Polynomial Articles - Wyzant Tutor Blogshttps://www.wyzant.com/resources/blogs/polynomial?f=activeThis is an aggregate of all of the Polynomial articles in Wyzant.com's Tutors' Blogs. Wyzant.com is your source for tutors and students.Mon, 01 Mar 2021 01:54:58 -0600https://www.wyzant.com/images/logos/wyzant-logo.pngPolynomial Articles - Wyzant Tutor Blogshttps://www.wyzant.com/resources/blogs/polynomial?f=activehttps://www.wyzant.com/resources/blogs/polynomial?f=active357281https://www.wyzant.com/resources/blogs/357281/taylor_expansion_and_approximating_roots_of_polynomials_over_the_rational_fieldCasey W.https://www.wyzant.com/resources/users/view/85739199Taylor Expansion and approximating roots of polynomials over the rational field...<div>I answered a nice question on WyzAnt about approximating roots to a quadratic equation that had real but irrational roots. </div>
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<div>Calculus enables us to find rational approximations to these irrational roots. I will re-present the solution to the problem, and continue to find a general expression to do this for an arbitrary polynomial.</div>
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<div>"Problem: approximate roots of -9x^2+8x+5.</div>
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<div>There is a nice approach using calculus to estimate/approximate a function without a square root and calculator.<br /> <br />We can use the concept of moments to get an approximation to a function. For this example, we have a quadratic function in (x) with coefficients, a=-9, b=8, and c=5, as indicated in a previous solution.<br /> <br />Thus f(x)=ax^2+bx+c.<br /> <br />First we need to find a general idea of where a root (an x where f(x)=0) is located. To do this we can check the value of the function for some easy numbers, x=0 and x=1 are always good choices:<br /> <br />f(1)=a+b+c=4<br />f(0)=c=5<br /> <br />We notice the function is increasing from 0 to 1 (the value decreased as x decreased). So if we try x>1 maybe we will find a 0...next evaluate f(2)<br /> <br />f(2)=a*4+b*2+c=-36+16+5 < 0<br /> <br />Awesome, since this is a continuous function we know there must be a root between 1 and 2...lets call this root x=1+\eps, for some little \eps that is between 0 and 1.<br /> <br />We are now ready to use a linear approximation and a first derivative to approximate where this root is!<br /> <br />Let (x_0,y_0)=(1,4), and we set \delta x=\epsilon.<br /> <br />We need to recall the derivative f'(x) denotes the slope of the function at a point, so we will approximate the derivative at the root using f'(1)...Since f'(x)=2ax+b...f'(1)=2a+b=-10...<br /> <br />So near to x=1, say like the root we seek at x=1+\epsilon, the slope of a line approximating this function should be about -10...thus when y=0 at x=1+\epsilon a line approximating the curve should be:<br /> <br />y-y_0=m(x-x_0)...0-4=-10(\epsilon)...so \epsilon=4/10=2/5...and the root should be close by to x=1.4.<br /> <br />Now evaluate f(1.4)= -9*1.4^2+8*1.4+5 = -9*1.96+11.2+5=-17.64+16.2 = -1.44...<br /> <br />So we overshot by a little, our \epsilon was too large, it made our y value decrease from 4 to something past 0, a negative number...thus the true \epsilon should of been smaller. We have a choice now, with given possible values above we can quickly see which choice has an x greater than 1 and less than 1.4 {1.31,-.42}.<br /> <br />Of course evaluating all the choices initially works too...but this allows us to apply calculus to get an approximation, in case no choices were available and a linear approximation would be acceptable.<br /> <br />We can use this new point (x_1,y_1)=(1.4,-1.44)=(7/5,-36/25) to find a better approximation.<br /> <br />Use x=7/5-\delta, for some small \delta between 0 and 1, and repeat the procedure:<br /> <br />y-y_1=m(x-x_1), where we now let m be the deviated at x_1=7/5...<br /> <br />Thus 0-(-36/25)=f'(7/5)*(-\delta)...and \delta=18/215...so x=1+\epsilon-\delta=1+2/5-18/215=(215+86-18)/215=283/215...doing long divison to a few decimal places yields 1.316.<br /> <br />This shows how we can by hand...using no machinery, calculators, excel sheets...produce a better approximation over the field of rational numbers than the choices given offers!"</div>
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<div>We see that in general for a function y=f(x) that we are looking for a root of...if we find an x_0 where f(x_0)>0 and f(x_0+1)<0 (or vice versa)...we know there exists an \epsilon in (0,1), with x=x_0+\epsilon and f(x)=0.</div>
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<div>We can find this by writing down the linear approximation:</div>
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<div>0-f(x_0)=f'(x_0)(\epsilon), and solve for \epsilon=-f(x_0)/f'(x_0)...this is the first moment, an approximation of first order:</div>
<div>root ~ x_0-f(x_0)/f'(x_0)</div>
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<div>We can now evaluate f at x_0+\epsilon=x_1 and find a new base point to approximate near....x_1=x_0-f(x_0)/f'(x_0)...</div>
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<div>We first need to check if f(x_1) is positive or negative and then proceed by adding or subtracting appropriately...but we end up with a telescoping sum of epsilons and deltas that get better and better approximations for our root...</div>
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<div>WLOG assume that f(x_1)<0, meaning our epsilon was too large and we should subtract off a \delta in (0,1), and let</div>
<div>x = x_1 - \delta = x_0 + \epsilon - \delta.</div>
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<div>We then have a linear approximation of this x by looking at the line her x_1=x_0+\epsilon:</div>
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<div>0-f(x_1)=f'(x_1)(-\delta)...so \delta = f(x_1)/f'(x_1)...and a second order approximation for the irrational root becomes:</div>
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<div>root ~ x_0+\epsilon-\delta = x_0-f(x_0)/f'(x_0)-f(x_1)/f'(x_1)</div>
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<div>Note how this is a sort of generating function for moments...substitution our formula for x_1 in terms of x_0, x_1=x_0-f(x_0)/f'(x_0), yields:</div>
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<div>root ~ x_0+\epsilon-\delta = x_0 - f(x_0)/f'(x_0) - f(x_0-f(x_0)/f'(x_0))/f'(x_0-f(x_0)/f'(x_0))</div>
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<div>Note the nested derivative in the second order term here.</div>
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<div>Untangling this using what we know about derivatives leads to a general taylor expansion for a function near a point x=a when the function is infinitely differentiable at x=a. </div>
<div> </div>Sat, 30 May 2015 14:35:49 -05002015-05-30T14:35:49-05:00264492https://www.wyzant.com/resources/blogs/264492/application_of_algebraic_polynomials_in_cost_accountancyWalter G.https://www.wyzant.com/resources/users/view/82731680Application of Algebraic Polynomials in Cost Accountancy<div>MP3 players example</div>
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<div>The profit in millions of dollars for an MP3 player can be done with the polynomial P=-4x^3 + 12X^2 + 16X. X represents the number of MP3 players produced annually. The company currently produces 3 million MP3 players and makes a profit of $ 48,000,000. To figure out the least amount of MP3 players the company can produce and still make the same profit we need to solve for P.</div>
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<div>Step 1</div>
<div>Set P=48 This represents the total profit. 48=-4X^3 + 12X^2 + 16X</div>
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<div>Step 2 subtract the 48 from 48 and from the end of the equation, like this 48=-4X^3 + 12X^2 + 16X - 48</div>
<div> -48</div>
<div>Step 3 Than you factor by using the Greatest Common Factor as show below.</div>
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<div>-4X^2(X-3) + 16 (X-3) Notice how it is now a binomial, which means two terms. It was previously a Trinomial.</div>
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<div>Step 5 Multiply the variables in each Binomial separately and set them equal to 0.If necessary, you might need to divide like the Binomial 2</div>
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<div>So -4X^2(X-3) turns to 4X^2+16=0 and 16(X-3) 16X/-48 than X-3=0</div>
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<div>Step 6</div>
<div>Subtract 16 from Equation 1 and add 3 to Equation 2 to get the results shown below</div>
<div>4X^2=-16 and X=3</div>
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<div>Step 7 Reduce the terms in equation 1 </div>
<div>16 can be reduced to 4, because 4*4=16, while 4x^2 can be reduced to X^2.</div>
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<div>Step 8 solve 4=X^2. The answer is 2 or 2 million MP3s can be manufactured make the same profit of 4 3 million.</div>
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<div>Step 9 Double check</div>
<div>48=-4(2)^2+12(2)^2+12(2)</div>
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<div>The company can produce 2 Million MP3s and still make the same profit as producing 3 million MP3s. The reason for this is that perhaps variable costs doesn't change if they produce 1 million less MP3s. It could be that their production facilities got more efficient at producing the MP3s. It now takes less time and money to make the MP3s, so they company can afford to produce 2 million units and still earn the same profit as making 3 million units.</div>
<div> </div>Sat, 08 Mar 2014 18:49:54 -06002014-03-08T18:49:54-06:00