You're right, I forgot to mention that. When k > 2 or k < 2, there are no solutions (because the amplitude is 2).
Additionally, could I approach this problem algebraically (without referring to a graph)?
Thanks.
http://www4c.wolframalpha.com/Calculate/MSP/MSP4251a45574034148cah00001164cg2h50a94c78?MSPStoreType=image/gif&s=5&w=300&h=183&cdf=RangeControl [Broken]
So if k:
is = -2, there will be 2 zeros
is between (-2,0] there will be 4 zeros
is between (0,2), there will be 2 zeros
is = 2, there...
Homework Statement
Homework Equations
The Attempt at a Solution
I can solve the first part of the question. -2 ≤ k ≤ 2 because -1 ≤ sin(x) ≤ 1. How do I solve the second part of the question? Thanks.
Thank you. That's what I thought too but at least now I've solved it. :) I'd love to find another way to solve it but I don't seem to understand your solution. :( I will look at it again.
Unfortunately, I don't understand why that is implied.
Is this a valid solution?
x_n = \frac{x_{n-1}^5 +1}{5x_{n-1}}\\
f(x) = \frac{x^5+1}{5x}
Find the minimum using the first derivative.
f '(x)=\frac{4x^5-1}{5x^2}\\
f '(x)=0\\
x=(\frac{1}{4})^\frac{1}{5}\\
f(x) =...
Never mind; silly me. It is the definition of a recursive sequence.
I don't understand how exactly to prove by induction. Can I do the following?
x_{n+1}> ((3/11)^5+ 1)/5(3/11)= 0.74> 3/11
The notation is confusing me. The problem is equating two different sequences, right? And the sequence on the left is defined as xn+1 from n=1.
So for n=1: xn+1 = x(1)+1 = x2
Isn't this the first term?
I meant to say x_{1+1}=\frac{x^5_1 + 1}{5x_1}=1
Since the first term of the sequence on the left is x_{((1)+1)}
won't it be x_2=1 and not x_1=1?
Thanks.
Homework Statement
Homework Equations
The Attempt at a Solution
This is what I have so far:
x_{n+1}=\frac{x^5_n + 1}{5x_n}=1
x_{n+2}=\frac{x^5_{n+1} + 1}{5x_{n+1}}=\frac{1^5+1}{5}
I think you have to do some sort of repeated substitution but I don't quite see it. Any help? Thanks.
96n + 88 = 4u^2 + 4v^2
24n + 22 = u^2 + v^2
(LHS= 2 mod 4 --> both u and v are odd)
24n + 22 = (2w+1)^2 + (2x+1)^2
6n + 5 = w^2 + w + x^2 + x
(LHS = 3 mod 4 --> ??)
If k^2 = 0 mod 4 then either k = 0 mod 4 or k = 2 mod 4
If k^2 = 1 mod 4 then either k = 1 mod 4 or k = 3 mod 4
Is...