Describing function - Revision history

en>Giraffedata: comprised of

2014-12-17T04:10:53Z

← Older revision		Revision as of 06:10, 17 December 2014
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	~~{{Multiple issues\|unreferenced = December 2011\|orphan = December 2011}}~~		Marvella is what you can contact her but it's not the most feminine name out there. Minnesota has always been his home but his spouse wants them to move. My working day job is a meter reader. To collect badges is what her family members and her appreciate.<br><br>Here is my site ... [http://nuvem.tk/altergalactica/MarissaepDonnt std home test]

	~~In [[mathematics]], a '''[[polynomial]]'''~~ is ~~a function of the form <math> a_0 + a_1 x + \cdots + a_{n} x^n </math>.~~
	~~This article will discuss various '''polynomial transformations'~~''.

	~~==Reciprocals of the roots==~~
	~~Suppose we have some polynomial <math> a_0 + a_1 x + \cdots + a_{n} x^n </math> and we are to find the sum of the [[Multiplicative inverse\|reciprocals]] of the [[nth root\|root]]~~s~~. Initially we may think that we are faced with~~ the ~~task of finding the roots and then their reciprocals and summing them, but~~ there ~~is an easier way~~.

	~~For every root <math>r_n</math> of f(x) the following holds,~~
	~~<math> f\left( \frac{1}{\frac{1}{r_n}} \right) = 0 </math>~~. ~~Looking at this statement further we see that plugging in x for <math> \frac{1}{r_n} </math> we get~~ a ~~polynomial whose roots are the reciprocals of the roots of f(x). Nevertheless this function will not be a polynomial, as its highest power will be <math> \frac{1}{x^n} </math>~~. To ~~turn this into a polynomial all we have to do~~ is multiply it by <math> x^n </math>. Note the order of the [[coefficient]]s, earlier the polynomial was <math> a_0 + a_1 x + \cdots + a_{n} x^n </math>, now the coefficients have simply reversed so we get <math> a_{n} + a_{n-1} x + \cdots + a_0 x^n </math>. The sum of the roots of this polynomial will be <math> a_1 </math>, which was the product of the roots taken n-1 at a time for the original polynomial.

	~~==Constant multiples of the roots==~~
	~~To find say a function that has k times the roots of f(x) we can either again find all the roots, multiply then by k~~ and ~~then multiply them together appropriately to find some function for which they are the roots or we can think of it the smart way~~. ~~We can make the substitution~~
	<~~math~~> ~~\frac{x}{2} = x~~ <~~/math~~> ~~into f(x)~~.

	~~==Roots that differ by a constant==~~

	~~Suppose we have some polynomial <math> a_0 + a_1 x + \cdots + a_{n-1} x^{(n-1)} </math> and we want to find some polynomial that has roots that are k units greater than the roots of f(x)~~. One obvious approach would be to find the roots of f(x) and then multiply out a polynomial that has k greater than those roots. Also notice that this can be easily accomplished by thinking of the graph of the function. If it has roots at <math> r_1, r_2 \ldots r_n </math> than we can simply shift the function over k units which will give us a function that has roots that are k units greater than the roots of f(x). so for ''x'' in ''f''(''x'') substutite ''x'' − ''k''.

	~~==The pattern==~~
	~~Suppose we have some one to one onto~~ [[Function (mathematics)\|function]] ''m''(''x'') and some function ''f''(''x'') that has roots <math> r_0, r_1, r_2, r_3, \ldots , r_n </math> and we are faced with the problem of finding a function that has roots

	:~~<math> m(r_0), m(r_1), m(r_2), \ldots, m(r_n). </math>~~

	~~Given that the function ''m''(''x'') is both one to one and onto we can find <math> m^{-1}(x) </math>. Now to find a function that has roots <math> m(r_0), m(r_1), m(r_2), \ldots, m(r_n) <~~/~~math> we only have to make the substitution <math> x= m^{-1}(x) <~~/~~math>~~. This function will have roots at <math> m(r_0), m(r_1), m(r_2), \ldots, m(r_n) </math> because ''f''(''x'') only has roots at <math> r_0, r_1, r_2, r_3, \ldots , r_n </math> and plugging in <math> m(r_n) </math> into the function will be the same as pluggiing in <math> r_n </~~math> into f(x) because <math> m^{-1}(x) <~~/~~math> is a function such that <math> m^{-1}(m(x)) = x </math>.~~

	~~[[Category:Algebra]~~]

en>Monkbot: Fix CS1 deprecated date parameter errors

2014-01-16T16:13:40Z

Fix CS1 deprecated date parameter errors

← Older revision		Revision as of 18:13, 16 January 2014
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	Hi there, ~~I am Alyson Boon even though~~ it ~~is not~~ the ~~name on my beginning certificate~~. The ~~favorite hobby~~ for ~~him~~ and ~~his kids is~~ to ~~play lacross and he'll be beginning~~ some ~~thing else along with~~ it. ~~For~~ a ~~while I've been in Alaska but I will~~ have to ~~move in~~ a ~~yr or two~~. ~~I am presently~~ a ~~journey agent~~.<br><br>~~my blog post~~: ~~free online tarot card readings~~ - ~~[http:~~//~~fashionlinked~~.~~com~~/~~index.php?do=~~/~~profile~~-~~13453~~/~~info~~/ ~~fashionlinked~~.~~com~~] -		{{Multiple issues\|unreferenced = December 2011\|orphan = December 2011}}

			In [[mathematics]], a '''[[polynomial]]''' is a function of the form <math> a_0 + a_1 x + \cdots + a_{n} x^n </math>.
			This article will discuss various '''polynomial transformations'''.

			==Reciprocals of the roots==
			Suppose we have some polynomial <math> a_0 + a_1 x + \cdots + a_{n} x^n </math> and we are to find the sum of the [[Multiplicative inverse\|reciprocals]] of the [[nth root\|root]]s. Initially we may think that we are faced with the task of finding the roots and then their reciprocals and summing them, but there is an easier way.

			For every root <math>r_n</math> of f(x) the following holds,
			<math> f\left( \frac{1}{\frac{1}{r_n}} \right) = 0 </math>. Looking at this statement further we see that plugging in x for <math> \frac{1}{r_n} </math> we get a polynomial whose roots are the reciprocals of the roots of f(x). Nevertheless this function will not be a polynomial, as its highest power will be <math> \frac{1}{x^n} </math>. To turn this into a polynomial all we have to do is multiply it by <math> x^n </math>. Note the order of the [[coefficient]]s, earlier the polynomial was <math> a_0 + a_1 x + \cdots + a_{n} x^n </math>, now the coefficients have simply reversed so we get <math> a_{n} + a_{n-1} x + \cdots + a_0 x^n </math>. The sum of the roots of this polynomial will be <math> a_1 </math>, which was the product of the roots taken n-1 at a time for the original polynomial.

			==Constant multiples of the roots==
			To find say a function that has k times the roots of f(x) we can either again find all the roots, multiply then by k and then multiply them together appropriately to find some function for which they are the roots or we can think of it the smart way. We can make the substitution
			<math> \frac{x}{2} = x </math> into f(x).

			==Roots that differ by a constant==

			Suppose we have some polynomial <math> a_0 + a_1 x + \cdots + a_{n-1} x^{(n-1)} </math> and we want to find some polynomial that has roots that are k units greater than the roots of f(x). One obvious approach would be to find the roots of f(x) and then multiply out a polynomial that has k greater than those roots. Also notice that this can be easily accomplished by thinking of the graph of the function. If it has roots at <math> r_1, r_2 \ldots r_n </math> than we can simply shift the function over k units which will give us a function that has roots that are k units greater than the roots of f(x). so for ''x'' in ''f''(''x'') substutite ''x'' − ''k''.

			==The pattern==
			Suppose we have some one to one onto [[Function (mathematics)\|function]] ''m''(''x'') and some function ''f''(''x'') that has roots <math> r_0, r_1, r_2, r_3, \ldots , r_n </math> and we are faced with the problem of finding a function that has roots

			:<math> m(r_0), m(r_1), m(r_2), \ldots, m(r_n). </math>

			Given that the function ''m''(''x'') is both one to one and onto we can find <math> m^{-1}(x) </math>. Now to find a function that has roots <math> m(r_0), m(r_1), m(r_2), \ldots, m(r_n) </math> we only have to make the substitution <math> x= m^{-1}(x) </math>. This function will have roots at <math> m(r_0), m(r_1), m(r_2), \ldots, m(r_n) </math> because ''f''(''x'') only has roots at <math> r_0, r_1, r_2, r_3, \ldots , r_n </math> and plugging in <math> m(r_n) </math> into the function will be the same as pluggiing in <math> r_n </math> into f(x) because <math> m^{-1}(x) </math> is a function such that <math> m^{-1}(m(x)) = x </math>.

			[[Category:Algebra]]

en>Kuznetsov N.V.: /* Caveats */

2012-07-16T20:19:30Z

Caveats

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Hi there, I am Alyson Boon even though it is not the name on my beginning certificate. The favorite hobby for him and his kids is to play lacross and he'll be beginning some thing else along with it. For a while I've been in Alaska but I will have to move in a yr or two. I am presently a journey agent.<br><br>my blog post: free online tarot card readings - [http://fashionlinked.com/index.php?do=/profile-13453/info/ fashionlinked.com] -