Question #5003d

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Question #5003d

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Answer 1 · 2015-01-01T21:50:22

I am not sure if I got the meaning of your question right but I'll try anyway:

$\int_{0}^{x} t^{2} d t = 2 (x - 1)$ $int_0^xt^2dt=2(x-1)$
$\frac{t^{3}}{3} ∣_{0}^{x} = 2 (x - 1)$ $t^3/3|_0^x=2(x-1)$
$\frac{x^{3}}{3} = 2 (x - 1)$ $x^3/3=2(x-1)$

Rearranging:
$x^{3} - 6 x + 6 = 0$ $x^3-6x+6=0$

To solve this cubic equation I used the Cardano-Tartaglia formula that you can find in most texts or on the internet such as in Wikipedia:

enter image source here

I supposed that you needed only real roots so I neglected the ones which are complex numbers.
You have only one real root (an irrational number) given as:
$x_{1} = - 2, 84732$ $x_1=-2,84732$

You can check it by substituting in:
$\frac{x^{3}}{3} = 2 (x - 1)$ $x^3/3=2(x-1)$

Probably there are simpler techniques but I do not know. Hope it helped.

Answer 2 · 2015-01-02T01:52:04

There is no value of $x$ $x$ that solves this problem.

By "greatest integer function of $t$ $t$ " I assume you mean $⌈ t ⌉$ $|~t~|$ , called the ceiling of $t$ $t$ or the ceiling function.

For every finite interval chosen as it's domain, the ceiling function and it's square $⌈ t ⌉^{2}$ $|~t~|^2$ are step functions , which means that there is a finite collection of intervals in which the function takes a constant value for each interval. Equivalently, $f$ $f$ is called a step function if, for a finite number $m$ $m$ of intervals $I_{k}$ $I_k$

$f (x) = c_{k} (b_{k} - a_{k}), x \in I_{k}$ $f(x) = c_k(b_k - a_k), x in I_k$

where $b_{k} - a_{k}$ $b_k - a_k$ is the length of the interval $I_{k}$ $I_k$

The integral of any step function is simply the sum of the lengths of all intervals multiplied by their respective constants:

$\int_{a_{1}}^{b_{m}} f (x) = m \sum k = 1 c_{k} (b_{k} - a_{k})$ $int_(a_1)^(b_m) f(x) = sum_(k=1)^m c_k (b_k - a_k)$

Now, the function $⌈ t ⌉^{2}$ $|~t~|^2$ is defined as

$⌈ t ⌉^{2} = n^{2}, x \in (n - 1, n)$ $|~t~|^2 = n^2, x in (n-1, n)$

where $n$ $n$ are integers.

It's integral is, observing that all intervals of the form $(n - 1, n)$ $(n-1, n)$ have length $1$ $1$ and that the last interval $(⌊ x ⌋, x)$ $( |__ x __|, x )$ has length $x - ⌊ x ⌋$ $x- |__ x __|$ , where $⌊ x ⌋$ $|__ x __|$ is the floor function:

$\int_{0}^{x} ⎡ ⎢ ⎢ ⎢ t {⎤ ⎥ ⎥ ⎥}^{2} = ⌊ x ⌋ \sum k = 1 k^{2} + ⎡ ⎢ ⎢ ⎢ x {⎤ ⎥ ⎥ ⎥}^{2} (x - ⌊ x ⌋)$ $int_(0)^(x) |~t~|^2 = sum_(k=1)^(|__ x __|) k^2 + |~x~|^2 (x-|__ x __|)$

Now, your question also gives us the condition that

$\int_{0}^{x} ⌈ t ⌉^{2} = 2 (x - 1)$ $int_(0)^(x) |~t~|^2 = 2(x-1)$

Wich means:

$⌊ x ⌋ \sum k = 1 k^{2} + ⎡ ⎢ ⎢ ⎢ x {⎤ ⎥ ⎥ ⎥}^{2} (x - ⌊ x ⌋) = 2 x - 2$ $sum_(k=1)^(|__ x __|) k^2 + |~x~|^2 (x-|__ x __|) = 2x-2$

Now, the sum of the first $q$ $q$ squares is $\frac{2 q^{3} + 3 q^{2} + q}{6}$ $(2q^3+3q^2+q)/6$ .

So,

$⌊ x ⌋ \sum k = 1 k^{2} = \frac{2 {(⌊ x ⌋)}^{3} + 3 {(⌊ x ⌋)}^{2} + ⌊ x ⌋}{6}$ $sum_(k=1)^(|__ x __|) k^2= (2(|__ x __|)^3+3(|__ x __|)^2+|__ x __|)/6$

With the additional condition given to us by the question:

$\frac{2 {(⌊ x ⌋)}^{3} + 3 {(⌊ x ⌋)}^{2} + ⌊ x ⌋}{6} + ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ x {⎤ ⎥ ⎥ ⎥ ⎥ ⎥}^{2} (x - ⌊ x ⌋) = 2 x - 2$ $(2(|__ x __|)^3+3(|__ x __|)^2+|__ x __|)/6 + |~x~|^2 (x-|__ x __|) = 2x-2$

Finding the solution to this equation is not simple. But noticing that the left side behaves itself similarly to a $3$ $3$ rd degree polynomial, as the value of $x$ $x$ gets bigger, it should grow faster than the right hand side.

Consider the left side equation for $x = 1$ $x=1$ . We have:

$\frac{2 {(⌊ 1 ⌋)}^{3} + 3 {(⌊ 1 ⌋)}^{2} + ⌊ 1 ⌋}{6} = \frac{2 \times 1 + 3 \times 1 + 1}{6} = 1$ $(2(|__ 1 __|)^3+3(|__ 1 __|)^2+|__ 1 __|)/6 = (2 times 1 + 3 times 1 +1)/6 = 1$

The right hand side:

$2 \times 1 - 2 = 0$ $2 times 1 - 2 =0$

So even where the function at the left side grows at it's slowest, the left side of the equation is greater than the right side. So, for $x > 0$ $x>0$ , there is no solution.

Another way of thinking about this problem is graphing the integral. For $x < 1$ $x<1$ , the graph of the integral is a line segment of slope $1$ $1$ passing through the origin. For $1 \geq x < 2$ $1 geq x < 2$ , it's a line segment of slope $4$ $4$ , and as $x$ $x$ get's bigger, so does the slope. For $x < 1$ $x<1$ , $2 x - 2 < 0$ $2x-2<0$ , so this line does not intersect the graph of the integral in this interval. But for $x > 1$ $x>1$ , the integral grows much faster then $2 x - 2$ $2x-2$ , so the two graphs will never intersect.

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Question #5003d

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