Unity+ Posted October 5, 2014 Share Posted October 5, 2014 (edited) Alright prove the following: Let [math]f(x) = a(x)b(x)[/math], where b(x) is the inverse of a(x) and a(x) is a linear function. Now, let us redefine the definition of derivatives: [math]f'(x) =lim_{h\rightarrow 0} \frac{f(x+d_{i}h) - f(d_{e}x)}{h}[/math] Where [math]d_{i}[/math] is equal to the coefficient of x while [math]d_{e}[/math] is a constant of the linear function. Prove(or disprove) that for all [math]d_{e} > 1[/math], the limit from both sides will not exist. Edited October 5, 2014 by Unity+ Link to comment Share on other sites More sharing options...

/backslash/ Posted October 5, 2014 Share Posted October 5, 2014 Too bad I'm only up to precalc. Link to comment Share on other sites More sharing options...

Bignose Posted October 5, 2014 Share Posted October 5, 2014 Seems very homeworkish. What steps have you done to attempt it? Link to comment Share on other sites More sharing options...

elfmotat Posted October 5, 2014 Share Posted October 5, 2014 (edited) Alright, so they tell you that: [math]a(x)=d_i x+d_e[/math] [math]b(x)=a^{-1}(x)=\frac{x-d_e}{d_i}[/math] [math]f(x) = a(x) b(x) = \frac{(d_i x + d_e) (d_i - d_e)}{x}[/math] This does seem a bit homework-ish, so I'll let you take it from here. Just plug and chug and see what you get. EDIT: corrected a^{-1}. Edited October 5, 2014 by elfmotat Link to comment Share on other sites More sharing options...

Unity+ Posted October 5, 2014 Author Share Posted October 5, 2014 (edited) Seems very homeworkish. What steps have you done to attempt it? It's not homework, actually. I just thought people might want to take a whack at it. I don't think my Calculus class even gets into the stuff when modifying the values within the derivative definition. Alright, so they tell you that: [math]a(x)=d_i x+d_e[/math] [math]b(x)=a^{-1}(x)=\frac{d_i-d_e}{x}[/math] [math]f(x) = a(x) b(x) = \frac{(d_i x + d_e) (d_i - d_e)}{x}[/math] This does seem a bit homework-ish, so I'll let you take it from here. Just plug and chug and see what you get. [math]b(x)=a^{-1}(x)=\frac{x-d_e}{d_{i}}[/math] How would plugging in values even prove anything? The proof has to apply to all numbers specified. My conjecture is that for all values of d_e above 1, the result of the limit will be none existent. Edited October 5, 2014 by Unity+ Link to comment Share on other sites More sharing options...

elfmotat Posted October 5, 2014 Share Posted October 5, 2014 [math]b(x)=a^{-1}(x)=\frac{x-d_e}{d_{i}}[/math] You're right, sorry about that. How would plugging in values even prove anything? The proof has to apply to all numbers specified. My conjecture is that for all values of d_e above 1, the result of the limit will be none existent. Evaluate it and see if any contradictions pop up. If they do then the derivative DNE. Link to comment Share on other sites More sharing options...

Unity+ Posted October 5, 2014 Author Share Posted October 5, 2014 You're right, sorry about that. Evaluate it and see if any contradictions pop up. If they do then the derivative DNE. The problem with that is the definition of the derivative is changed in the conjecture. Therefore, regular tests won't work, I don't think. Link to comment Share on other sites More sharing options...

elfmotat Posted October 5, 2014 Share Posted October 5, 2014 Alright, I'll take your word for it that it's not homework. We have up to first order in h (which is all that's necessary if h->0): [math]f(x+d_i h)=x^2+2d_i hx - d_e x-d_i d_e h+r x +d_e h-rd_e[/math] [math]f(d_e x)=d_e x^2+rx-d_e x - r[/math] where r=d_{e}/d_{i}. So: [math]\frac{f(x+d_i h)-f(d_e x)}{h}=2d_i x +d_e (1-d_i)+\frac{(x^2+r)(1-d_e)}{h}[/math] So if the equation holds for all x, then the limit as h->0 DNE because the last term goes to infinity. However if d_e=1 then the last term drops out and the limit, and therefore the derivative, exists. 2 Link to comment Share on other sites More sharing options...

Unity+ Posted October 5, 2014 Author Share Posted October 5, 2014 (edited) Alright, I'll take your word for it that it's not homework. We have up to first order in h (which is all that's necessary if h->0): [math]f(x+d_i h)=x^2+2d_i hx - d_e x-d_i d_e h+r x +d_e h-rd_e[/math] [math]f(d_e x)=d_e x^2+rx-d_e x - r[/math] where r=d_{e}/d_{i}. So: [math]\frac{f(x+d_i h)-f(d_e x)}{h}=2d_i x +d_e (1-d_i)+\frac{(x^2+r)(1-d_e)}{h}[/math] So if the equation holds for all x, then the limit as h->0 DNE because the last term goes to infinity. However if d_e=1 then the last term drops out and the limit, and therefore the derivative, exists. Alright, that seems to be a stable proof. You see, I am trying to prove the Collatz conjecture and I feel there is something down this road that will lead to something. EDIT: Also, if the case is that d_e must not be bigger than 1, it also cannot be less than 1. Edited October 5, 2014 by Unity+ Link to comment Share on other sites More sharing options...

fiveworlds Posted October 5, 2014 Share Posted October 5, 2014 (edited) Also, if the case is that d_e must not be bigger than 1, it also cannot be less than 1. This function is never bigger or less than 1 ((sin(a*pi)=cos(y*pi-(pi/2))+1)/2=z Edited October 5, 2014 by fiveworlds -1 Link to comment Share on other sites More sharing options...

Unity+ Posted October 5, 2014 Author Share Posted October 5, 2014 This function is never bigger or less than 1 ((sin(a*pi)=cos(y*pi-(pi/2))+1)/2=z And can you say why that is important? Link to comment Share on other sites More sharing options...

fiveworlds Posted October 5, 2014 Share Posted October 5, 2014 And can you say why that is important? You can specify mathematical limits within a function without say lim a > anything. Link to comment Share on other sites More sharing options...

Unity+ Posted October 6, 2014 Author Share Posted October 6, 2014 You can specify mathematical limits within a function without say lim a > anything. I can't understand what you are saying. Can you explain more? Link to comment Share on other sites More sharing options...

fiveworlds Posted October 6, 2014 Share Posted October 6, 2014 (edited) I can't understand what you are saying. Can you explain more? You wanted to specify that d_e cannot be greater than or less than 1. Therefore ((sin(a*pi)=cos(y*pi-(pi/2))+1)/2=+-d_e does exactly that. There is also other graphs you could use similar for. Say x=y but you only want to select between +-1 and +-2 you could say (((sin(a*pi)=cos(y*pi-(pi/2))+1)/2)+1=+-x=y. This makes graphs with gaps in them much easier. Edited October 6, 2014 by fiveworlds -1 Link to comment Share on other sites More sharing options...

John Posted October 6, 2014 Share Posted October 6, 2014 (edited) If you think about this geometrically, then the result makes fairly good sense.Consider that when taking the standard limit to find the derivative, we're looking at the value of the slope of a line between two points, f(x) and some other point f(x + h), as the two points are brought closer and closer together along the curve in question. Thus, in the limit, we find the slope when the two points are "the same," i.e. the slope of the tangent line at that point. Now look at the modifications you've introduced to the function arguments in the numerator. Since d_{i}h still goes to zero, f(x + d_{i}h) still approaches f(x). However, if d_{e} is anything other than 1, then f(d_{e}x) will not be equal to f(x). Thus f(x + d_{i}h) - f(d_{e}x) will be non-zero. Since h in the denominator still goes to zero, the result is "infinity," i.e. the limit does not exist.Perhaps you've already thought along these lines, but just in case you hadn't, I thought I'd post. Edited October 6, 2014 by John 2 Link to comment Share on other sites More sharing options...

Unity+ Posted October 6, 2014 Author Share Posted October 6, 2014 (edited) If you think about this geometrically, then the result makes fairly good sense. Consider that when taking the standard limit to find the derivative, we're looking at the value of the slope of a line between two points, f(x) and some other point f(x + h), as the two points are brought closer and closer together along the curve in question. Thus, in the limit, we find the slope when the two points are "the same," i.e. the slope of the tangent line at that point. Now look at the modifications you've introduced to the function arguments in the numerator. Since d_{i}h still goes to zero, f(x + d_{i}h) still approaches f(x). However, if d_{e} is anything other than 1, then f(d_{e}x) will not be equal to f(x). Thus f(x + d_{i}h) - f(d_{e}x) will be non-zero. Since h in the denominator still goes to zero, the result is "infinity," i.e. the limit does not exist. Perhaps you've already thought along these lines, but just in case you hadn't, I thought I'd post. Well, the problem is without having x+d_ih, the results aren't the same. For example, without having the modification the result is 6x-2/3. However, after adding it, it becomes 6x-2. This applies to other linear functions. EDIT: Yeah, it is kind of self-explanatory, I guess. Edited October 6, 2014 by Unity+ Link to comment Share on other sites More sharing options...

imatfaal Posted October 6, 2014 Share Posted October 6, 2014 ! Moderator Note Fiveworld Do not introduce spurious and nonsensical branches into threads. Start a new thread and PM the OP if you want to present a correlated idea. DO NOT respond to this moderation within the thread. Link to comment Share on other sites More sharing options...

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