Recent content by sunnybrooke

Never mind. Thank you Michael & piercebeatz.

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 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 will be 1...

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.

Thank you so much haruspex and SammyS for breaking it down and guiding me through the problem. It helped me a lot :)

w(w+1)=odd*even=even --> w mod 2 = 0. same goes for x^2 + x. So the RHS is even but the LHS is odd. therefore it's not possible. Is that it?

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...

Yes, they'd both have to be even. Should I substitute, a = 2u, b = 2v ?

96 is divisible by 4 (remainder = 0). 88 is also divisible by 4 (remainder = 0). So sum of remainders (0 + 0) = 0.