benji93:
goofyone,
you are right brother but you misstated. (1 + x)^1/x = e as x approaches 0

(1 + 1/x)^x =e as x approaches infinity.
so as u rightly did we substitute 1/2h for x
(1+2x)^1/x = (1 + 1/h)^2h = ((1 + 1/h)^h)^2 = e^2 as h approaches infinity
but as h approaches infinity x approaches 0, therefore the limit is e^2
but you have really done well by solving all these, God bless this forum

Thanks for pointing that out. Corrected.

guys, Please what do think it cool as elective course??

Complex Analysis or abstract Algebra.

Please candid opinion will be appreciated .

Tolzeal:
guys, Please what do think it cool as elective course??

Complex Analysis or abstract Algebra.

Please candid opinion will be appreciated .

They are both interesting. You should choose the one you find easier. If you do want to experience a whole new world in mathematics though, then choose abstract algebra. If you are a physics major and need extra tools for your physics, then complex analysis is the way to go. Goodluck.

goofyone:

They are both interesting. You should choose the one you find easier. If you do what to experience a whole new world in mathematics though, then choose abstract algebra. If you are a physics major and need extra tools for your physics, then complex analysis is the way to go. Goodluck.

Thaanka...

The gurus are resurfacing ...

nice work guys..., wish I have enough time...will try sha..


BTW: Benkuks=agentofchange1

@ sir richiez I sight thee boss.

The gurus are resurfacing ...

nice work guys..., wish I have enough time...will try sha..


BTW: Benkuks=agentofchange1

@ sir richiez I sight thee boss.

agentofchange1:
The gurus are resurfacing ...


nice work guys..., wish I have enough time...will try sha..



BTW: Benkuks=agentofchange1


@ sir richiez I sight thee boss.

Sup Benbuks?

naturalwaves:
Sup Benbuks?

Am cool.

Tolzeal:
guys, Please what do think it cool as elective course??

Complex Analysis or abstract Algebra.

Please candid opinion will be appreciated .

To add to what goofyone wrote. Abstract Algebra has more of theorems, lemma and proves. It deals with algebraic structures rather than the usual number systems we all know, unlike complex analysis.

goofyone:


If -i is a root, i will also be a root following the complex conjugate theorem. I can prove that if you want, but i don't think it's relevant to do that here.

Hence, two roots already, ± i

(x+i)(x-i) = x² -i.i = x² + 1

x⁴ + 4x³ + 6x² + 4x + 5 divided by x² + 1 to get the other factor.

Division produces x² + 4x + 5

Solving with quadratic equation: ¹/₂(-4 ± √(16 - 20))
x = -2 ± i

please boss check the division ,i have been trying to follow up the solution but couldn't come up with same answer as ur's don't knw if i'm doing the wrong thing

Soneh:

please boss check the division ,i have been trying to follow up the solution but couldn't come up with same answer as ur's don't knw if i'm doing the wrong thing

Let me help you....

naturalwaves:

Let me help you....

thanks .if you don't mind can u pls help with the explanation to this

3. Prove that the equation mx(x2 + 2x + 3) = x2 – 2x – 3 has exactly one real root if m = 1 and exactly 3 real roots if m= -2/3



For this question, you can use Descartes's rule of signs. You should actually use the more detailed Sturm's extension of this rule, but i think the former could be okay here.
Descartes's rule of signs says if v is the variation of sign in a polynomial p(x) and the number of positive real zeros (roots) is n, then
n<=v and v-n is even.

Same applies to number of negative real zeros with p(-x)

With m=1, you get
x3 + x2 + 5x + 3 = 0

p(x) = x3 + x2 + 5x + 3
No sign variation, so v=0.
n will also be 0 so there are no positive real roots.

p(-x) = -x3 + x2 - 5x + 3

Number of sign variation is 3 (- to + to - to +)
n will be 1, so that 3-1 is even. So there's a negative real root.

Equation thus has one negative real root and two complex roots which will be conjugates.


With m= -2/3

p(x) = -(2/3)x3 - (7/3)x2 + 3

Number of sign variation is 1. n will be 1 so that 1-1=0 which is even.
Hence, one positive real root.

p(-x) = (2/3)x3 - (7/3)x2 + 3.

Number of sign variation here is 2. n cannot be 1, cos v-n should be even. So n is 2. And 2-2=0.
So we have two negative real roots.

Hence, three real roots (one positive and two negative)

I FELT LIKE CRYING WHEN I SAW THIS FORUM: MATHEMARICS. MY BEST SUBJECT. AM SO FOLLOWING, IF WORK WILL ALLOW ME.

Soneh:

thanks .if you don't mind can u pls help with the explanation to this

3. Prove that the equation mx(x2 + 2x + 3) = x2 – 2x – 3 has exactly one real root if m = 1 and exactly 3 real roots if m= -2/3



For this question, you can use Descartes's rule of signs. You should actually use the more detailed Sturm's extension of this rule, but i think the former could be okay here.
Descartes's rule of signs says if v is the variation of sign in a polynomial p(x) and the number of positive real zeros (roots) is n, then
n<=v and v-n is even.

Same applies to number of negative real zeros with p(-x)

With m=1, you get
x3 + x2 + 5x + 3 = 0

p(x) = x3 + x2 + 5x + 3
No sign variation, so v=0.
n will also be 0 so there are no positive real roots.

p(-x) = -x3 + x2 - 5x + 3

Number of sign variation is 3 (- to + to - to +)
n will be 1, so that 3-1 is even. So there's a negative real root.

Equation thus has one negative real root and two complex roots which will be conjugates.


With m= -2/3

p(x) = -(2/3)x3 - (7/3)x2 + 3

Number of sign variation is 1. n will be 1 so that 1-1=0 which is even.
Hence, one positive real root.

p(-x) = (2/3)x3 - (7/3)x2 + 3.

Number of sign variation here is 2. n cannot be 1, cos v-n should be even. So n is 2. And 2-2=0.
So we have two negative real roots.

Hence, three real roots (one positive and two negative)

To be honest, I do not know much about those principles but I could read through his solution and try and break it down for you in my spare time.

Wow! I cant believe where i just landed... Nice meeting you all Sirs/Mas. I look forward to learning more than a lot from you.
Please help me.
6Px = 30.

Where P = permutation

donriddo:
Wow! I cant believe where i just landed... Nice meeting you all Sirs/Mas. I look forward to learning more than a lot from you.
Please help me.
6Px = 30.

Where P = permutation

The answer is 2. See solution attached below......

donriddo:
Wow! I cant believe where i just landed... Nice meeting you all Sirs/Mas. I look forward to learning more than a lot from you.
Please help me.
6Px = 30.

Where P = permutation

⁶P_x = 30,

6! / (6 - x)! = 30,

30 x 4! = 30(6 - x)!,

(6 - x)! = 4!,

Obviously,

x = 2 Q.E.D

dejt4u:
⁶P_x = 30, 6! / (6 - x)! = 30, 30 x 4! = 30(6 - x)!, (6 - x)! = 4!, Obviously, x = 2 Q.E.D

naturalwaves:
The answer is 2. See solution attached below......

Thanks my ßosses. I really appreciate.

naturalwaves:

The answer is 2. See solution attached below......

Good solution, naturalwaves. But I don't think you can mathematically cancel "!" signs like that. It's just sufficient to say (6-x)=4, I believe.

Can someone figure out an analytical solution to this equation?

4^x + 3^x = 5^x

^^^

The answer is 2. But I can only give you graphical solution!

DatechMan:
Can someone figure out an analytical solution to this equation?

4^x + 3^x = 5^x

It's much easier to prove that x cannot be less than 2, and x cannot also be greater than 2. Hence x = 2 for all real numbers.

1) prove that every metric space is a hausdorff space.....2) show that the empty intersection of sets (the finite intersection of empty sets) is the universal set....

goofyone:

Good solution, naturalwaves. But I don't think you can mathematically cancel "!" signs like that. It's just sufficient to say (6-x)=4, I believe.

thanks sir, what you have up there and mine are the same just that you cancelled yours and didn't indicate while I showed mine. We both cancelled it. I thought about it too and came up with the conclusion that it can be cancelled provided that the l.h.s equates to the r.h.s. For example, if x! = y!, then x must be = y. If x! = (y+z)!, then x must be equal to (y+z) and so on and that was why I cancelled it. I wouldn't mind any example that disproves these facts. Thanks.

DatechMan:
Can someone figure out an analytical solution to this equation?

4^x + 3^x = 5^x

4^x + 3^x = 5^x
divide bothsides by 4^x
1 + (3/4)^x = (5/4)^x
1 + 0.75^x = 1.25^x
1 + (1-0.25)^x = (1+0.25)^x
Approximating linearly
1 + 1 - 0.25x = 1 + 0.25x
2 - 0.25x = 1 + 0.25x
0.5x = 1
x = 2

@goofyone, please help Soneh with the explanation of that Descarte's principle and how you used it to solve that problem,he doesn't seem to understand it. Thanks

naturalwaves:

thanks sir, what you have up there and mine are the same just that you cancelled yours and didn't indicate while I showed mine. We both cancelled it. I thought about it too and came up with the conclusion that it can be cancelled provided that the l.h.s equates to the r.h.s. For example, if x! = y!, then x must be = y. If x! = (y+z)!, then x must be equal to (y+z) and so on and that was why I cancelled it. I wouldn't mind any example that disproves these facts. Thanks.

Yeah, i understand you. It's okay, bro.

Thank God you guys are back to level I cn comprehend.. The fear of madam Jackpot's questions is d beginning of learning..lol

Integrate cos^4x dx

Nwiboazubuike:
Integrate cos^4x dx

Made a mistake in the last line it's sin 2x over 4 not over 8

Nwiboazubuike:
Integrate cos^4x dx

^^^^^^^^SolutioN^^^^^^^^^^
let's rewrite as
$(cos² x )²dx
known from trig. ( Pythagorean) identities ,
cos² =0.5(cos2x+1)
thus ,=> $[(0.5(cos2x+1)]^2 dx
=> 1/4$((cos^2 (2x) +2cos2x +1))dx we can also
have cos^2(2x) = 0.5(cos4x +1)
thus , get have 1/4$[0.5(cos4x +1) +2cos2x +1] dx
now on integrating we get
=> 1/4[ 1/2*1/4*sin4x + x/2 +sin2x + x ] +k
=> 1/32* sin4x +1/4*sin2x+ 3x/8 + k

naturalwaves:

I think I meant to say Stennon. She is the one studying physics edu right?

thanks, STENON is in the mood of her final exams.
Good morning,
Hope you are good?