₦airaland Forum

Calculusf(x):
.hmmm,here is my own approach to it...sin2x+sinx=1...2sinxcosx+sinx=1...2sinx.sqrt(1-sin^2x)+sinx=1...{let sinx be a...(i)}...2a.sqrt(1-a^2)+a=1...2a.sqrt(1-a^2)=1-a...square both sides...4a^2(1-a^2)=1-2a+a^2...4a^2-4a^4=1-2a+a^2...4a^4-3a^2-2a+1=0,it's obvious that a=1(one of the four roots) then we divide 4a^4-3a^2-2a+1 by a-1 to give 4a^3+4a^2+a-1=0...then transform the equation to reduced cubic equation to get by representing a as y-1/3...(ii) to give y^3-y/12-7/27...you can use any method for solving cubic polynomial to do it but for me,i used tartaglia's method for finding real root of a cubic equation and got y as 0.68114(approximated),don't forget that a=y-1/3 and by substituting that a=0.34781(the first root of the cubic polynomial)...to get the other two roots...from relationship between the co-efficients and roots of a polynomial...a1+a2+a3=-1 a1.a2.a3=0.25(from the polynomial 4a^3+4a^2+a-1=0)...let a1 be 0.34781,by solving a2+a3=-1.34781 and a2.a3=0.7188 and by solving the simultaneous equation a={-1.34781±j1.0287}/2...with all we have done so far,the two real roots got were a=1 and a=0.34781,don't forget that sinx=a...let a be 1,then sinx=1 and x=90 and if a=0.34781,sinx=0.34781 and x=20.35...i rest my case

Calculusf(x):
To square anything which comes between 25-50...say 46...just say 46 minus 25...to give 21,keep that,say 50 minus 46 to give 4(square the 4 to give 16)...take the 16 to 21 and write together...so that answer for 46^2 is 2116...just other ones and ask me questions on the ones that seems to be difficult between 25-50...but note this,it works for all numbers in that range...

benbuks: Mayb sm of u kw dis
multplyin any integer by 6 without using calculator

e.g 6 x 3 =18

put a zero in front of '3'
(30) then divide by '2' (15) then add the number (3) to the answer you got (15+3)=18

also
6x4 =40/2 =20+4 =24
6x5=50/2=25+5=30
6x6=60/2=30+6=36


nb: (works for all integers)

hence
6x n= n0/2 =m+n,

.just my lit2 discovery

rhydex 247: Here is my question.
1. Suppose G is an abelian group. Show dat any factor group G/H is also abelian.

2. Suppose v=(1,3,5,7). Find the projection of v onto W or, in other words, find w€W that minimizes //v-w//, where W is d subspance of R^4 spanned by.
a) u1=(1,1,1,1) and u2=(1,-3,4,-2),
b) v1=(1,1,1,1) and v2=(1,2,3,2).
All is well.

rhydex 247: Here is my question.
1. Suppose G is an abelian group. Show dat any factor group G/H is also abelian.

2. Suppose v=(1,3,5,7). Find the projection of v onto W or, in other words, find w€W that minimizes //v-w//, where W is d subspance of R^4 spanned by.
a) u1=(1,1,1,1) and u2=(1,-3,4,-2),
b) v1=(1,1,1,1) and v2=(1,2,3,2).
All is well.

Ortarico: Hmmm. . . If x is all the values satisfying the interval 0 theta 360 then;
sin2x + sinx = 1
2sinxcosx + sinx - 1 = 0
sinx(2cosx + 1) = 0
sinx = 0
x = arcsin(0)
x = 0, 180, 360

2cosx = -1
cosx = -1/2
x = arccos(-1/2)
x = 120, 240

Laplacian: ...integrate cos(k@)/cos(@-pi/4)....
W.r.t @...where k is a whole number...limit of integration runs from 0 to pi/2....

jauntee02: this angle is great on the whole circle of nairaland.!

Nice job mathematicians.

Just joining the house but find it highly interestng.

Someone should please get this done:

for some fixed number k,let W be a set of all real valued functns for which f(6)=k.
Are there any values of k for which W will be a subspace of F[a,b] with a=< 6 =<b?

smurfy: Thought as much. Despite that, I had material advantage, and my constant attacks must have been paralyzing for you. Guess this is what happens when you play a pro like me

smurfy: @ Laplacian
I told you you'll get smurfed. This is the part where you click the Resign button.

jauntee02: this angle is great on the whole circle of nairaland.!

Nice job mathematicians.

Just joining the house but find it highly interestng.

Someone should please get this done:

for some fixed number k,let W be a set of all real valued functns for which f(6)=k.
Are there any values of k for which W will be a subspace of F[a,b] with a=< 6 =<b?

jackpot: for n>=6, subject of the formula may not be possible.

d citizen:
@jackpot,

Look closely at ur rationalisation, u will see an error there

-1+1/1-(cos@+isin@)

It is wrongly equated and rationalised.

Again, in ur elementary sum to infinity,

The r is not a modulus. It is an ordinary r dat is less than 1 not its absolute value

Let me give u an example if u have r= -1 as ur common ratio, then d modulus, is equal to 1, which go contrary to the principle of geometric series to infinity.
. I will like the person dat pose the questions to confirm if my answer is wrong. Although, i am solving d problem with my fone. I am certain dat the answer is right

jackpot: Hi Sir d citizen, first of all, I must start by acknowledging your mathematical prowess and ability to think critically.

But I hope you know that the formula for the sum to infinity of a GP is valid if only absolute value of the common ratio is less than 1; i.e., |r|<1.

But it is easy to see that the GP
e^i@, e^2i@, e^3i@, . . .
has common ratio r= e^i@
and its absolute value is
|r|=|e^i@|=|cos@+i sin@|
=sqrt{cos²@+sin²@}=1
and as such, applying the sum to infinity of a GP there is extremely fallacious.

Consequently, I'm afraid that your solution is wrong.
Granted, your solution is already wrong. Now your rationalization(as shown in the bolded) here is also wrong since
^{e^i@}/_(1-e^i@)= -1+¹/_(1-e^i@)
=-1 + ¹/_{{(1-cos@)-i sin@}}
=-1+^{{(1-cos@)+i sin@}}/_{{(1-cos@)²+sin²@}}
=-1+^{{(1-cos@)+i sin@}}/_{2(1-cos@)}
and if you take the real part of this, you should be able to get
-1+ ^(1-cos@)/_2(1-cos@)
=-1+¹/₂
=-¹/₂


My Regards



I remain
your humble servant-ress

JACKPOT

rhydex 247: Here is my question.
1). The linear map H:R^3---->R^3 defined by H(x,y,z)=(x+y-2z, x+2y+z, 2x+2y-3z). Is H non singular?. Find the formula for H^-1 and hence evaluate H^-1(1,2,3).

Mbahchiboy: bross biko just finish d no 1 to fulfil all righteousness.....
to d no two how did u get dat answer?EXPLAIN!

jackpot: Hi Sir Laplacian,

I think you should change the underlined to minus sign (-)Explain!

rhydex 247: I assume we are all familiar with Maclaurin series expansion:
f(x)=f(0)+xf'(0)+x^2f"(0)/2!+---
Apply this to cosx, cos2x, then cos3x, etc. and sum them. What do you get?
By the way, I would point out that this series does not converge, but rather with increasing values of 'n' cycles about 0 with an amplitude that depends on the value of @. I'm sure dis helps.

rhydex 247: @ laplacian.
Here is the soln.
a^(x^a) implies
(a.x)a-(a.a)x. Where x=(x1,x2,x3) let a=(a1,a2,a3).
a.a=(a1,a2,a3)(a1,a2,a3).
a.a=a^21^a^22+a^23.
a.x=(a1,a2,a3)(x1,x2,x3).
a.x=a1x1+a2x2+a3x3.
a^(x^a)=(a1x1+a2x2+a3x3)a-(a^21+a^22+a^23)x. Recall that a=(a1,a2,a3) nd x=(x1,x2,x3).
a^(x^a)=(a1,a2,a3)(a1x1+a2x2+a3x3)-(x1,x2,x3)(a^21+a^22+a^23). When u expand nd open d bracket u get 0.
Hence a^(x^a)=0. Buh 4rm d questn we av x+a^(x^a)=b. Put a^(x^a)=0 in d eqn we av
x+0=b. x=b. Where x=x1,x2,x3. Hence x=(x1,x2,x3)=b. Note ^ means cap but ^ in a^21+a^22+a^23 means raise to power. All is well.

Alpha Maximus: ...I believe this is a simple combination matter, let's get to it!!!
(A) if each couple is to sit together, then each pair is considered as a single entity in the computation , thus
8C4=8*7*6*5/1*2*3*4=1680/24=70 ways
(B)if each couple except the freshly married novices(excuse my free language) is to sit together, then this would be interpreted as 5 seperate entities in the following computation, thus
8C5=8*7*6*5*4/1*2*3*4*5=6720/120=56 ways
(C)if a particular couple decided to head home over the rented Okokomaiko apartment, then there would be three couples left, translating to 3 seperate entities, thus(and you didn't state the conditions as to which the remaining couples would be seated, so I'm assuming they are together)
8C3=8*7*6/1*2*3 =336/6=56 ways
P.S: if I was wrong, blame it on the mathematics fever, Euler is on the matter!!

busuyem: Please house, help me solve these questions with explanation:

(1) Convert 123.53 base 10 to base 3
(2)0.36 base 10 to base 8
(3) 23.123 base 10 to base 5

Note: I did it by dividing the two parts by the required bases but the answers are different. I'm pussled. Help me out.

rhydex 247: Here is my question.
1. Solve for x. If x+a^(x^a)=b. Where x=(x1,x2,x3). Note that ^ means cap.

2. Let W be a subspace of real space R^3. Give a geometrical describtion of W in terms of its dimension.

smurfy: Solve the simultaneous equations
(10^x)(4^y) = 1
8^x = 10^(y+1)

jackpot: in the form a+ib
z=-16+0i
[s]sorry, Mathematicians dont write the imaginary part with j like Physicists and Engineers [/s]

now, for the fourth root, i'll prefer working with the exponential form to avoid the impossible task of typing out square roots here.

z=-16+0i=16e^i(pi)=w⁴

Using the nth root of a complex number formula

w=nth root of |z| times e^i[(Argz+2k pi)/n] for k= 0, 1, 2 ,. . ., n-2, n-1

Thus, the fourth roots are simply

w= 2e^i(pi/4), 2e^i(3pi/4), 2e^i(5pi/4), 2e^i(7pi/4)