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ch4aika [34]
4 years ago
14

When do you use the divergence test, comparison test, p test, geometric, integral, comparison, limit comparison, and alternating

series?
Mathematics
1 answer:
Anvisha [2.4K]4 years ago
8 0

Step-by-step explanation:

Divergence Test: use this when the limit as n approaches infinity of a sequence isn't 0.

If lim(n→∞) an ≠ 0, then an diverges.

(Note this only tests divergence, not convergence.)

P Test: use this when the series is a p-series.

For an = 1 / nᵖ, if p > 1, then the series converges.  Otherwise, it diverges.

Geometric Test: use this when the series is a geometric series.

For an = a₁ (r)ⁿ, if -1 < r < 1, then the series converges.  Otherwise, it diverges.

Integral Test: use this when the sequence can be easily integrated.

If ∫₁°° f(x) dx converges, then ∑₁°° f(n) converges.

If ∫₁°° f(x) dx diverges, then ∑₁°° f(n) diverges.

Comparison Test: use this when a sequence is similar to a p-series or a geometric series.

If bn > an and bn converges, then an converges.

If bn < an and bn diverges, then an diverges.

Otherwise, inconclusive.

Limit Comparison Test: use this when comparison test is inconclusive.

If an ≥ 0 and bn > 0, and lim(n→∞) an/bn > 0 and finite, then an and bn either both converge or both diverge.

Alternating Series Test: use this when the series is alternating.  This usually includes (-1)ⁿ or (-1)ⁿ⁺¹, but might use trig functions instead.

If an = (-1)ⁿ bn or (-1)ⁿ⁺¹ bn, where bn ≥ 0, and if lim(n→∞) bn = 0, and bn is decreasing, then an converges.

(Notice this only tests for convergence, not divergence.)

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Answer:

\frac{7\pi}{24} and \frac{31\pi}{24}

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\sqrt{3} \tan(x-\frac{\pi}{8})-1=0

Let's first isolate the trig function.

Add 1 one on both sides:

\sqrt{3} \tan(x-\frac{\pi}{8})=1

Divide both sides by \sqrt{3}:

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The first ratio I have can be found using \frac{\pi}{6} in the first rotation of the unit circle.

The second ratio I have can be found using \frac{7\pi}{6} you can see this is on the same line as the \frac{\pi}{6} so you could write \frac{7\pi}{6} as \frac{\pi}{6}+\pi.

So this means the following:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}

is true when x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi

where n is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

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Add \frac{\pi}{8} on both sides:

x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi

Find common denominator between the first two terms on the right.

That is 24.

x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi

x=\frac{7\pi}{24}+n \pi (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval [0,2\pi).

So if \sqrt{3} \tan(x-\frac{\pi}{8})-1=0 and we let u=x-\frac{\pi}{8}, then solving for x gives us:

u+\frac{\pi}{8}=x ( I just added \frac{\pi}{8} on both sides.)

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Then 0 \le u+\frac{\pi}{8}.

Subtract \frac{\pi}{8} on both sides:

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Simplify:

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So we want to find solutions to:

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-\frac{\pi}{8}\le u

That's just at \frac{\pi}{6} and \frac{7\pi}{6}

So now adding \frac{\pi}{8} to both gives us the solutions to:

\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}} in the interval:

0\le x.

The solutions we are looking for are:

\frac{\pi}{6}+\frac{\pi}{8} and \frac{7\pi}{6}+\frac{\pi}{8}

Let's simplifying:

(\frac{1}{6}+\frac{1}{8})\pi and (\frac{7}{6}+\frac{1}{8})\pi

\frac{7}{24}\pi and \frac{31}{24}\pi

\frac{7\pi}{24} and \frac{31\pi}{24}

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