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Question Number 25822    Answers: 0   Comments: 0

answer to the question of p(X)= (1+iX)^n −(1−iX)^n key of solutionafter resolving p(X)=0 the roots of p(X) are x_k =tan(kπ/n) with k in [[0.n−1]] so p(X)= ∝Π_(k=0) ^(k=n−1) (X−x_k^ ) let searsh ∝ by using binome formula p(X)= 2iΣ_(p=0) ^ (−1)^(p ) C_n ^(2p+1) X^(2p+1) so ∝= 2i(−1)^([n−1/2]) C_n ^ case1 n=2N p(X)=∝ Π_(k=0) ^(k=2N−1) (X−tan(kπ/2N)) and ∝=4in(−1)^(N−1) case2 n=2N+1 p(X)=∝Π_(k=0) ^(k=2N) (X−tan(kπ/2N+1) and ∝=2i(−1)^N

$${answer}\:{to}\:{the}\:{question}\:{of}\:{p}\left({X}\right)=\:\left(\mathrm{1}+{iX}\right)^{{n}} −\left(\mathrm{1}−{iX}\right)^{{n}} \\ $$$${key}\:{of}\:{solutionafter}\:{resolving}\:{p}\left({X}\right)=\mathrm{0}\:\:{the}\:{roots}\:{of}\:{p}\left({X}\right) \\ $$$${are}\:\:{x}_{{k}} ={tan}\left({k}\pi/{n}\right)\:{with}\:{k}\:\:{in}\:\left[\left[\mathrm{0}.{n}−\mathrm{1}\right]\right]\:{so} \\ $$$${p}\left({X}\right)=\:\propto\prod_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \left({X}−{x}_{{k}^{} } \right)\:{let}\:{searsh}\:\propto\:{by}\:{using}\:{binome}\:{formula} \\ $$$${p}\left({X}\right)=\:\mathrm{2}{i}\sum_{{p}=\mathrm{0}} ^{} \left(−\mathrm{1}\right)^{{p}\:} {C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} {X}^{\mathrm{2}{p}+\mathrm{1}} {so} \\ $$$$\propto=\:\mathrm{2}{i}\left(−\mathrm{1}\right)^{\left[{n}−\mathrm{1}/\mathrm{2}\right]} \:{C}_{{n}} ^{} \\ $$$${case}\mathrm{1}\:{n}=\mathrm{2}{N}\:\:\:{p}\left({X}\right)=\propto\:\prod_{{k}=\mathrm{0}} ^{{k}=\mathrm{2}{N}−\mathrm{1}} \left({X}−{tan}\left({k}\pi/\mathrm{2}{N}\right)\right) \\ $$$${and}\:\propto=\mathrm{4}{in}\left(−\mathrm{1}\right)^{{N}−\mathrm{1}} \\ $$$${case}\mathrm{2}\:\:{n}=\mathrm{2}{N}+\mathrm{1}\:\:\:\:{p}\left({X}\right)=\propto\prod_{{k}=\mathrm{0}} ^{{k}=\mathrm{2}{N}} \left({X}−{tan}\left({k}\pi/\mathrm{2}{N}+\mathrm{1}\right)\right. \\ $$$${and}\:\propto=\mathrm{2}{i}\left(−\mathrm{1}\right)^{{N}} \\ $$

Question Number 25787    Answers: 0   Comments: 2

2^x =x^(2 ) ,hence x=2 . prove

$$\mathrm{2}^{{x}} ={x}^{\mathrm{2}\:} ,{hence}\:{x}=\mathrm{2}\:.\:{prove} \\ $$

Question Number 25709    Answers: 0   Comments: 1

Q. 25444 (another solution) z^ +1 =iz^2 +∣z∣^2 let z=x+iy , ⇒ x−iy+1=i(x^2 −y^2 )−2xy+x^2 +y^2 or (x−y)^2 =x+1 ,and (y^2 −x^2 )=y let y−x=u and y+x=v ⇒ 2u^2 =v−u+2 ...(i) 2uv=u+v ....(ii) ⇒ v=(u/(2u−1)) , substituting in (i) 2u^2 +u=(u/(2u−1))+2 u=1 satisfies this , so ⇒ (2u+3)(u−1)=(u/(2u−1))−1 ⇒ (2u+3)(u−1)=(((1−u))/(2u−1)) ⇒ u=1 or (2u+3)(2u−1)+1=0 or 4u^2 +4u−2=0 or 2u^2 +2u−1=0 ⇒ u=((−2±(√(4+8)))/4) =((−1±(√3))/2) for u_1 =1 , v_1 =(u_1 /(2u_1 −1))=1 so x_1 =((v_1 −u_1 )/2)=0 and y_1 =((v_1 +u_1 )/2) =1 hence z_1 =i for u_2 =(((√3)−1)/2) , v_2 =(((√3)−1)/(2((√3)−1−1))) =−((((√3)−1)(2+(√3)))/2) =−((((√3)+1))/2) x_2 =((v_2 −u_2 )/2)=−((√3)/2) y_2 =((v_2 +u_2 )/2) =−(1/2) hence z_2 =−(1/2)((√3)+i) for u_3 =−((((√3)+1))/2) v_3 =(u_3 /(2u_3 −1)) =((((√3)+1))/(2((√3)+2))) =((((√3)+1)(2−(√3)))/2) =(((√3)−1)/2) x_3 =((v_3 −u_3 )/2)=((√3)/2) and y_3 =((v_3 +u_3 )/2)=−(1/2) ⇒ z_3 =(1/2)((√3)−i) So to summarize, z_1 =i , z_2 =−(1/2)((√3)+i), and z_3 =(1/2)((√3)−i) .

$${Q}.\:\mathrm{25444}\:\:\left({another}\:{solution}\right) \\ $$$$\:\:\bar {{z}}+\mathrm{1}\:={iz}^{\mathrm{2}} +\mid{z}\mid^{\mathrm{2}} \\ $$$${let}\:\:{z}={x}+{iy}\:,\:\Rightarrow \\ $$$${x}−{iy}+\mathrm{1}={i}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)−\mathrm{2}{xy}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$${or}\:\:\left({x}−{y}\right)^{\mathrm{2}} ={x}+\mathrm{1}\:,{and}\: \\ $$$$\:\:\:\:\:\:\:\left({y}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)={y} \\ $$$${let}\:\:\:{y}−{x}={u}\:\:{and}\:\:{y}+{x}={v} \\ $$$$\Rightarrow\:\:\mathrm{2}{u}^{\mathrm{2}} ={v}−{u}+\mathrm{2}\:\:\:...\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{uv}={u}+{v}\:\:\:\:\:\:\:\:....\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\:{v}=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}\:\:,\:{substituting}\:{in}\:\left({i}\right) \\ $$$$\:\:\:\:\:\mathrm{2}{u}^{\mathrm{2}} +{u}=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}+\mathrm{2} \\ $$$${u}=\mathrm{1}\:{satisfies}\:{this}\:,\:{so} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left({u}−\mathrm{1}\right)=\frac{{u}}{\mathrm{2}{u}−\mathrm{1}}−\mathrm{1} \\ $$$$\Rightarrow\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left({u}−\mathrm{1}\right)=\frac{\left(\mathrm{1}−{u}\right)}{\mathrm{2}{u}−\mathrm{1}} \\ $$$$\Rightarrow\:\:{u}=\mathrm{1}\:{or}\:\:\left(\mathrm{2}{u}+\mathrm{3}\right)\left(\mathrm{2}{u}−\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\mathrm{4}{u}^{\mathrm{2}} +\mathrm{4}{u}−\mathrm{2}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\mathrm{2}{u}^{\mathrm{2}} +\mathrm{2}{u}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\:{u}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{8}}}{\mathrm{4}}\:=\frac{−\mathrm{1}\pm\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${for}\:{u}_{\mathrm{1}} =\mathrm{1}\:\:,\:{v}_{\mathrm{1}} =\frac{{u}_{\mathrm{1}} }{\mathrm{2}{u}_{\mathrm{1}} −\mathrm{1}}=\mathrm{1} \\ $$$${so}\:\:{x}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} −{u}_{\mathrm{1}} }{\mathrm{2}}=\mathrm{0}\:\:\:{and}\: \\ $$$$\:\:\:\:\:{y}_{\mathrm{1}} =\frac{{v}_{\mathrm{1}} +{u}_{\mathrm{1}} }{\mathrm{2}}\:=\mathrm{1} \\ $$$${hence}\:\:{z}_{\mathrm{1}} ={i} \\ $$$${for}\:{u}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}\:,\:\:{v}_{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}−\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=−\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:{x}_{\mathrm{2}} =\frac{{v}_{\mathrm{2}} −{u}_{\mathrm{2}} }{\mathrm{2}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:{y}_{\mathrm{2}} =\frac{{v}_{\mathrm{2}} +{u}_{\mathrm{2}} }{\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${hence}\:\:\:{z}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}+{i}\right) \\ $$$${for}\:\:{u}_{\mathrm{3}} =−\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:{v}_{\mathrm{3}} =\frac{{u}_{\mathrm{3}} }{\mathrm{2}{u}_{\mathrm{3}} −\mathrm{1}}\:=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}} \\ $$$${x}_{\mathrm{3}} =\frac{{v}_{\mathrm{3}} −{u}_{\mathrm{3}} }{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:{and} \\ $$$$\:{y}_{\mathrm{3}} =\frac{{v}_{\mathrm{3}} +{u}_{\mathrm{3}} }{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{z}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−{i}\right) \\ $$$${So}\:\:{to}\:{summarize}, \\ $$$$\boldsymbol{{z}}_{\mathrm{1}} =\boldsymbol{{i}}\:\:,\:\:\boldsymbol{{z}}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}+\boldsymbol{{i}}\right),\:{and} \\ $$$$\:\boldsymbol{{z}}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−\boldsymbol{{i}}\right)\:. \\ $$

Question Number 25680    Answers: 0   Comments: 0

let p(X) = (1 +iX )^(n) − ( 1 − iX )^(n) if p(X ) =∝Σ ( X −xk) find xk and α

$${let}\:{p}\left({X}\right)\:=\:\left(\mathrm{1}\:+{iX}\:\overset{{n}} {\right)}\:−\:\left(\:\mathrm{1}\:−\:{iX}\:\overset{{n}} {\right)}\:\:{if}\:\:{p}\left({X}\:\right)\:=\propto\Sigma\:\left(\:{X}\:−{xk}\right)\:\:{find}\:{xk}\:{and}\:\alpha \\ $$

Question Number 25638    Answers: 0   Comments: 1

Question Number 25619    Answers: 1   Comments: 0

Question Number 25589    Answers: 0   Comments: 3

possible or not ? a^a +b^b > a^b +b^a with below conditions: case 1) a>b>1 case 2) 0<b<a<1

$$\boldsymbol{{possible}}\:\boldsymbol{{or}}\:\boldsymbol{{not}}\:? \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{a}}^{\boldsymbol{{a}}} +\boldsymbol{{b}}^{\boldsymbol{{b}}} >\:\boldsymbol{{a}}^{\boldsymbol{{b}}} +\boldsymbol{{b}}^{\boldsymbol{{a}}} \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{below}}\:\boldsymbol{{conditions}}: \\ $$$$\left.\:\boldsymbol{{case}}\:\mathrm{1}\right)\:\:\boldsymbol{{a}}>\boldsymbol{{b}}>\mathrm{1} \\ $$$$\left.\boldsymbol{{case}}\:\mathrm{2}\right)\:\:\:\:\mathrm{0}<\boldsymbol{{b}}<\boldsymbol{{a}}<\mathrm{1} \\ $$

Question Number 25536    Answers: 1   Comments: 0

If x = 2sec(3t) and y = 4tan(3t), Show that y = 4(x^2 − 4)

$$\mathrm{If}\:\:\:\mathrm{x}\:=\:\mathrm{2sec}\left(\mathrm{3t}\right)\:\:\mathrm{and}\:\:\mathrm{y}\:=\:\mathrm{4tan}\left(\mathrm{3t}\right),\:\:\:\mathrm{Show}\:\mathrm{that}\:\:\:\:\mathrm{y}\:=\:\mathrm{4}\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4}\right) \\ $$

Question Number 25533    Answers: 1   Comments: 0

Three sets of beans l, m, and n are sold for5 for $5.00, $15.00 and $17.50 respectively per kg. If they are mixed in the ratio of 1:3:2, what is the cost per kg of the mixture?×

$${Three}\:{sets}\:{of}\:{beans}\:{l},\:{m},\:{and}\:{n}\:{are}\:{sold}\:{for}\mathrm{5} \\ $$$${for}\:\$\mathrm{5}.\mathrm{00},\:\$\mathrm{15}.\mathrm{00}\:{and}\:\$\mathrm{17}.\mathrm{50}\:{respectively} \\ $$$${per}\:{kg}.\:{If}\:{they}\:{are}\:{mixed}\:{in}\:{the}\:{ratio}\:{of} \\ $$$$\mathrm{1}:\mathrm{3}:\mathrm{2},\:{what}\:{is}\:{the}\:{cost}\:{per}\:{kg}\:{of}\:{the}\:{mixture}?× \\ $$

Question Number 25482    Answers: 1   Comments: 3

Question Number 25457    Answers: 1   Comments: 0

Question Number 25444    Answers: 1   Comments: 8

Question Number 25425    Answers: 0   Comments: 0

Sum of series 1 + 2x + 7x^2 + 20x^3 + ... up to n terms when x = −1 is

$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{series}\:\mathrm{1}\:+\:\mathrm{2}{x}\:+\:\mathrm{7}{x}^{\mathrm{2}} \:+\:\mathrm{20}{x}^{\mathrm{3}} \:+\:... \\ $$$$\mathrm{up}\:\mathrm{to}\:{n}\:\mathrm{terms}\:\mathrm{when}\:{x}\:=\:−\mathrm{1}\:\mathrm{is} \\ $$

Question Number 25462    Answers: 1   Comments: 0

Let S_n , n = 1, 2, 3... be the sum of infinite geometric series whose first term is n and the common ratio is (1/(n + 1)). Then lim_(n→∞) ((S_1 S_n + S_2 S_(n−1) + S_3 S_(n−2) ... + S_n S_1 )/(S_1 ^2 + S_2 ^2 + ... + S_n ^2 )) is

$$\mathrm{Let}\:{S}_{{n}} ,\:{n}\:=\:\mathrm{1},\:\mathrm{2},\:\mathrm{3}...\:\mathrm{be}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{infinite}\:\mathrm{geometric}\:\mathrm{series}\:\mathrm{whose}\:\mathrm{first} \\ $$$$\mathrm{term}\:\mathrm{is}\:{n}\:\mathrm{and}\:\mathrm{the}\:\mathrm{common}\:\mathrm{ratio}\:\mathrm{is} \\ $$$$\frac{\mathrm{1}}{{n}\:+\:\mathrm{1}}.\:\mathrm{Then} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{S}_{\mathrm{1}} {S}_{{n}} \:+\:{S}_{\mathrm{2}} {S}_{{n}−\mathrm{1}} \:+\:{S}_{\mathrm{3}} {S}_{{n}−\mathrm{2}} \:...\:+\:{S}_{{n}} {S}_{\mathrm{1}} }{{S}_{\mathrm{1}} ^{\mathrm{2}} \:+\:{S}_{\mathrm{2}} ^{\mathrm{2}} \:+\:...\:+\:{S}_{{n}} ^{\mathrm{2}} } \\ $$$$\mathrm{is} \\ $$

Question Number 25480    Answers: 0   Comments: 0

Question Number 25390    Answers: 1   Comments: 0

sin 90

$$\mathrm{sin}\:\mathrm{90} \\ $$

Question Number 25381    Answers: 1   Comments: 0

The first term of a sequence is 1, the second is 2 and every term is the sum of the two preceding terms. The n^(th) term is.

$$\mathrm{The}\:\mathrm{first}\:\mathrm{term}\:\mathrm{of}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{1},\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{is}\:\mathrm{2}\:\mathrm{and}\:\mathrm{every}\:\mathrm{term}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{preceding}\:\mathrm{terms}.\:\mathrm{The}\:{n}^{\mathrm{th}} \:\mathrm{term} \\ $$$$\mathrm{is}. \\ $$

Question Number 25378    Answers: 1   Comments: 0

If log x, log y, log z (x,y,z > 1) are in GP then 2x+log(bx), 3x+log(by), 4x+log(bz) are in A.P. True/False?

$${If}\:\mathrm{log}\:{x},\:\mathrm{log}\:{y},\:\mathrm{log}\:{z}\:\left({x},{y},{z}\:>\:\mathrm{1}\right)\:{are}\:{in} \\ $$$${GP}\:{then}\:\mathrm{2}{x}+\mathrm{log}\left({bx}\right),\:\mathrm{3}{x}+\mathrm{log}\left({by}\right), \\ $$$$\mathrm{4}{x}+\mathrm{log}\left({bz}\right)\:{are}\:{in}\:{A}.{P}. \\ $$$$\boldsymbol{{True}}/\boldsymbol{{False}}? \\ $$

Question Number 25317    Answers: 2   Comments: 0

solvd for x:((√(2+(√3))))^x +((√(2−(√3))))^x =4

$${solvd}\:{for}\:{x}:\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{{x}} +\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right)^{{x}} =\mathrm{4} \\ $$

Question Number 25283    Answers: 2   Comments: 1

Question Number 25246    Answers: 1   Comments: 0

8x^(3/(2n)) −8x^((−3)/(2n)) =63

$$\mathrm{8x}^{\frac{\mathrm{3}}{\mathrm{2n}}} −\mathrm{8x}^{\frac{−\mathrm{3}}{\mathrm{2n}}} \:=\mathrm{63} \\ $$

Question Number 25278    Answers: 0   Comments: 4

Find the number of solutions of log∣x∣ = e^x

$${Find}\:{the}\:{number}\:{of}\:{solutions}\:{of} \\ $$$$\mathrm{log}\mid{x}\mid\:=\:{e}^{{x}} \\ $$

Question Number 25226    Answers: 1   Comments: 0

Show that if x=3−(√3).Show that x^2 +((36)/x^2 )=24

$${Show}\:{that}\:{if}\:{x}=\mathrm{3}−\sqrt{\mathrm{3}}.{Show}\:{that}\:{x}^{\mathrm{2}} +\frac{\mathrm{36}}{{x}^{\mathrm{2}} }=\mathrm{24} \\ $$

Question Number 25215    Answers: 0   Comments: 1

Question Number 25173    Answers: 1   Comments: 0

Show that for all nεN−{0} 7^(2n+1) +1 is an integer multiple of 8.

$${Show}\:{that}\:{for}\:{all}\:{n}\epsilon{N}−\left\{\mathrm{0}\right\}\: \\ $$$$\mathrm{7}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1}\:{is}\:{an}\:{integer}\:\:{multiple}\:{of} \\ $$$$\mathrm{8}. \\ $$

Question Number 25171    Answers: 1   Comments: 0

100n>n^2 for integral n>100

$$\mathrm{100}{n}>{n}^{\mathrm{2}} \:{for}\:{integral}\:{n}>\mathrm{100} \\ $$$$ \\ $$

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