Please check if my method is correct. Solution : Let $$f(x) = x^8 - x^7 + x^2 -x +15 $$ Now, let $g(x)= x^8 -x^7=x^7(x - 1)$ and $h(x)= x^2 -x=x(x-1)$. Thus, $$f(x)=g(x) + h(x) + 15$$  On analyzing $g(x)$ and $h(x)$ we see that both the functions are negative between $0$ and $1$  and the maximum negative value of both the functions do not exceed $-1$. Thus, the minimum value of $f(x)$ is not less than $13$. Hence, $f(x)$ has no real root. Q.E.D
Let $ P(x)=x^8 - x^7 +x^2 -x +15 $, Descartes' Rule of Signs tells us that the polynomial has 4 positive real roots , but if we group the terms as $$ P(x)= x(x-1)(x^6+1) +15 $$ we find that $ P(x) $ does not have any positive roots.  Where did I err ?
Importing users that will be installed by programs I was answering a question on Stack Overflow and required to use a trailing space in one of my code snippets.  I formatted it like so:  `&lt;a `   And it renders without the trailing space: &lt;a  Using the &lt;code&gt; tag does account for trailing spaces, though in my case it busts because of the &lt;. Here's my formatting with the &lt;code&gt; tag:  &lt;code&gt;&lt;a &lt;/code&gt;   Here's how it renders:  (Yes, you should see nothing there).  This is a pretty annoying bug. Can we get it fixed? I don't want to do this through trial and error, and the best way I have found so far was to start dividing from 1.     $n! = \text {a really big number}$   Ex. $n! = 9999999$   Is there a way to approximate n or solve for n through a formula of some sort?  Update (Here is my attempt):  Stirling's Approximation: $$n! \approx \sqrt{2 \pi n} \left( \dfrac{n}{e} \right ) ^ n$$  So taking the log:     $(2\cdot \pi\cdot n)^{1/2} \cdot n^n \cdot e^{-n}$       $(2\cdot \pi)^{1/2} \cdot n^{1/2} \cdot n^n \cdot e^{-n}$       $.5\log (2\pi) + .5\log n + n\log n \cdot -n \log e$       $.5\log (2\pi) + \log n(.5+n) - n$   Now to solve for n:     $.5\log (2\pi) + \log n(.5+n) - n = r$       $\log n(.5+n) - n = r - .5 \log (2\pi)$   Now I am a little caught up here. How do I copy Team Fortress 2 to other computers? Selecting a front derailleur: parameters The spec sheet for the Shimano Alivio FD-M430 front derailleur contains a number of points that I don't understand.   What does "Maximum capacity: 22 teeth" mean? Why are "Top gear teeth" constrained to "44/48T"? Is there anything wrong with 46? What is "Cable Routing: dual-pull type"? And most puzzling, what is "Chain Stay Angle  63-66 / 66-69" and how do I measure mine? Does time freeze at Absolute Zero? Time has many definitions per se, but the basic idea being it's "the measurement of change" so as we know, all matter looses it's ability of changing with the loss of kinetic energy. and the where it becomes zero is the absolute zero (−273.15°C). so my question would be, will time stop/freeze at Absolute zero? and does time get slow while reaching there? Find the exact value of the infinite sum $\sum_{n=1}^\infty \big\{\mathrm{e}-\big(1+\frac1n\big)^{n}\big\}$ How can we find the exact value of the infinite sum  $$ \displaystyle\sum_{n=1}^\infty \left\{\mathrm{e}-\Big(1+\frac1n\Big)^n\right\}? $$  This problem appears in:  , p.114. Complement of maximal multiplicative set is a prime ideal I have an unusual problem with Chrome when I try to open PDFs. If I have already downloaded a PDF and try to download it again, I can open it by tapping its name; in the screenshot below, I could tap "pdf.pdf" to open the file.   The issue is that tapping this link actually opens RollerCoaster Tycoon Classic! If I choose to download again, Chrome redownloads and opens the PDF correctly, so it's not a huge issue, but still an inconvenience.  In Settings, it says that RollerCoaster Tycoon Classic has no defaults, and I am unable to change the "open supported links" setting.    How do I stop RollerCoaster Tycoon  trying to be a PDF reader? Does borrowing from my 401(k) make sense in my specific circumstance? If a creature with Multiattack is affected by Haste, can it make two Multiattacks? It's a fairly straightforward question.   A creature has a Multiattack option, and it also can cast the Haste spell. While under the effect of Haste, can it use the second Action to take another Multiattack Action?  My main concern is the clause of the Haste spell that limits the number of attacks made from taking the Attack Action. This is a different Action, though.   Then there is the concern of being able to use Multiattack at all, because Haste says what kinds of Actions you can take. Every time I touch an object a full-render starts I don't know what setting I switched, or what happened, but for some reason now every time I move, scale or rotate an object in the 3d View - blender starts a render. I am not talking about the 3d View being in "Rendered" mode. I am talking about the same behavior as when you click the "Render" button (switch to image editor, begin full render). This is driving me crazy and I can't seem to figure out how to turn it off or what I did to start it. It happens whether I use the mouse or keys. If I open a new document the behavior goes away so it is probably a preference setting(?) If you can help, I will be eternally grateful.  Thanks in advance. I am running Xubuntu 11.10 and everything works nicely. I have selected a nice Xfce theme and everything looks beautiful, except Libreoffice.  While other programmes (Gimp, Inkscape, Thunar etc.) integrate with the selected Appearance &amp; Window Manager theme, Libreoffice looks like a Windows 98 programme.  Does anyone know how to fix this? I have already tried this solution, without success:   Thanks for your help! Inspired by  I was wondering why the Half Blood Prince left his book as a class copy. One would think that it was precious to him, since he invested so much in it. He made up potions and spells up by his own. I would expect him to keep it to himself. Both, because of the personal value of it, and to prevent others to (as happened at last) use his work (misguided and/or against him).  Is there any canon explanation why his book was left as class copy? Which direction will the yoyo move? How is Faulhaber's formula derived? I have been wanting to understand how to find the sum of this series.   $$1^p + 2^p + 3^p +{\dots} + n^p$$  I am familiar with Gauss' diagonalised adding trick for the sum of the first $n$ natural numbers.  I can prove the formulas for  $$\begin{align}   \sum_{1}^{n} k^2 &amp;= \frac{n(n+1)(2n+1)}{6}\\  \sum_{1}^{n} k^3 &amp;= \frac{n^2(n+1)^2}{4}\\ \sum_{1}^{n} k^4 &amp;= \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} \\ \end{align}$$  With mathematical induction. But, beyond that even proofs with mathematical induction are difficult.  I'm interested in learning the theory and the proof behind Faulhaber's formula. What is the knowledge required to understand this proof ? Is the prefix "pre-" meaningless in the terms "pre-heated" and "pre-board"? This question poses a paradox of meaning. The general question is whether, if two sentences (x and y) can be used in the same situation, with the same literal meaning, and x and y only differ in that x has morphological affixes that y doesn't, aren't the surplus affixes "meaningless" by the alternation test?   Take this example:  I put the chicken inside a preheated oven.    Wouldn't it make more sense to say:  I put the chicken inside a heated oven.      There is only two states an oven can possibly be in—heated or unheated.      Another example: Does first-class board the plane, or pre-board the plane?  What does that mean—to get on before you get on? How to copy files that have too long of a filepath in Windows? How do I generate the upper triangular indices from a list? The square roots of different primes are linearly independent over the field of rationals I need to find a way of proving that the square roots of a finite set    of different primes are linearly independent over the field of    rationals.    I've tried to solve the problem using elementary algebra  and also using the theory of field extensions, without success. To  prove linear independence of two primes is easy but then my problems  arise. I would be very thankful for an answer to this question.