| 
November 1st, 2009, 02:03 PM
| | Member | | Join Date: Apr 2009
Posts: 189
Country:  Thanks: 21
Thanked 0 Times in 0 Posts
| |  mx=sin x
Find roots of the equation mx=sin x considering different values of m
I have figured out that 0<m<1 or else the graph wont intersect other than at x=0. But I cant calculate the roots generally, unless i plug in different values of m.
|
November 2nd, 2009, 04:32 AM
| | MHF Contributor | | Join Date: Apr 2005
Posts: 3,499
Thanks: 328
Thanked 1,214 Times in 1,115 Posts
| |
What more do you want? There is no algebraic method to solve mx= sin x.
In general, there is no algebraic method to solve x= f(x) where f is any transcendental function. |
November 2nd, 2009, 04:37 AM
| | Member | | Join Date: Apr 2009
Posts: 189
Country: Thanks: 21
Thanked 0 Times in 0 Posts
| |
Quote:
Originally Posted by HallsofIvy  What more do you want? There is no algebraic method to solve mx= sin x.
In general, there is no algebraic method to solve x= f(x) where f is any transcendental function. | That's all the question said, so I am guessing that we are just meant to consider that there are 3 solutions if the gradient is less than 1, or else there will be just 1. I wasn't sure what else we could do with it.
There was also another question on the sheet which was similar that I thought couldn't be solved: Solve a^b=b^a for all real a and b
but it might be solvable through logs?
Last edited by Aquafina; November 2nd, 2009 at 08:14 AM.
|
November 2nd, 2009, 07:49 AM
|  | Senior Member | | Join Date: May 2009 Location: ALGERIA
Posts: 415
Country:  Thanks: 115
Thanked 23 Times in 19 Posts
| |
Hello
You have 3 cases :
1 ) 
2 ) 
3 ) 
__________________
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
|
November 2nd, 2009, 07:50 AM
|  | MHF Contributor | | Join Date: Dec 2008 Location: South Coast of England
Posts: 1,659
Country: Thanks: 111
Thanked 928 Times in 808 Posts
| |
Hello Aquafina Quote:
Originally Posted by Aquafina That's all the question said, so I am guessing that we are just meant to consider that there are 3 solutions if the gradient is less than 1, or else there will be just 1... | Are you sure about that? See the attached diagram. Quote:
There was also another question on the sheet which was similar that I thought couldn't be solved:
Solve ab=ba for all real a and b
but it might be solvable through logs?
| You obviously can't mean  . Do you mean  ?
Grandad |
November 2nd, 2009, 08:16 AM
| | Member | | Join Date: Apr 2009
Posts: 189
Country: Thanks: 21
Thanked 0 Times in 0 Posts
| |
Quote:
Originally Posted by Grandad Hello AquafinaAre you sure about that? See the attached diagram. | Hi Grandad! How does that help? Both the gradients are less than 1, so it doesn't disprove what I said? Yes sorry typo, I have corrected it now to a^b = b^a. Any ideas with this? |
November 2nd, 2009, 09:00 AM
| | MHF Contributor | | Join Date: Dec 2008 Location: South Coast of England
Posts: 1,659
Country: Thanks: 111
Thanked 928 Times in 808 Posts
| |
Hello Aquafina
 clearly intersects  at 7 points. So how can you say there are only 3 solutions?
Grandad
|
November 2nd, 2009, 09:24 AM
| | Senior Member | | Join Date: May 2009 Location: ALGERIA
Posts: 415
Country: Thanks: 115
Thanked 23 Times in 19 Posts
| |
Hello Aquafina: y = x/100 clearly intersects y=sinx at 63 points conclusion 63 solutions look here:
__________________
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
| | Thread Tools | | | | Display Modes |  Linear Mode | 
Posting Rules
| You may not post new threads You may not post replies You may not post attachments You may not edit your posts
HTML code is Off | | | All times are GMT -7. The time now is 12:23 AM. | | |
 | |  |
