MHB What Properties of Fourier Series Are Revealed by This Equation Transformation?

nacho-man
Messages
166
Reaction score
0
Is there some properties I should be aware of?

after making the relevant substitutions, I ended up with

$2 = 1 + \sum\nolimits_{m=0}^\infty \frac{4}{(2m+1)\pi}\sin(\frac{(2m+1)\pi}{2})$

but I can't get past this
 

Attachments

  • asdasd.png
    asdasd.png
    21.1 KB · Views: 99
Physics news on Phys.org
nacho said:
Is there some properties I should be aware of?

after making the relevant substitutions, I ended up with

$2 = 1 + \sum\nolimits_{m=0}^\infty \frac{4}{(2m+1)\pi}\sin(\frac{(2m+1)\pi}{2})$

but I can't get past this

Rearrange to:

$$\frac{\pi}{4}=\sum_{m=0}^\infty \frac{1}{2m+1}\sin\left(\frac{(2m+1)\pi}{2}\right)$$

and ask yourself what values does the sine of odd half multiples of $$\pi $$ take?

.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
28
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K