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3 years ago

Evaluate the limit by first recognizing the sum as the Riemann sum for a function defined on the interval [0,1]: lim as n approa

ches infinity (1/n)((sqrt(1/n)+(sqrt(2/n)+(sqrt(3/n)+...............+(sqrt(n/n))

Mathematics

1 answer:

mina [271]3 years ago

3 0

$\lim_{n \to +\infty} \left ( \dfrac{1}{n} \cdot \left ( \sqrt{ \dfrac{1}{n} } + \sqrt{ \dfrac{2}{n} } + \sqrt{ \dfrac{3}{n} } + \cdots + \sqrt{ \dfrac{n}{n} }} \right ) \right ) =$

$\dfrac{1}{n} \sum_{i = 1}^n \left ( \sqrt{ \dfrac{i}{n} } \right ) = \int\limits^1_0 { \sqrt{x} } \, dx = \left [ \frac{2}{3}x^{ \frac{3}{2} } \right ]^{1}_{0} = \dfrac{2}{3} \cdot 1^{ \frac{3}{2} } - \dfrac{2}{3} \cdot 0^{ \frac{3}{2} } = \boxed{ \dfrac{2}{3} }$

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(5/10)^4 in exponential form

Whitepunk [10]

Answer:

$6.25 \times {10}^{ - 2}$

Step-by-step explanation:

${ \bigg( \frac{5}{10} \bigg) }^{4} \\ \\ = {(0.5)}^{4} \\ \\ = 0.0625 \\ \\ = 6.25 \times {10}^{ - 2}$

6 0

3 years ago

Read 2 more answers

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$