- #1
myshadow
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I just ran into this problem and have no idea how to solve it. Basically I'm trying to prove that all orders of derivative of the given function is bounded by the function on the right. I'm pretty sure the inequality is true, but I really have no clue on how to prove it. I thought about using Leibniz Rule (I actually didn't know about it before), but I got stuck. I think I might be making it more complicated than it actually is.
Prove
|d^n_x(\frac{e^{-(a+b)x}}{3+\cos(e^{-bx})})|≤\frac{C}{(1+x)^d}
for any n, any d>0, and all 0≤x<∞,
where a and b are constants greater than zero and C is a constant greater than zero that depends on d and n.
General Leibniz Rule (relevant?)
f and g are n differentiable functions then the nth derivative is of their product is
(f\cdot g)^n=\sum_{k=0}^n \begin{pmatrix}n\\k\end{pmatrix}f^{(k)} g^{(n-k)},
where \begin{pmatrix}n\\k\end{pmatrix} is the binomial coefficient.
I don't know if any of this is correct...
The function is analytical for all x and therefore smooth for all x. Hence, for n=0, the left side exponentially decays to zero. From here, for n=0, I don't know how to prove that the inequality is true for all d>0.
For n>0, since the function is smooth, I can apply Leibniz Rule as follows:
f=e^{-(a+b)x} and g=(3+\cos(e^{-bx}))^{-1}.
Then,
f^{(k)}=(-(a+b))^k e^{-(a+b)x}.
Therefore,
|d^n_x(\frac{e^{-(a+b)x}}{3+\cos(e^{-bx})})|=\sum_{k=0}^n \begin{pmatrix}n\\k\end{pmatrix}(-(a+b))^k (e^{-(a+b)x})g^{(n-k)}
Lastly, I guess I can argue, since the left side of the inequality is smooth and always multiplied by an exponentially decaying function, the left side is bounded.
I'm pretty sure my last step is wrong because I didn't show what the behaviour of the g^(n-k) is. Also, again I don't know what to do for the "any d>0" part. I'm really stuck and I have no idea what to do. I'm a mechanical engineer so my math skills aren't super advanced. Thanks in advance for any help.
Homework Statement
Prove
|d^n_x(\frac{e^{-(a+b)x}}{3+\cos(e^{-bx})})|≤\frac{C}{(1+x)^d}
for any n, any d>0, and all 0≤x<∞,
where a and b are constants greater than zero and C is a constant greater than zero that depends on d and n.
Homework Equations
General Leibniz Rule (relevant?)
f and g are n differentiable functions then the nth derivative is of their product is
(f\cdot g)^n=\sum_{k=0}^n \begin{pmatrix}n\\k\end{pmatrix}f^{(k)} g^{(n-k)},
where \begin{pmatrix}n\\k\end{pmatrix} is the binomial coefficient.
The Attempt at a Solution
I don't know if any of this is correct...
The function is analytical for all x and therefore smooth for all x. Hence, for n=0, the left side exponentially decays to zero. From here, for n=0, I don't know how to prove that the inequality is true for all d>0.
For n>0, since the function is smooth, I can apply Leibniz Rule as follows:
f=e^{-(a+b)x} and g=(3+\cos(e^{-bx}))^{-1}.
Then,
f^{(k)}=(-(a+b))^k e^{-(a+b)x}.
Therefore,
|d^n_x(\frac{e^{-(a+b)x}}{3+\cos(e^{-bx})})|=\sum_{k=0}^n \begin{pmatrix}n\\k\end{pmatrix}(-(a+b))^k (e^{-(a+b)x})g^{(n-k)}
Lastly, I guess I can argue, since the left side of the inequality is smooth and always multiplied by an exponentially decaying function, the left side is bounded.
I'm pretty sure my last step is wrong because I didn't show what the behaviour of the g^(n-k) is. Also, again I don't know what to do for the "any d>0" part. I'm really stuck and I have no idea what to do. I'm a mechanical engineer so my math skills aren't super advanced. Thanks in advance for any help.
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