Homework Statement
Question: what rate does the water lose gravitational potential energy?
Data:
I have a pipe that water is flowing through and the pipe has 2 sections.
In section 1:
- the pipe is 11.9m above section 2 so, h = 11.9m
- the velocity of water is $$v_1 = 0.3240ms^{-1}$$...
Ok, yeah I got no idea.
I'm pretty sure 1/x < 1/(lnx)^9, I just can't prove it like I can for 1/x < 1/lnx.
And since 1/x diverges so does 1/(lnx)^9.
I'm not sure how the facts that lnx < x iff x < e^x (is this ever false?) and limit of e^x / x^n -> infinity helps.
Ok, so we can say that ## \ln x < e^x## so ## (\ln x)^9 < e^9x ## and ## 1/e^9x < 1/(lnx)^9##
Is that the idea? (those x's should be in the exponent as well)
Oh right,
lnx < \sqrt{x}
\sqrt{x} - lnx > 0
\frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{x}-lnx)=\frac{1}{2\sqrt{x}}-\frac{1}{x} > 0
when x >= 2.
so if that's true then so is this:
(lnx)^2 < x
Is that right?
Well, lnx <= 0 whenever 0 < x <= 1 so that bit is obvious I guess.
Also, whenever x > 1, lnx < x because it's a monotonically increasing function with first derivative:
1 - 1/x > 0.
For that to be true x > 1.
\sin x\leqslant 1
\frac{\sin x}{x^{2}}\leqslant \frac{1}{x^{2}}
\frac{\sin^2 x}{x^{2}}\leqslant \sin(\frac{1}{x^{2}})
Am I on the right track here?
Sorry, I'm not really sure how sin(x)/x relates to sin^2(1/x) in terms of which is greater than the other. How can I figure that out?
Hmm, what about this, if:
\sin x\approx x
then
\sin^2 x\approx x^{2}
Since I'm just squaring both sides it should still be true right?
Then I can go:
a_{x}=\sin ^2\frac{1}{x}
b_{x}=\frac{1}{x^2}
It does work out this way, but what function did you have in mind to compare it to?
Homework Statement
Use a comparison test to determine whether this series converges:
\sum_{x=1}^{\infty }\sin ^2(\frac{1}{x}) Homework EquationsThe Attempt at a Solution
At small values of x:
\sin x\approx x
a_{x}=\sin \frac{1}{x}
b_{x}=\frac{1}{x}
\lim...
Homework Statement
\sum_{x=2}^{\infty } \frac{1}{(lnx)^9}
Homework EquationsThe Attempt at a Solution
x \geqslant 2
0 \leqslant lnx < x
0 < \frac{1}{x} < \frac{1}{lnx}
From this we know that 1 / lnx diverges and I wanted to use this fact to show that 1 / [(lnx) ^ 9] diverges but at k...
So it goes kinda likes this:
For n > 1 we have n^2>n hence, 2n^2>n So:
2n^2+4n-1>5n-1>5n>n
And in a similar way:
For n > 1 we have 3 - 8n > 3 - 8 = -5 So:
\mid \frac{-5}{n} \mid < \epsilon
and,
n > 5 / \epsilon
Is that what you mean?