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Question-182459

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Question Number 182459 by HeferH last updated on 09/Dec/22
Commented by HeferH last updated on 09/Dec/22
how is this calculator gettin these complex values ?
$${how}\:{is}\:{this}\:{calculator}\:{gettin}\:{these}\:{complex} \\ $$$$\:{values}\:? \\ $$
Commented by MJS_new last updated on 09/Dec/22
It′s wrong. It′s not calculating the root, it′s instead solving x^(16) =2^(15) . You should get a better calculator.
$$\mathrm{It}'\mathrm{s}\:\mathrm{wrong}.\:\mathrm{It}'\mathrm{s}\:{not}\:\mathrm{calculating}\:\mathrm{the}\:\mathrm{root},\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{instead}\:\mathrm{solving}\:{x}^{\mathrm{16}} =\mathrm{2}^{\mathrm{15}} .\:\mathrm{You}\:\mathrm{should}\:\mathrm{get}\:\mathrm{a} \\ $$$$\mathrm{better}\:\mathrm{calculator}. \\ $$
Commented by Acem last updated on 10/Dec/22
What should be the answer?
$${What}\:{should}\:{be}\:{the}\:{answer}? \\ $$
Commented by MJS_new last updated on 10/Dec/22
1.915... (2^(15) )^(1/(16)) =((32768))^(1/(16)) ∈R^+
$$\mathrm{1}.\mathrm{915}… \\ $$$$\sqrt[{\mathrm{16}}]{\mathrm{2}^{\mathrm{15}} }=\sqrt[{\mathrm{16}}]{\mathrm{32768}}\in\mathbb{R}^{+} \\ $$
Commented by Acem last updated on 10/Dec/22
Yes Sir , but^(16) (√(32 768)) = 1.915... so what′s the issue?
$${Yes}\:{Sir}\:,\:{but}\:^{\mathrm{16}} \sqrt{\mathrm{32}\:\mathrm{768}}\:=\:\mathrm{1}.\mathrm{915}… \\ $$$$\:{so}\:{what}'{s}\:{the}\:{issue}? \\ $$
Commented by Acem last updated on 10/Dec/22
Commented by Acem last updated on 10/Dec/22
what′s the problem with this result?
$$\:{what}'{s}\:{the}\:{problem}\:{with}\:{this}\:{result}? \\ $$
Commented by MJS_new last updated on 10/Dec/22
no problem with this result, but the calculator shows a complex solution too which is wrong
$$\mathrm{no}\:\mathrm{problem}\:\mathrm{with}\:{this}\:\mathrm{result},\:\mathrm{but}\:\mathrm{the}\:\mathrm{calculator} \\ $$$$\mathrm{shows}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{solution}\:\mathrm{too}\:\mathrm{which}\:\mathrm{is}\:\mathrm{wrong} \\ $$
Commented by mr W last updated on 10/Dec/22
this is not a problem of the app. it displays all roots, because in this case the user has chosen to let it display all roots in complex mode. if this information confuses you, you should not let the app to display it. in settings you only need to select “Only Principal Root” instead of “All Roots”.
$${this}\:{is}\:{not}\:{a}\:{problem}\:{of}\:{the}\:{app}. \\ $$$${it}\:{displays}\:{all}\:{roots},\:{because}\:{in}\:{this} \\ $$$${case}\:{the}\:{user}\:{has}\:{chosen}\:{to}\:{let}\:{it} \\ $$$${display}\:{all}\:{roots}\:{in}\:{complex}\:{mode}. \\ $$$${if}\:{this}\:{information}\:{confuses}\:{you}, \\ $$$${you}\:{should}\:{not}\:{let}\:{the}\:{app}\:{to}\:{display} \\ $$$${it}.\:{in}\:{settings}\:{you}\:{only}\:{need}\:{to}\:{select} \\ $$$$“{Only}\:{Principal}\:{Root}''\:{instead}\:{of} \\ $$$$“{All}\:{Roots}''. \\ $$
Commented by mr W last updated on 10/Dec/22
Commented by mr W last updated on 10/Dec/22
you get with “Only Principal Root”:
$${you}\:{get}\:{with}\:“{Only}\:{Principal}\:{Root}'': \\ $$
Commented by mr W last updated on 10/Dec/22
Commented by mr W last updated on 10/Dec/22
you get with “All Roots”:
$${you}\:{get}\:{with}\:“{All}\:{Roots}'': \\ $$
Commented by mr W last updated on 10/Dec/22
Commented by mr W last updated on 10/Dec/22
by default the app doesn′t use complex mode and doesn′t display all roots. so i don′t understand those people who explicitly want the app use complex mode and display all roots, and then are confused when the app indeed does this.
$${by}\:{default}\:{the}\:{app}\:{doesn}'{t}\:{use}\: \\ $$$${complex}\:{mode}\:{and}\:{doesn}'{t}\:{display} \\ $$$${all}\:{roots}.\:{so}\:{i}\:{don}'{t}\:{understand} \\ $$$${those}\:{people}\:{who}\:{explicitly}\:{want}\:{the}\: \\ $$$${app}\:{use}\:{complex}\:{mode}\:{and}\:{display}\: \\ $$$${all}\:{roots},\:{and}\:{then}\:{are}\:{confused}\:{when} \\ $$$${the}\:{app}\:{indeed}\:{does}\:{this}. \\ $$
Commented by MJS_new last updated on 10/Dec/22
I hereby give up. people started to mix up x=(√4) and x^2 =4 and now the app makers also do this. use it as you like, but it′s against the definitions. if (2^(15) )^(1/(16)) ∉R then you also have to accept (√4)=−2 with all kinds of weird logical problems. i.e. ∣z∣≥0∀z∈C and ∣a+bi∣=(√(a^2 +b^2 )) but what if z=((16))^(1/4) +((81))^(1/4) i? how many values is z at once or which value is the “main value”? and (√((((16))^(1/4) )^2 +(((81))^(1/4) )^2 )) can not only be <0 but also be a complex number??? or simply: (√4)+(√4)=0 proof: (√4)=−2∨(√4)=2 ⇒ (√4)+(√4)=−2+2=0 I′ll keep following the rules I learned 35 years ago, everybody else is free to use whatever they want.
$$\mathrm{I}\:\mathrm{hereby}\:\mathrm{give}\:\mathrm{up}. \\ $$$$\mathrm{people}\:\mathrm{started}\:\mathrm{to}\:\mathrm{mix}\:\mathrm{up}\:{x}=\sqrt{\mathrm{4}}\:\mathrm{and}\:{x}^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{and}\:\mathrm{now}\:\mathrm{the}\:\mathrm{app}\:\mathrm{makers}\:\mathrm{also}\:\mathrm{do}\:\mathrm{this}.\:\mathrm{use}\:\mathrm{it} \\ $$$$\mathrm{as}\:\mathrm{you}\:\mathrm{like},\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{against}\:\mathrm{the}\:\mathrm{definitions}. \\ $$$$\mathrm{if}\:\sqrt[{\mathrm{16}}]{\mathrm{2}^{\mathrm{15}} }\:\notin\mathbb{R}\:\mathrm{then}\:\mathrm{you}\:\mathrm{also}\:\mathrm{have}\:\mathrm{to}\:\mathrm{accept} \\ $$$$\sqrt{\mathrm{4}}=−\mathrm{2}\:\mathrm{with}\:\mathrm{all}\:\mathrm{kinds}\:\mathrm{of}\:\mathrm{weird}\:\mathrm{logical} \\ $$$$\mathrm{problems}. \\ $$$$\mathrm{i}.\mathrm{e}.\:\mid{z}\mid\geqslant\mathrm{0}\forall{z}\in\mathbb{C}\:\mathrm{and}\:\mid{a}+{b}\mathrm{i}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\mathrm{but}\:\mathrm{what}\:\mathrm{if}\:{z}=\sqrt[{\mathrm{4}}]{\mathrm{16}}+\sqrt[{\mathrm{4}}]{\mathrm{81}}\mathrm{i}?\:\mathrm{how}\:\mathrm{many}\:\mathrm{values} \\ $$$$\mathrm{is}\:{z}\:\mathrm{at}\:\mathrm{once}\:\mathrm{or}\:\mathrm{which}\:\mathrm{value}\:\mathrm{is}\:\mathrm{the}\:“\mathrm{main}\:\mathrm{value}''? \\ $$$$\mathrm{and}\:\sqrt{\left(\sqrt[{\mathrm{4}}]{\mathrm{16}}\right)^{\mathrm{2}} +\left(\sqrt[{\mathrm{4}}]{\mathrm{81}}\right)^{\mathrm{2}} }\:\mathrm{can}\:\mathrm{not}\:\mathrm{only}\:\mathrm{be}\:<\mathrm{0}\:\mathrm{but} \\ $$$$\mathrm{also}\:\mathrm{be}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}??? \\ $$$$\mathrm{or}\:\mathrm{simply}: \\ $$$$\sqrt{\mathrm{4}}+\sqrt{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{proof}: \\ $$$$\sqrt{\mathrm{4}}=−\mathrm{2}\vee\sqrt{\mathrm{4}}=\mathrm{2} \\ $$$$\Rightarrow\:\sqrt{\mathrm{4}}+\sqrt{\mathrm{4}}=−\mathrm{2}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{keep}\:\mathrm{following}\:\mathrm{the}\:\mathrm{rules}\:\mathrm{I}\:\mathrm{learned}\:\mathrm{35}\:\mathrm{years} \\ $$$$\mathrm{ago},\:\mathrm{everybody}\:\mathrm{else}\:\mathrm{is}\:\mathrm{free}\:\mathrm{to}\:\mathrm{use}\:\mathrm{whatever} \\ $$$$\mathrm{they}\:\mathrm{want}. \\ $$
Commented by MJS_new last updated on 10/Dec/22
if these “new rules” are used ((x+3))^(1/4) −(√(x+4))=3 has 2 solutions: x_1 ≈−.816094359 x_2 ≈23.8367388 someone prove the opposite...
$$\mathrm{if}\:\mathrm{these}\:“\mathrm{new}\:\mathrm{rules}''\:\mathrm{are}\:\mathrm{used} \\ $$$$\sqrt[{\mathrm{4}}]{{x}+\mathrm{3}}−\sqrt{{x}+\mathrm{4}}=\mathrm{3} \\ $$$$\mathrm{has}\:\mathrm{2}\:\mathrm{solutions}: \\ $$$${x}_{\mathrm{1}} \approx−.\mathrm{816094359} \\ $$$${x}_{\mathrm{2}} \approx\mathrm{23}.\mathrm{8367388} \\ $$$$\mathrm{someone}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{opposite}… \\ $$
Commented by mr W last updated on 10/Dec/22
i totally agree with sir! basically it is clear and 100% correct what you said. but as about the app, i don′t think that the app makes an error. it states clearly that it displays all roots of a number in complex plane. in complex mode this app means with (3)^(1/3) simply the roots of the number 3 in complex plane.
$${i}\:{totally}\:{agree}\:{with}\:{sir}!\:{basically}\:{it} \\ $$$${is}\:{clear}\:{and}\:\mathrm{100\%}\:{correct}\:{what}\:{you}\: \\ $$$${said}. \\ $$$${but}\:{as}\:{about}\:{the}\:{app},\:{i}\:{don}'{t}\:{think}\:{that} \\ $$$${the}\:{app}\:{makes}\:{an}\:{error}.\:{it}\:{states} \\ $$$${clearly}\:{that}\:{it}\:{displays}\:{all}\:{roots}\:{of} \\ $$$${a}\:{number}\:{in}\:{complex}\:{plane}. \\ $$$${in}\:{complex}\:{mode}\:{this}\:{app}\:{means}\: \\ $$$${with}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:{simply}\:{the}\:{roots}\:{of}\:{the}\:{number} \\ $$$$\mathrm{3}\:{in}\:{complex}\:{plane}. \\ $$
Commented by Acem last updated on 10/Dec/22
Yes I agree with you Sir MJS
$${Yes}\:{I}\:{agree}\:{with}\:{you}\:{Sir}\:{MJS} \\ $$
Commented by Acem last updated on 10/Dec/22
The wrong is not with the application but with how it is used. The option CPLX must chosen just for R^( −)
$${The}\:{wrong}\:{is}\:{not}\:{with}\:{the}\:{application}\:{but}\:{with} \\ $$$$\:{how}\:{it}\:{is}\:{used}.\:{The}\:{option}\:{CPLX}\:{must}\:{chosen}\:{just} \\ $$$$\:{for}\:\mathbb{R}^{\:−} \: \\ $$
Commented by Acem last updated on 10/Dec/22

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