Just try to post my first problem
For each positive integer n , let
f n ( x ) = sin ( f n − 1 ( x ) ) ,
with f 0 ( x ) = x .
Let y ( x ) = lim n → ∞ f n ( x ) .
Determine y ′ ( 0 ) .
The answer is 0.00.
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3 solutions
Moderator note:
This solution is incorrect. It did not justify interchanging of the limits with taking of the derivative.
This solution is incorrect, because you cannot interchange the derivative with the limit without further justification.
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I'm so sorry for the wrong solution. The solution is the one given to me by my teacher after i finished the test and i thought that it was right. I'm so sorry. I'm just level 2 in calculus. Hopefully i can learn from this mistake :)
Let y=f(x)= sin(sin(...sin(x)...)) Then, Left hand derivative (LHD) of f(x) at x=0 is f ′ ( 0 ) = lim h → 0 + ( 0 − h ) − 0 f ( 0 − h ) − f ( 0 ) = lim h → 0 + ( − h ) f ( − h ) ( s i n c e f ( 0 ) = 0 ) = lim h → 0 + s i n ( . . . s i n ( − h ) . . . ) s i n ( s i n ( . . . s i n ( − h ) . . . ) ) . . . . . s i n ( − h ) s i n ( s i n ( − h ) × ( − h ) s i n ( − h ) = 1 (multiplying and dividing with the parameter in each of sine function to obtain standard limit of : lim x → 0 x sin x which is equal to 1) Therefore LHD of f(x) at x=0 is 1 Similarly, for RHD of f(x) at x=0 f ′ ( 0 ) = lim x → 0 + 0 + h − 0 f ( 0 + h ) − f ( 0 ) = lim x → 0 + h f ( h ) = 1 (same explanation as above) Therefore, Derivative of f(x) at x=0 exist and is equal to 1
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And yes the f(x) is continuous at all points which can also be proven
Can you clarify if you were intending for there to be a finite number of sin θ , or an infinite number of them? Did I change the question significantly from what you intended?
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Yes i were intending for there to be an infinite number of sin θ . I think the changes is about what i'm intending to. Thank you for the correction. I just wrote the question like the one i ever saw in my test, sorry for it.
y is a sine operation applied a finite number of times to x. It doesn't matter whether it's applied 100 times or 1000 times, so why can't we just say it's applied once?
y = sin x y ′ ( 0 ) = cos 0 = 1
There is no logic in your solution. For example, take f ( x ) = x 2 . Then clearly f ′ ( x ) = 2 x while f 2 ( x ) = x 4 and so it's derivative is 4 x 3 .
Also, the interpretation of the problem is not that the function is applied finitely many times,
Wow . . . I don't know about that theorem, thanks for telling me. But i still don't understand. Can it be applied to any value of x? As example :
if y = sin ( sin ( sin ( sin . . . ( sin ( x ) ) . . . )
I don't think that we can just say it as y = sin x when x = 3 0 . Could you please explain to me about that? Thank you.
No offense here, i'm just want to know more :)
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It's definitely not true in the general sense. Of course, sin x = sin ( sin x ) . But the problem provided was deliberately vague about the number of times sine was applied, so in order for it to be solvable, I assumed the number of times simply didn't matter.
So it isn't a mathematical argument; I assumed the problem was solvable, and the problem doesn't tell us how many sine's there are, so therefore the number of sine's shouldn't be important for our solution.
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Okay, now i get it. Thank's :D
Actually it doesn't matter how many sine's are there. Because the differential will be continously as :
y ′ = cos ( sin ( sin ( . . . . ( sin ( x ) . . . ) ) ) cos ( sin ( sin ( . . . . ( sin ( x ) . . . ) ) ) . . . . .
that's why i added the x as : 0 so it will be easily solvable. Because everything in the ( sin ( sin ( . . . . ( sin ( x ) . . . ) ) ) will become 0 . Therefore the equation would be simple :
y ′ = cos 0 . cos 0 . cos 0 . . . . . . = 1 . 1 . 1 . . . . . . = 1
i don't know if i am correct or not: but still what i think: sin30=1/2 and sin 1/2=0.0087 and sin(sin(sin(30)))=0.00015. use the calculator. Thus we see answer is constantly decreasing. and it will never become zero. But it would be so small that this function y=sin(sin(sin...(sin(x))....)) tends to 0. so derivative of it will be lim y tends to 0 cos(y).cos(y).cos(y)...
Now see: as the value of y was not zero and just greater than it, cos(y) will not be 1 just smaller than it(i.e. 0.999999999...)
And now the point is lim x tends to 1(-) [ that means from left side of number line] x^infinity =0 because it will rapidly decrease with increment in power maybe, Calvin Lin can tell us the actual method.
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It is generally assumed that trigonometric operations are performed on angles in terms of radians, not degrees (which is an arbitrary scale, compared with radians). When computing the sine of a positive number close to 0, one gets a number very close to that number, but slightly lower. I am not in 100% agreement with the solution given (1), and this is more fully explained in my post in the "disputes" section. The problem clearly implies an infinite number of sine operations, and if this is the case, the argument can be made that y=0 for all real x, giving us a differential of 0, not 1.
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@Brian Galebach – For the given nested sine function one can show that the derivative of the give function exist and is 1 using the method of differentiability of a function at a point which I have tried to show above.
Yeah, maybe Calvin Lin is needed here, but i don't know how to ask him to join this discussion. By the way, that's an interesting explanation, thank you.
and this question was interesting than the original one :)
y = sin ( sin ( sin . . . . . sin ( 0 ) ) . . . ) ) y ′ = cos ( sin ( sin . . . sin ( 0 ) ) . . ) ) cos ( sin ( sin . . . sin ( 0 ) ) . . ) ) . . . y ′ = cos 0 cos 0 cos 0 . . . y ′ = 1