in your “objective function”.
Check it against my function.
C^{(i)} is not constant with respect to w and b.
I see. I should have substituted y0/x0 with to w and b terms. I will work on it and let you know if I find something.
I got this after doing the calculations:
- \sum_{i=1}^{m} \frac{(w^2x^{(i)} - w(y^{(i)} - b))(2wx^{(i)} - y^{(i)} + b) + (2w^2b - wx^{(i)} + y^{(i)} + b)(4wb - x^{(i)})}{\sqrt{(...)^2 + (...)^2}} = 0
- \sum_{i=1}^{m} \frac{(w^2x^{(i)} - w(y^{(i)} - b))(w) + (2w^2b - wx^{(i)} + y^{(i)} + b)(2w^2 + 1)}{\sqrt{(...)^2 + (...)^2}} = 0
It is not possible to solve for w and b by hand to get a general formula. To find the actual numerical values, you would need a dataset of (x^{(i)}, y^{(i)}) values and then use a numerical optimization (e.g. Gradient Descent).
Hope this helps!
I can’t view all of this on my iPhone - can you post an image instead.