X Xeris New member Joined Jul 19, 2005 Messages 10 Jul 19, 2005 #1 Can anyone help me with this problem I just cant seem to figure it out. Its been to long since Ive done this stuff. Find f(n^r/4) if f( X) = (-2 sin (2X))/3 + (5 cos (4X))/6 Thanks.
Can anyone help me with this problem I just cant seem to figure it out. Its been to long since Ive done this stuff. Find f(n^r/4) if f( X) = (-2 sin (2X))/3 + (5 cos (4X))/6 Thanks.
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,325 Jul 20, 2005 #2 Xeris said: Find f(n^r/4) if f( X) = (-2 sin (2X))/3 + (5 cos (4X))/6 Click to expand... First, just plug it in. That's what the notation means. If f(x) = (-2 sin (2x))/3 + (5 cos (4x))/6, then f(n^r/4) = (-2 sin (2(n^r/4)))/3 + (5 cos (4(n^r/4)))/6 Simplify that as you wish.
Xeris said: Find f(n^r/4) if f( X) = (-2 sin (2X))/3 + (5 cos (4X))/6 Click to expand... First, just plug it in. That's what the notation means. If f(x) = (-2 sin (2x))/3 + (5 cos (4x))/6, then f(n^r/4) = (-2 sin (2(n^r/4)))/3 + (5 cos (4(n^r/4)))/6 Simplify that as you wish.
X Xeris New member Joined Jul 19, 2005 Messages 10 Jul 20, 2005 #3 how would you simplify it? Thats the part that in stuck on. It has to break down to -3/2 and thats what I cant seem to get.
how would you simplify it? Thats the part that in stuck on. It has to break down to -3/2 and thats what I cant seem to get.
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,325 Jul 20, 2005 #4 First, I'm not sure what you mean by n^r/4. Is that n^(r/4) or (n^r)/4 Feel free to use parentheses to clarify your meaning.
First, I'm not sure what you mean by n^r/4. Is that n^(r/4) or (n^r)/4 Feel free to use parentheses to clarify your meaning.
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,325 Jul 20, 2005 #6 You sure are you of that argument? Your f(X) achieves -3/2 at pi/4 + k*pi, for k an Integer. I'm really not seeing how ¼n<sup>r</sup> does that.
You sure are you of that argument? Your f(X) achieves -3/2 at pi/4 + k*pi, for k an Integer. I'm really not seeing how ¼n<sup>r</sup> does that.
X Xeris New member Joined Jul 19, 2005 Messages 10 Jul 21, 2005 #7 Maybe it was the othre way. I must have just interpreted it wrong. But the way that I wrote it down was the exact way they gave it to me.
Maybe it was the othre way. I must have just interpreted it wrong. But the way that I wrote it down was the exact way they gave it to me.
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,325 Jul 21, 2005 #8 Well, when you find the solution, be sure to let us know. Are you sure that 'n' is really an 'n', or is it a 'pi'-symbol?
Well, when you find the solution, be sure to let us know. Are you sure that 'n' is really an 'n', or is it a 'pi'-symbol?
X Xeris New member Joined Jul 19, 2005 Messages 10 Jul 21, 2005 #9 It was an "n" but maybe they ment to put a pi sign. If it was a pi sign would it be able to reduce to -3/2?
It was an "n" but maybe they ment to put a pi sign. If it was a pi sign would it be able to reduce to -3/2?
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,325 Jul 21, 2005 #10 Not a clue. I was hoping n = pi was just a sign that there were other things wrong with the problem statement. I appeal to the masses for additional ideas. I'm fresh out.
Not a clue. I was hoping n = pi was just a sign that there were other things wrong with the problem statement. I appeal to the masses for additional ideas. I'm fresh out.
