A Tough Calculus Question - Data Science Central2019-07-16T16:54:28Zhttps://www.datasciencecentral.com/forum/topics/a-tough-calculus-question?feed=yes&xn_auth=no2i*3i = 6i^2 = -6tag:www.datasciencecentral.com,2017-11-18:6448529:Comment:6508732017-11-18T17:45:49.583ZSteven Craigheadhttps://www.datasciencecentral.com/profile/StevenCraighead
2i*3i = 6i^2 = -6
2i*3i = 6i^2 = -6 MATLAB's sqrt returns only th…tag:www.datasciencecentral.com,2016-11-06:6448529:Comment:4854672016-11-06T15:15:42.838ZMark L. Stonehttps://www.datasciencecentral.com/profile/MarkLStone
<p>MATLAB's sqrt returns only the principal value, which under MATLAB's convention is defined to be</p>
<p>"For negative and complex numbers <code>z = u + i*w</code>, the complex square root <code>sqrt(z)</code> returns</p>
<p><code>sqrt(r)*(cos(phi/2) + 1i*sin(phi/2))</code></p>
<p>where <code>r = abs(z)</code> is the radius and <code>phi = angle(z)</code> is the phase angle on the closed interval <code>-pi <= phi <= pi</code>."</p>
<p><br></br> <cite>Michael Brock…</cite></p>
<p>MATLAB's sqrt returns only the principal value, which under MATLAB's convention is defined to be</p>
<p>"For negative and complex numbers <code>z = u + i*w</code>, the complex square root <code>sqrt(z)</code> returns</p>
<p><code>sqrt(r)*(cos(phi/2) + 1i*sin(phi/2))</code></p>
<p>where <code>r = abs(z)</code> is the radius and <code>phi = angle(z)</code> is the phase angle on the closed interval <code>-pi <= phi <= pi</code>."</p>
<p><br/> <cite>Michael Brock said:</cite></p>
<blockquote cite="http://www.datasciencecentral.com/forum/topics/a-tough-calculus-question?id=6448529%3ATopic%3A386002&page=2#6448529Comment392254"><div><div class="xg_user_generated"><p>According to Matlab, the answer is -6. But you probably already know that, so I'm wondering if you might be posing a more general question. I work with complex numbers (digital filter design).</p>
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</blockquote> Amen, bro. Maiia Bakhova sai…tag:www.datasciencecentral.com,2016-11-06:6448529:Comment:4854642016-11-06T15:09:29.238ZMark L. Stonehttps://www.datasciencecentral.com/profile/MarkLStone
<p>Amen, bro.<br></br> <br></br> <cite>Maiia Bakhova said:</cite></p>
<blockquote cite="http://www.datasciencecentral.com/forum/topics/a-tough-calculus-question?utm_content=buffer28857&utm_medium=social&utm_source=linkedin.com&utm_campaign=buffer#6448529Comment386648"><div><div class="xg_user_generated"><p>Hello, </p>
<p>If you consider complex numbers then both <span>SQRT(-4) and SQRT(-9) are not single-valued unless it is specified that we're to use only their primary…</span></p>
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<p>Amen, bro.<br/> <br/> <cite>Maiia Bakhova said:</cite></p>
<blockquote cite="http://www.datasciencecentral.com/forum/topics/a-tough-calculus-question?utm_content=buffer28857&utm_medium=social&utm_source=linkedin.com&utm_campaign=buffer#6448529Comment386648"><div><div class="xg_user_generated"><p>Hello, </p>
<p>If you consider complex numbers then both <span>SQRT(-4) and SQRT(-9) are not single-valued unless it is specified that we're to use only their primary roots. <br/></span></p>
<p><span>SQRT(-4) can be 2i or -2i, and SQRT(-9) can be 3i or -3i. Only the first values in each pair are primary roots.</span></p>
<p><span>Hence their product is not single valued as well. It can be 6 or -6.</span></p>
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<p><span>Look at the reference at Brilliant: it is implied that the answer is only one. Therefore they consider primary roots only, and it must be 6.</span></p>
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</blockquote> A lot of people are just sayi…tag:www.datasciencecentral.com,2016-11-03:6448529:Comment:4844232016-11-03T05:50:15.666ZMichael Turchinhttps://www.datasciencecentral.com/profile/MichaelTurchin
A lot of people are just saying +6 - shouldn't it be +6 or -6?
A lot of people are just saying +6 - shouldn't it be +6 or -6? This calcul is not posed as i…tag:www.datasciencecentral.com,2016-06-27:6448529:Comment:4403692016-06-27T09:10:37.235ZGeorges Bressangehttps://www.datasciencecentral.com/profile/Bressange
This calcul is not posed as it should: the notation with radical SQRT is reserved for real numbers where the square root as an unique value for positive numbers because in real numbers, every positive number has an unique square root. But in complex set C, every non zero number has two "square roots": the square roots of -9 are 3i and -3i and those of -4 are 2i and -2i so the product can be either 6 or -6. The problem should have been written: "in C, if we multiply a square root of -9 by a…
This calcul is not posed as it should: the notation with radical SQRT is reserved for real numbers where the square root as an unique value for positive numbers because in real numbers, every positive number has an unique square root. But in complex set C, every non zero number has two "square roots": the square roots of -9 are 3i and -3i and those of -4 are 2i and -2i so the product can be either 6 or -6. The problem should have been written: "in C, if we multiply a square root of -9 by a square root of -4, what could be the result ?" For the question about the de…tag:www.datasciencecentral.com,2016-02-27:6448529:Comment:3924862016-02-27T06:16:44.868ZMichael Brockhttps://www.datasciencecentral.com/profile/MichaelBrock
<p>For the question about the decimals of pi. Am assuming that you mean pi expressed in base 2 digits, composed of the numbers zero and one only, and the numbers going to infinity. I recall hearing from a colleague that the digit sequence can be predicted for the base 2 case, and it is infinite, so evidently, the numerals zero and one appear infinitely many times. </p>
<p>For the question about the decimals of pi. Am assuming that you mean pi expressed in base 2 digits, composed of the numbers zero and one only, and the numbers going to infinity. I recall hearing from a colleague that the digit sequence can be predicted for the base 2 case, and it is infinite, so evidently, the numerals zero and one appear infinitely many times. </p> According to Matlab, the answ…tag:www.datasciencecentral.com,2016-02-27:6448529:Comment:3922542016-02-27T05:55:56.493ZMichael Brockhttps://www.datasciencecentral.com/profile/MichaelBrock
<p>According to Matlab, the answer is -6. But you probably already know that, so I'm wondering if you might be posing a more general question. I work with complex numbers (digital filter design).</p>
<p>According to Matlab, the answer is -6. But you probably already know that, so I'm wondering if you might be posing a more general question. I work with complex numbers (digital filter design).</p> No limits = not calculus / an…tag:www.datasciencecentral.com,2016-02-16:6448529:Comment:3879792016-02-16T10:23:20.582ZJoe Hillinghttps://www.datasciencecentral.com/profile/JoeHilling
<p>No limits = not calculus / analytic. Square the expression and take the +ve and -ve branches.</p>
<p>No limits = not calculus / analytic. Square the expression and take the +ve and -ve branches.</p> Hello,
If you consider compl…tag:www.datasciencecentral.com,2016-02-12:6448529:Comment:3866482016-02-12T00:29:24.177ZMaiia Bakhovahttps://www.datasciencecentral.com/profile/MaiiaBakhova
<p>Hello, </p>
<p>If you consider complex numbers then both <span>SQRT(-4) and SQRT(-9) are not single-valued unless it is specified that we're to use only their primary roots. <br></br></span></p>
<p><span>SQRT(-4) can be 2i or -2i, and SQRT(-9) can be 3i or -3i. Only the first values in each pair are primary roots.</span></p>
<p><span>Hence their product is not single valued as well. It can be 6 or -6.</span></p>
<p></p>
<p><span>Look at the reference at Brilliant: it is implied that the answer…</span></p>
<p>Hello, </p>
<p>If you consider complex numbers then both <span>SQRT(-4) and SQRT(-9) are not single-valued unless it is specified that we're to use only their primary roots. <br/></span></p>
<p><span>SQRT(-4) can be 2i or -2i, and SQRT(-9) can be 3i or -3i. Only the first values in each pair are primary roots.</span></p>
<p><span>Hence their product is not single valued as well. It can be 6 or -6.</span></p>
<p></p>
<p><span>Look at the reference at Brilliant: it is implied that the answer is only one. Therefore they consider primary roots only, and it must be 6.</span></p>
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<p></p> What he said.. Stephen MacMi…tag:www.datasciencecentral.com,2016-02-11:6448529:Comment:3864432016-02-11T21:57:32.778ZMichael S. Pukishhttps://www.datasciencecentral.com/profile/MichaelSPukish
<p>What he said..<br></br> <br></br> <cite>Stephen MacMinn said:</cite></p>
<blockquote cite="http://www.datasciencecentral.com/forum/topics/a-tough-calculus-question?commentId=6448529%3AComment%3A386627&xg_source=msg_com_forum#6448529Comment386627"><div><div class="xg_user_generated"><p>Hi Marc,</p>
<p>Unfortunately, the square root function is not necessarily distributive for negative reals. Richard Fowler's answer above is the correct one.<br></br> <br></br> <cite>Marc Borowczak…</cite></p>
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<p>What he said..<br/> <br/> <cite>Stephen MacMinn said:</cite></p>
<blockquote cite="http://www.datasciencecentral.com/forum/topics/a-tough-calculus-question?commentId=6448529%3AComment%3A386627&xg_source=msg_com_forum#6448529Comment386627"><div><div class="xg_user_generated"><p>Hi Marc,</p>
<p>Unfortunately, the square root function is not necessarily distributive for negative reals. Richard Fowler's answer above is the correct one.<br/> <br/> <cite>Marc Borowczak said:</cite></p>
<blockquote cite="http://www.datasciencecentral.com/forum/topics/a-tough-calculus-question#6448529Comment386425"><div><div class="xg_user_generated"><p>In fact, this is not a calculus, but an algebra problem.<br/>It is correct to transform the question as:<br/>X = SQRT(-4) * SQRT(-9) or, since SQRT(a) * SQRT(b) = SQRT(a*b), after rearranging the equation:</p>
<p>X - SQRT(36) = 0, which we can transform by squaring both sides as:<br/>X^2 - 36 = 0, and by factorization, since a^2 - b^2 = (a+b) * (a-b),<br/>(X-6) * (X+6) = 0.<br/>This clearly shows the 2 roots, i.e:<br/>X= - 6. and X= +6 both verify the relationship.</p>
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<p>Hope that helps...</p>
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