Recent content by rhdinah

Yes, thank you so much! I looked at my original diagram and realized that I had the trig wrong! Shame on me! I was mislead a bit by the ref thread. I should have trusted my original diagram and not listened to the banter on that thread. To correct this: ##Largest \left|\frac{d...

Thank you Orodruin! Let me put your ideas in motion! ##Largest \left|\frac{d Component}{Component}\right| = \left|\frac {d F_1}{F_1}\right| + \left|\frac {cos(\theta)}{sin(\theta)} d \theta\right| = \frac{1}{500}+\frac{cos(\frac{\pi}{3})}{sin(\frac{\pi}{3})} * \frac{0.5\pi}{180} = .002 +...

Homework Statement I am working on this problem and having difficulty getting the required answer. It is the exact problem as here , but I’m still not getting it. BTW this is problem 10, Section 4, Chapter 4 Partial Differentiation from M. Boas’s book Mathematical Methods in the Physical...

Professor Boas was kind enough to give me some pointers on this: $$z=x^2+2y^2=x^2+2(r^2-x^2)=x^2+2(x^2sec^2\theta-x^2)=x^2-2x^2+2x^2sec^2\theta=-x^2+2x^2sec^2\theta$$ $$\frac{d}{d\theta}(-x^2+2x^2sec^2\theta)=4x^2tan\theta sec^2\theta=4 \frac{x^2}{cos^2\theta}tan\theta=4r^2tan\theta\hspace{10...

Thank you Mr. Fresh_42 for your help! I'm not sure where the ##x## in your solution would have come from as ##x## is supposed to be held constant in the differentiation. Perhaps as a side effect of the ##y## partial. I'm going to try to ask for help from Mary Boas's son as he is monitoring the...

Thank you! Okay, let's give it a go ... been thinking about this all day ... :-) ##z=x^2+2y^2=(x^2+y^2)+y^2=r^2+r^2sin^2(\theta)=r^2(1+sin^2(\theta))## ##\Big(\frac{∂z}{∂\theta}\Big)=\Big(\frac{∂}{∂\theta}\Big)r^2(1+sin^2(\theta))=2r^2sin(\theta)cos(\theta)=r^2sin(2\theta)## Now by *not*...

Homework Statement If ## z=x^2+2y^2 ##, find the following partial derivative: \Big(\frac{∂z}{∂\theta}\Big)_x Homework Equations ## x=r cos(\theta), ~y=r sin(\theta),~r^2=x^2+y^2,~\theta=tan^{-1}\frac{y}{x} ## The Attempt at a Solution I've been using Boas for self-study and been working on...

Thank you for this interview. Griffiths has interesting things to say. I particularly like his statement to "Learn the math" ... a view that I wholeheartedly hold. It is where the wheat is separated from the chaff ... so to speak ...

Okay so then $$-\sum_{n=1}^{\infty}\frac{1}{n 2^n}=-1/2-1/8-1/24-1/64...=-ln(2),$$which makes$$\sum_{n=1}^{\infty}\frac{1}{n 2^n}=1/2+1/8+1/24+1/64...=ln(2). \hspace{1cm}\square$$ Thank you all very much! I learned a lot from this including the term Mercator series to signify ln(1+x)

I hear you, but [1] is $$\sum_{n=1}^{\infty}\frac{1}{n 2^n}$$ not $$-\sum_{n=1}^{\infty}\frac{1}{n 2^n}$$

I get my desired series $$-1/2-1/8-1/24-1/64,$$ but with the signs negative ... the negative of what I need. Some progress ...

If you mean $$x=1/2,$$ then we get $$ln(3/2)=ln(3)-ln(2)$$

As I mentioned above ... -ln(2).

Well it is only allowed to be in the interval -1<x<=1 ... picking a negative value in this range takes [2] to all negative terms ... and in fact choosing x=-1/2 makes the series converge to -ln(2) ... the opposite sign. But the idea here is not to start from the answer and work backward ...

Thank you micromass! I've been through that already with no insight. For example the series we desire is [1]. Manipulating [3] we easily get the first term (1-1/2), but starting on the next term we need 1/8 to come out ... but to do this with a -1/8 in [3] it seems clear that stuff must add up...