+9628 【Calculus】
\(f(x)=\dfrac{\cos\left(x + \dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\cdot\cdot\cdot+\dfrac{x^{101}}{101!}+\sin x\right)}{\sin\left(1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\cdot\cdot\cdot+\dfrac{x^{100}}{100!}+\cos x\right)}\)
Find \(\displaystyle\int^{\pi/2}_{-\pi/2}f^{(2017)}(x) \;dx\)
Where \(f^{(2017)}(x) \) denotes the 2017th derivative of f(x)
Even the smartest cookie in the world don't have any idea of solving this so I am seeking help here :(
Btw This looks pretty symmetrical :D
~The smartest cookie in the world
+33616 f(x) is an odd function (i.e. f(-x) = - f(x)) and its derivatives are alternately even and odd. This means that if you were to count f(x) as f(1)(x) then f(2017)(x) would also be odd and the integral would be zero.
However, if you really mean f(2017)(x) is the 2017th derivative of f(x) then it is an even function and I don't see a simple way of determining it (that's not to say there isn't a way; just that I can't see it!).
I KNOW/believe the answer is Zero.
The function f is an even function.
The functions fn alternate as odd and even.
f2017 is an odd function since 2017 is odd.
The definite integral of an odd function over a symmetrical interval about zero is precisely Zero. 
Maybe someone can take it from here...
Notice that the series on the numerator is 1/2(e^x -e^(-x) ) and on the denominator it's
1/2( e^x + e^(-x) )
At 100 or so terms, the function e^(-x) becomes so small that we can discount it and the function tends to
f(x) =cos(1/2 e^x + sin(x) ) / sin( 1/2e^x + cos (x) )
We are obviously not expected to differentiate it two thousand times.
The other thing I noticed is that ,under the Quotient rule, (u/v)' the denominator, v^2 will start appearing like this
f (1) - first derivative v^2 = sin^2(1/2 e^x + cos (x) )
f(2) - second " " v^2= sin^4 (1/2 e^x + cos (x) )
f(3) = sin^8 ( " " " )
f(4) = sin^16( " " " )
f(n) = sin^ (2^n) ( " " " ) ie will numerically tend to zero
After a couple of thousand derivatives,the numerator will be such a long chain of sine and cosine products that it will numerically tend to zero and I agree with Alan that the integral is zero. (probably!)
