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DanielleElmas [232]
2 years ago
11

Write the prime factorization of 10. Use exponents when appropriate and order the factors from least to greatest (for example, 2

235).
7th math

Mathematics
2 answers:
Goryan [66]2 years ago
8 0

Answer:

1,2,5,10

Step-by-step explanation:

satela [25.4K]2 years ago
4 0

Answer: 1,2,5,10 are factors of 10

Step-by-step explanation:

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marusya05 [52]

Answer:

See below

(B) and (C) are correct.

Step-by-step explanation:

We have the following limit

$\lim \limits_{n\rightarrow \infty} \left(\frac{n^n(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right)}{n!(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right)^{\dfrac{x}{n} }, \forall x>0$

I am not sure about methods concerning the quotient, but in this type of question I would try to convert this limit into integration.

Considering the numerator, we have

$(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right) = \prod_{k=1}^n  \left(x+\dfrac{n}{k} \right)$

- I didn't forget about n^n

Considering the denominator, we have

$(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right) = \prod_{k=1}^n  \left(x^2+\dfrac{n^2}{k^2} \right)$

- I didn't forget about n!

Therefore,

$\left(\frac{n^n(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right)}{n!(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right)^{\dfrac{x}{n} } = \left(\dfrac{n^2 \prod_{k=1}^n  \left(x+\dfrac{n}{k} \right)}{ n!\prod_{k=1}^n  \left(x^2+\dfrac{n^2}{k^2} \right)} \right)$

$= \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}$

Now we have

$\lim \limits_{n\rightarrow \infty}  \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}, \forall x>0$

This is just the notation change so far.

What I want to do here is apply definite integrals using Riemann Integrals (We will write the limit as an definite integral). A nice way to do it is using logarithms. Therefore, we can apply the natural logarithm in both sides.

Now, recall two properties of logarithms:

\boxed{\log_a mn = \log_a m + \log_a n}

\boxed{\log_a m^p = p\log_a m}

\boxed{\log_a  \left(\dfrac{m}{n} \right) = \log_a m- \log_a n}

Thus,

$\ln f(x) = \lim \limits_{n\rightarrow \infty}  \ln \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}} $

$= \lim \limits_{n\rightarrow \infty}   \dfrac{x}{n}\ln\left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right) $

$=  \lim \limits_{n\rightarrow \infty}  \dfrac{x}{n} \left[\ln  \left(n^n \prod_{k=1}^n  \left(x+\dfrac{n}{k} \right)  \right)-\ln  \left( n!\prod_{k=1}^n  \left(x^2+\dfrac{n^2}{k^2} \right)  \right) \right]$

$=  \lim \limits_{n\rightarrow \infty}  \dfrac{x}{n} \left[\ln  n^n + \prod_{k=1}^n  \ln \left(x+\dfrac{n}{k}  \right)-\ln  n! -\prod_{k=1}^n  \ln\left(x^2+\dfrac{n^2}{k^2} \right)  \right]$

Considering

$\lim \limits_{n\rightarrow \infty} \frac{x}{n} (\ln n^n - \ln  n! ) = \lim \limits_{n\rightarrow \infty} \frac{x}{n} (n\ln n - \ln  n! )= \lim \limits_{n\rightarrow \infty} \frac{x \cdot\ln\frac{n^n}{n!} }{n} $

Using Stirling's formula

$\dfrac{n^n}{n!}\underset{\infty}{\sim} \dfrac{n^n}{\sqrt{2n \pi}\left(\dfrac{n}{e}\right)^n}=\dfrac{e^n}{\sqrt{2n \pi}}$

then

$\ln\left(\frac{n^n}{n!}\right)\underset{\infty}{=}n\ln\left(e\right)-\frac{1}{2}\ln\left(2n\pi\right)+o\left(1\right)$

$\implies \frac{\ln\left(\frac{n^n}{n!}\right)}{n}=1-\frac{\ln(2n\pi)}{2n}+o\left(1\right)$

This shows our limit equals 1 as $\frac{\log(2\pi n)}{2n} \rightarrow 0$ and \ln(e)=1

Employing a Riemann sum in the main limit, we have

$= \lim \limits_{n\rightarrow \infty}  \dfrac{x}{n} \left[ \sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)  - \sum_{k=1}^n\ln \left(x^2+\dfrac{n^2}{k^2} \right)  \right]$

Now dividing the terms inside the parenthesis by \dfrac{n}{k} in \sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)

we have

$\sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)  = \sum_{k=1}^n \ln \left(\frac{kx}{n} +1\right) $

Now dividing the terms inside the parenthesis by \dfrac{n^2}{k^2} in \sum_{k=1}^n \ln \left(x^2+\dfrac{n^2}{k^2} \right)

we have

$\sum_{k=1}^n \ln \left(x^2+\dfrac{n^2}{k^2} \right)  = \sum_{k=1}^n \ln \left(\frac{(kx)^2}{n^2} +1\right) $

Therefore

$= \frac xn\sum_{k=1}^n\ln\dfrac{z+1}{z^2+1}$

for \dfrac{kx}{n}  = z

Using Riemann Integral,

$\lim \limits_{n\rightarrow \infty}  \int_0^1\ln\frac{z+1}{z^2+1}dz$

From

$\frac{f'(x)}{f(x)}=\ln\frac{z+1}{z^2+1}$

We can see that the function is increasing for , but because of the denominator, it is negative for .

Therefore,

(A) is false because \dfrac{1}{2} < 1

(B) is true because

(C) is true the slope is negative at that point

(D) is false, just consider $\ln\frac{z+1}{z^2+1}$ for z=1 and z=2

7 0
2 years ago
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