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ser-zykov [4K]
3 years ago
8

Find the smallest integer n for an O(xn) estimate of order of the function f(x)

Mathematics
1 answer:
alexdok [17]3 years ago
6 0

Answer with explanation:

f(x)=\frac{(x^3+x^2+5)(x^4-3x^2)}{(2x^2+2x-3)(4x^2+7)}\\\\f(x)=\frac{x^3\times (x^4-3x^2)+x^2 \times (x^4-3x^2)+5 \times (x^4-3x^2)}{2x^2\times (4x^2+7)+2x(4x^2+7)-3\times (4x^2+7)}\\\\f(x)=\frac{x^7-3x^5+x^6-3x^4+5x^4-15x^2}{8x^4+14x^2+8x^3+14 x-12x^2-21}\\\\f(x)=\frac{x^7+x^6-3x^5+2x^4-15x^2}{8x^4+8x^3+2x^2+14 x-21}

The largest degree of numerator is 7 , while the largest degree of denominator is 4.So, as we know

 \frac{x^a}{x^b}=x^{a-b}\\\\\frac{x^7}{x^4}=x^{7-4}\\\\=x^3

→So, order of the function is the highest degree in the function raised to power.Highest degree is 3,when you will reduce the function in Expression form.

So, Degree=3

Order=1

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How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used
MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the <em>n</em>-th term of the sequence, <em>a</em> (<em>n</em>), is given recursively by

• <em>a</em> (1) = 1

• <em>a</em> (<em>n</em>) = <em>a</em> (<em>n</em> - 1) + 21 … … … for <em>n</em> > 1

We can find the explicit rule for the sequence by iterative substitution:

<em>a</em> (2) = <em>a</em> (1) + 21

<em>a</em> (3) = <em>a</em> (2) + 21 = (<em>a</em> (1) + 21) + 21 = <em>a</em> (1) + 2×21

<em>a</em> (4) = <em>a</em> (3) + 21 = (<em>a</em> (1) + 2×21) + 21 = <em>a</em> (1) + 3×21

and so on, with the general pattern

<em>a</em> (<em>n</em>) = <em>a</em> (1) + 21 (<em>n</em> - 1) = 21<em>n</em> - 20

Now, we're told that the sum of some number <em>N</em> of terms in this sequence is 2332. In other words, the <em>N</em>-th partial sum of the sequence is

<em>a</em> (1) + <em>a</em> (2) + <em>a</em> (3) + … + <em>a</em> (<em>N</em> - 1) + <em>a</em> (<em>N</em>) = 2332

or more compactly,

\displaystyle\sum_{n=1}^N a(n) = 2332

It's important to note that <em>N</em> must be some positive integer.

Replace <em>a</em> (<em>n</em>) by the explicit rule:

\displaystyle\sum_{n=1}^N (21n-20) = 2332

Expand the sum on the left as

\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332

and recall the formulas,

\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n

\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2

So the sum of the first <em>N</em> terms of <em>a</em> (<em>n</em>) is such that

21 × <em>N</em> (<em>N</em> + 1)/2 - 20<em>N</em> = 2332

Solve for <em>N</em> :

21 (<em>N</em> ² + <em>N</em>) - 40<em>N</em> = 4664

21 <em>N</em> ² - 19 <em>N</em> - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

<em>N</em> = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer <em>N</em> for which the first <em>N</em> terms of the sum add up to 2332.

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Read 2 more answers
3x−8 ≤ 23 OR −4x+26≥6 find x
ElenaW [278]

Answer:

3x-8<=23

3x<=23+8

3x< =31

X< = 10.3

or

-4x+26>=6

-4x>=-20

X>=5

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Nastasia [14]

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