yungryce: Na dem b dis
Prove by induction that for any two nos a & b, & for any nEN that
(a+b)^n = a^n + na^(n-1)b + [n(n-1)/2!]a^(n-2)b^(2) + ...+ [n!/(n-k)!k!]a^(n-k)b^k + ... + b^(k)

2. prove d existence theorem of laplace transform
3. show L[f"(t)] = s^2F(s) - sf(0) - f'(0)
4. using the methods of contour intergration & convolution theorem, evaluate
L^-1{1/(s+1)(s^2+1)} &
L^-1{1/(s-2)(s-1)^2}

Question 2

If f(t) is a function dened for all, then it's Laplace transform is dened as=>

F(s) = L(f) = ∫^∞₀ e^(-st)f(t)dt

Assuming f(t) is such that the integral exists, i.e has some finite value. Note that F(s) and the above operation which yields F(s) from f(t) are both called Laplace transform.

Also, if f(t) is defined and continuous on every finite interval on the semi-axis and satisfies for some constants M and k, then the Laplace transform exists for all s > k .

And for any linear functions f(t) and g(t) whose transforms exist and any constants a and b the transform of af(x) + bg(x) exists =>

L[af(x) + bg(x)] = aL{f(x)} + bL{g(x)}

Question 3

L[f"(t)] = s^2F(s) - sf(0) - f'(0)

If f'(t) is continuos, then by definition of laplace transform =>

L[f'(t)] = ∫^∞₀ e^(-st)f'(t) dt

Applying integration by parts =>

L[f'(t)] = {e^(-st)f(t)}^∞₀ + s∫^∞₀ e^(-st)f(t) dt
L[f'(t)] = sL[(f(t)] - f(0) ......(1)

Using the same method,

L[f"(t)] = s L[f'(t)] - f'(0) .....(2)

Now, substituting (1) in (2) yields =>
L[f"(t)] = s[sL(f(t)) - f(0)] - f'(0)
L[f"(t)] = s^2L[f(t)] - sf(0) - f'(0)
Since L[f(t)] = F(s) = laplace transform for f(t), replacing L[f(t)] with F(s) yields =>
L[f"(t)] = s^2F(s) - sf(0) - f'(0)
:. LHS = RHS, proven!

Still working on the other questions!

sorry guys was out

good job

Youngsage:
x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
log(x+y) = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.

ur answer is right but the step is very wrong.....u mis interpreted d law of log its impossible to have log(x^x+y^y)=logx^x+logy^y....it should be log(x^x* y^y)...steps matter in mat nt answers

join ds facebook group to share ideas with like minds in the field of mathematics.......mathematically mad

Just when i was looking to post the solutions to youngryce's questions

hi richez,am dan and am 17years i'm really bad at maths,i dunno maybe its due to my laziness,cause i find it very difficult to do simple additions and subtractions talkless of multiplying,pls i need ur help i would be writing,waec and jamb this year,and i want yu to be part of my maths succes this year...i need ur help badly pls,cause i believe am not a dullard and i can become a maths guru like yu...help me pls.i beg of yu

Richiez: Just when i was looking to post the solutions to youngryce's questions

We are a team bruv! You started this thread and you've helped so many people here, I'm only giving the little I have to support. So it doesn't really matter where some solutions come from. All we need is 1luv and 1heart to keep the thread alive. Greetings General Richiez!

doubleDx: We are a team bruv! You started this thread and you've helped so many people here, I'm only giving the little I have to support. So it doesn't really matter where some solutions come from. All we need is 1luv and 1heart to keep the thread alive. Greetings General Richiez!

yes bro, u sure deserve the rank of a thread marshall, i think we've been playing the role of thread marshalls here by keeping the thread alive...lol, anyways keep it up bro

pls double dx and and the other maths gurs am really desperate,i posted earlier pls do read my post,i really wann be a maths guru tooo..

chintzs: hi richez,am dan and am 17years i'm really bad at maths,i dunno maybe its due to my laziness,cause i find it very difficult to do simple additions and subtractions talkless of multiplying,pls i need ur help i would be writing,waec and jamb this year,and i want yu to be part of my maths succes this year...i need ur help badly pls,cause i believe am not a dullard and i can become a maths guru like yu...help me pls.i beg of yu

okay bro, for the purpose of waec preparation, i'd advice you do lots of past questions. but becoming a math guru is a slow and gradual process

here are a fewtips;
1. you have to develop a great interest for maths, and have that strong desire to become a maths guru.
2. Realize that there's nothing magic about mathematics.
3. make friends with people who appreciate mathematics.
4. practise as many maths questions as possible, starting from the very simple ones to more complex ones. develop the urge to solve them on your own, but you can always seek for assistance whenever you get hooked.
5. do not forget any new thing you have learnt, keep reviewing them.
6. learn to memorize some basic maths formulas as this will build your confidence.
GOOD LUCK

yungryce: Na dem b dis
Prove by induction that for any two nos a & b, & for any nEN that
(a+b)^n = a^n + na^(n-1)b + [n(n-1)/2!]a^(n-2)b^(2) + ...+ [n!/(n-k)!k!]a^(n-k)b^k + ... + b^(k)


2. prove d existence theorem of laplace transform
3. show L[f"(t)] = s^2F(s) - sf(0) - f'(0)
4. using the methods of contour intergration & convolution theorem, evaluate
L^-1{1/(s+1)(s^2+1)} &
L^-1{1/(s-2)(s-1)^2}

Question 4.a

L^-1{1/(s+1)(s^2+1)}

To evaluate the inverse Laplace transform of the above, we have to locate the singularities of F(s) = 1/[(s + 1)(s^2 + 1)] by solving the denominator for the values of s =>

(s + 1)(s^2 + 1)
= (s + 1)(s - i)( s + i)
s1 = -1, s2 = i and s3 = -i. where i and -i are complex numbers.

We now see that => F(s) = 1/(s+1)(s^2+1) = 1/(s + 1)(s - i)(s + i) has simple poles at s1 = -1, s2 = i and s3 = -i

We can now compute the inverse Laplace transform using Bromwich integral =>

L^-1{1/(s+1)(s^2+1)} = 1/2π lim α=>β ∫^{(α + iβ)}_{(α - iβ)} e^(st) ds/[(s + 1)(s^2 + 1)]
= 1/2π ∫^{(α + iβ)}_{(α - iβ)} e^(st) F(s)ds.

Applying Cauchy's Residue Theorem, yields =>

= Res[(s - s1)F(s)e^(s1t), s1] + Res[(s - s2)F(s)e^(s2t), s2] + Res[(s - s3)F(s)e^(s3t), s3]
= Res[(s - s1)F(s)e^(s1t), -1] + Res[(s - s2)F(s)e^(s2t), i] + Res[(s - s3)F(s)e^(s3t), -i]

For the simple pole at s = -1, the residue Res[(s - s1)F(s)e^(s1t), s = -1] can be evalauted as follows=>

Res[(s - s1)F(s)e^(s1t)]
lim s=>-1, (s + 1)F(s)e^(s1t)
= lim s=>-1, (s + 1)e^(-t)/[(s +1)(s^2+ 1)]
= lim s=>-1, e^(-t)/(s^2 + 1)
= e^(-t)/((-1)^2 + 1)
= e^(-t)/2

For the simple pole at s2 = i, the residue Res[(s - s2)F(s)e^(s2t), s2 = i] can be evalauted as follows=>

= Res[(s - s2)F(s)e^(s2t)]
= lim s=>i (s - s2)F(s)e^(it)
= lim s=>i (s - i)e^(it)/[(s +1)(s - i)(s + i)]
= lim s=>i, e^(it)/[(s +1)(s + i)]
= e^(it)/[(i + 1)(i + i)]
= e^(it)/[(i + 1)(2i)]

For the simple pole at s3 = -i, the residue Res[(s -s3)F(s)e^(s3t), s3 = -i] can be evalauted as follows=>

Res[(s - s3)F(s)e^(s3t)
= lim s=>i (s + i)F(s)e^(s3t)
= lim s=>i (s + i)e^(-it)/[(s +1)(s - i)(s + i)]
= lim s=>i e^(-it)/[(s +1)(s - i)]
= e^(-it)/[(-i + 1)(-i - i)]
= e^(-it)/[(i - 1)(2i)]

Remember that=>

L^-1{1/(s+1)(s^2+1)} = Res[(s - s1)F(s)e^(s1t), -1] + Res[(s - s2)F(s)e^(s2t), i] + Res[(s - s3)F(s)e^(s3t), -i]

Thus,
L^-1{1/(s+1)(s^2+1)} = e^(-t)/(2) + e^(it)/[(i + 1)(2i)] + e^(-it)/[(i - 1)(2i)]
= [i(i - 1)(i + 1)e^(-t) + (i - 1)e^(it) + (i + 1)e^(-it)]/[2i(i - 1)(i + 1)]

^
We know that=> e^(ix) = cos x + isin x, e^(-ix) = cos x - isin x, coshx = 1/2(e^(x) + e^(-x)) and sinhx = 1/2(e^(x) - e^(-x)).

If you simplify further and substitute the above expressions, it yields =>

1/2(sin t - cos t - sinh t + cosh t)

Hence=>
L^-1{1/(s+1)(s^2+1)} = 1/2(sin t - cos t - sinh t + cosh t)

I'll type out and post the other solutions later. Hope this helps!

Richiez:
yes bro, u sure deserve the rank of a thread marshall, i think we've been playing the role of thread marshalls here by keeping the thread alive...lol, anyways keep it up bro

Lol...Alright, let's keep it moving Chief 1luv

Ok, do you drool at those guys above? Do you think they are a special breed of genetically modified autobots with extremely fat brains? No, they are not. If you can make it to school and back, you are genius already. Do you fucking know how many zillions of calculations you churned out while negotiating those bends and bearings? Men, it's a whole lot, like scary lot.

Now, you can see maths from different angles but it boils down to understanding that maths has three main parts;
langusge of mathematics (beauty of expression)
mathematical model (abstraction of physical and non-physical problems)
mathematics itself (rules and regulations)
....
Lol, nature calls, I will be back...

ultranet: pls can u solve dis question: 3^x=9x...(interested in d working)

solution x=3 and x=0.128

ultranet: pls can u solve dis question: 3^x=9x...(interested in d working)

solution x=3 and x=0.128

If you want to be a Maths guru, then enrol into any university and pick Mathematics as your choice course of study and then you can see the beauty of maths.

When you have contended with areas like Analysis (eg complex, real and numerical), Algebra (eg theory of numbers & Galois theory, groups and rings) Mechanics (eg fluid & solid, composite and non - classical) , Tensor and Topology, differential equations (ordinary & partial), then you can refer yourself as a Mathematician or maths gurus if you like.
Secondary school mathematics is hardly enough. It's only just an elementary. You can't call yourself a maths guru if had A1 in WAEC or you offered something other than Maths in the university.

^^^ Really! You can blend sex and maths it's really potent. Use to chill with a looooong bleep after a session of mathematical exercise, you got to balance the intelectual with the emotional to achieve harmony, tks tks tks

Johnpaul88: If you want to be a Maths guru, then enrol into any university and pick Mathematics as your choice course of study and then you can see the beauty of maths.

When you have contended with areas like Analysis (eg complex, real and numerical), Algebra (eg theory of numbers & Galois theory, groups and rings) Mechanics (eg fluid & solid, composite and non - classical) , Tensor and Topology, differential equations (ordinary & partial), then you can refer yourself as a Mathematician or maths gurus if you like.
Secondary school mathematics is hardly enough. It's only just an elementary. You can't call yourself a maths guru if had A1 in WAEC or you offered something other than Maths in the university.

Really? I find it very difficult to agree with the bolded part! An A1 in WAEC maths is a very good foundation (That's, if it's NOT arranged ). Are you also saying only folks who studied mathematics at BS level and above are good mathematians? Lol!

ositadima1: ^^^ Really! You can blend sex and maths it's really potent. Use to chill with a looooong bleep after a session of mathematical exercise, you got to balance the intelectual with the emotional to achieve harmony, tks tks tks

Lol. I agree , welcome back bruv!

@Johnpaul88, you should be helping us out here sometime when you are free, you know? Are you a math grad. or still a stud.?

doubleDx:

Really? I find it very difficult to agree with the bolded part! An A1 in WAEC maths is a very good foundation (That's, if it's NOT arranged ). Are you also saying only folks who studied mathematics at BS level and above are good mathematians? Lol!

I do not disagree that an A1 in Maths Waec is a good foundation. I'm only saying that it is only premature to see such a person as a maths guru because at his level, he's yet to encounter those areas of maths that i highlighted above and without them , i don't know what you can say you know in mathematics.

As for one who read perhaps engineering or economics or computer science or physics as his first degree, though they have good knowledge of maths since they touched some of those areas, i'm still reluctant to see them as maths gurus. Their knowledge of these areas they touched may not be deep enough. That's how i see it.

doubleDx: @Johnpaul88, you should be helping us out here sometime when you are free, you know? Are you a math grad. or still a stud.?

A finalist.
The problem is i have difficulty posting maths stuff with phone. The options are so limited or non existent. I even have difficulty understanding some of the ways solutions are posted here. Most of them are with phone and they can look so muddled up.
But i 'll try one of these days and see.

2nioshine: @doubleDX once again gr8 wuk....any who can should try this...2^x=4x.... question2
x+y+z=2, x^2+y^2+z^2=26,
x^3+y^3+z^3=38....plz no graphical soln..ans=4....Q2ans=4,-3 &1

question1 has 2 solutions: 4 and 0.3099

Johnpaul88: I do not disagree that an A1 in Maths Waec is a good foundation. I'm only saying that it is only premature to see such a person as a maths guru because at his level, he's yet to encounter those areas of maths that i highlighted above and without them , i don't know what you can say you know in mathematics.

As for one who read perhaps engineering or economics or computer science or physics as his first degree, though they have good knowledge of maths since they touched some of those areas, i'm still reluctant to see them as maths gurus. Their knowledge of these areas they touched may not be deep enough. That's how i see it.

Ok, I understand you now man. But Engineering? Well, there are a lot of Engineering courses with little or no Math/calculations, but EE ain't one of them....ask people who know about it! My younger cousin is a finalist too, Math BSc, FUT Minna! Kudos man, math is an interesting course!

Math is interesting, easy and straightforward compare to some Engineering courses! A few of my course mates back in college were first class math graduates before enrolling for Engineering but it still wasn't easy for them to graduate with a 2.1. Some Engineering courses are rugged bruv!

Johnpaul88: A finalist.
The problem is i have difficulty posting maths stuff with phone. The options are so limited or non existent. I even have difficulty understanding some of the ways solutions are posted here. Most of them are with phone and they can look so muddled up.
But i 'll try one of these days and see.

Yeah, you are right. I use mobile sometimes too and typing ain't easy!

dp

doubleDx: Lol...Alright, let's keep it moving Chief 1luv

certainly we are a good team

doubleDx: Yeah, you are right. I use mobile sometimes too and typing ain't easy!

same here oh but sha it wont stop us from helping those who need our assistance

@doubleDx
Are you a grad? Of maths? Which school?
Thanks.

^Nah, I did my Bachelors in Engineering Science; not in Nigeria though!

Kindly help with dis questions....1) write down d first four terms of d bionomial series expansion in ascending power of x of d functions (1-x)^-1/2 and hence show dat root 10= 3(1+1/20+1*3/20*40+1*3*5/20*40*60+........) No(2) find d common ratio of d geometric sequence sin2€, -sin€cos2€, sin2€cos^2(2€),.....prove dat for @<€<₹/2...the series sin2€-sin2€cos2€ + sin2€sin^2(2€) +....has a sum to infinity and show dat d sum to infinity is tan€... Where €,₹ and @ are alpha, pie and teetha respectively.... Tanks