MHB Sum of Infinite Series: Find 1/sqrt(2)

Click For Summary
The discussion centers on finding the sum of an infinite series that equals 1/sqrt(2). The original poster attempted to use the ratio test but found it unhelpful. A response clarifies that the series in question is the power series expansion for cos(π/4). By comparing it to the standard power series for cos(x), the correct result can be derived. This highlights the importance of recognizing series forms in solving such problems.
Umar
Messages
36
Reaction score
0
Hey guys! I just have a question regarding finding the sum of an infinite series. Attached is the image of the question. I've tried to use the ratio test but it doesn't give me the result I need which happens to be 1/sqrt(2). I feel like this is one of those power series questions, but I'm not quite sure how to deal with this. If anyone could help out, that would be greatly appreciated!

View attachment 6430
 

Attachments

  • find the sum of the following series.png
    find the sum of the following series.png
    1.5 KB · Views: 135
Physics news on Phys.org
Umar said:
Hey guys! I just have a question regarding finding the sum of an infinite series. Attached is the image of the question. I've tried to use the ratio test but it doesn't give me the result I need which happens to be 1/sqrt(2). I feel like this is one of those power series questions, but I'm not quite sure how to deal with this. If anyone could help out, that would be greatly appreciated!
This is just the power series expansion for $\cos\frac\pi4.$
 
This is \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}\left(\frac{\pi}{4}\right)^{2n}

Compare that to the power series for cos(x), \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n} to get Opalg's result.
 
Last edited by a moderator:
Thread 'Problem with calculating projections of curl using rotation of contour'
Hello! I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem. Given: ##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0## ##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1## ##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0## I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...

Similar threads

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