No
That's what I intended to do.
But my problem is with the one that can be x∈(0,δ) (when are δ>0) so that f'(x)=0 and then f(0)=f(x)=0 for ∀x∈(0,δ) And then what I have to prove is wrong.
Do you understand my problem?
What I wanted to do was set f(x)=##x^2##/2 - xcosx+sinx And show that f(x)>0.
f'(x)=x(1+sinx)
First I wanted to prove that f(x)<0 in the interval (0,∞)
0≤1+sinx≤2
And thus for all x> 0 f'(x)≥0 and therefore f(x)≥f(0)=0
And it doesn't help me much because I need to f(x)>0
I meant what you wrote here.
If you don't understand what I wrote, tell me and I'll upload a photo of the inequality.
Because I can't post it with the forum tools
I'm sorry I didn't write the problem to the end.
Yes I meant it #<## a_n+1##<##\frac a_n a_1 ##
Prove that if for every n is Happening ## a_n+1##<##\frac a_n a_1 ## then a1<0
I need help only in section 3
I have some kind of solution but I'm not sure because it seems too short and too simple.
We showed in section 1 that an> 0 per n.
it Given that a_n + 1 <0 and a_n+1<\frac a_n a_1 In addition therefore a_1 <0 is warranted
I need help only in section 3
I have some kind of solution but I'm not sure because it seems too short and too simple.
We showed in section 1 that an> 0 per n.
Given that an + 1 <0 and an + 1 = an / a1 therefore a1 <0 is warranted
Ok, I sort of understand what you want me to do.
You want me to divide the interval [a,b] Into two parts With the help of a midpoint c.
And say that For any t that is between f(a) and f(c) Exists x1∈[a,c] so that f(x1)=t , Then switch the point f(a) whit f(b) And say that ∃x2∈[c,b] so that...
It all makes sense to me, but I don't know how to formalize it nicely.
I wanted to divide it into two cases.
First case where f is fixed in the segment.
And a second case where f is not fixed in the segment.
But I don't know how to prove it for the case where f i is not fixed