WebsiteDesigner: @Richiez, pls can u solve a Von Smoluchowski equation. Let me know if u can.

Richiez:

Mathematical induction is a method of mathematically proving that a statement is true for all natural numbers i.e positive integers
it usually involves 2 basic steps;
1. showing that the statement is true when the lowest natural number is used e.g n=0 or n=1 (preferably)
2. showing that if the statement is true for any natural number n=k, then it is true for the next natural number n=k+1

let us do some examples;

1. prove by mathematical induction 1+2+3+4+.......+n+(n+1) = n(n+1)/2
SOLUTION
note that n represents the number of elements in the series
step 1---we have to show that the statement is true for n=1
1= 1(1+1)/2
1= 2/2
1=1 therefore since L.H.S = R.H.S the statement is true for n=1
we can even go ahead to show that the statement is true for any other small value of n, say n= 4
1+2+3+4 = 4(4+1)/2
10= 4(5)/2
10= 20/2
10= 10 therefore it is also true for n=4
next is step 2
here we shall consider an arbitrary value of n, say n=k and assume it is true
i.e 1+2+3+4+....+k = k(k+1)/2 ................(1)
if the above statement is true, then the final step is to show that it is also true for n= k+1
1+2+3+4+....+k+(k+1) = (k+1)(k+1+1)/2
1+2+3+4+....+k+(k+1) = (k+1)(k+2)/2 ...............(2)

to show that eqn 2 is true, let us add (k+1) to both sides of eqn 1
1+2+3+4+....+k+(k+1) = k(k+1)/2 + (k+1).............(3)
the job here is to make the R.H.S of eqn 3 look like that of eqn 2
taking l.c.m as 2
1+2+3+4+....+k+(k+1) = [k(k+1)+2(k+1)]/2
since (k+1) is a common factor in the R.H.S we factorize
1+2+3+4+....+k+(k+1) = (k+1)(k+2)/2

This is same as eqn 2 above, hence the statement 1+2+3+4+....n+(n+1) = n(n+1)/2 is true for all natural numbers.

feel free to post more exercises.

olasesi:

Naaaaa its.the same

x+y=5. Is the same as (x+y)=5

Try the same value for x and y in both equations.....
What do u get.....try 2 and 3

Calebsworld: @ Richiez,my question,is how can I become a mathematician,as in having distinctions in the subject.God bless u.

olasesi:

It is very essential to understand the meaning of "=" (equal to" ) in mathematics......
If AB=CD,
Then AB+F=CD+F

Therefore if (x+y)=5
only log (x+y) must be equal to log5

Log is another operation on its own

Ehinafe: Richiez:

nice try bro, although you made some some mistakes especially at the point where you took log of both sides. e.g it ought to be log(x+y)= log5 and not logx + logy = log5. but i must admit, you did a very good job.

here's another approach;

x + y = 5.......(1)
x^x + y^y =31.......(2)

from eqn(1), x=5-y, hence we substitute this value for x in eqn(2).
(5-y)^(5-y) + y^y = 31.......(3)
now, we have reduced the problem to what value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3
i.e (5-3)^(5-3) + 3^3
2^2 + 27
4+27=31
now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x
recall that x+y=5
therefore; x+3=5 or x=5-3=2
hence, x=2 when y=3

Richiez, how good are you in solving Dynamic Programming mathematical algorithm? I need to know because I have one here that I am not sure you can get the answer right.

Richiez:

yep, but don't be scared, any level of maths is accepted here

solomon111: the answer obeyed the equation.
X+y=5
2+3=5.
X^x+y^y=31
2^2+3^3=31
(equation proved)

solomon111: I dread statistics.
That is one area of mathematics,i have not mastered.
As far as am concerned,statisticians are the real mathematicians.

olasesi:

Naaaaaa. 6.2 was approximated to 6...,...
Why didn't u use 6.2....or must u approximate.....

2 and 3 are the answers but it was dubiously gotten

solomon111: I dread statistics.
That is one area of mathematics,i have not mastered.
As far as am concerned,statisticians are the real mathematicians.

Richiez:

nice try bro, although you made some some mistakes especially at the point where you took log of both sides. e.g it ought to be log(x+y)= log5 and not logx + logy = log5. but i must admit, you did a very good job.

here's another approach;

x + y = 5.......(1)
x^x + y^y =31.......(2)

from eqn(1), x=5-y, hence we substitute this value for x in eqn(2).
(5-y)^(5-y) + y^y = 31.......(3)
now, we have reduced the problem to what value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3
i.e (5-3)^(5-3) + 3^3
2^2 + 27
4+27=31
now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x
recall that x+y=5
therefore; x+3=5 or x=5-3=2
hence, x=2 when y=3

PortHarcourtBoy: Una say na "Mathematics" thread meanwhile all of una dey speak "English"

Some dey solve Indices, others dey solve Simultaneous equations...very soon another person go come solve Quadratic equations...abi na SS1 students dey here?

Wey the Engineers dem make we solve some First Order Differential Equations...and apply am to Fluid transfer and Thermodynamics

Meanwhile wey my Gala and La Casera?

Eze Promoe: I LOVE THIS THREAD AND I'LL LOVE THE MATHS GURU TO SOLVE THIS EQUATION FOR ME!

A boy, X sleeped with my girlfriend, Y and produced 2 kids (2XY), using the onyonyo meter I saw that the pikins is not my own shildren. If Y is 22 years old and her name is Gracie
i)find X?
ii)find X's father, X^2
iii)can X beat me, Z if we fight?

Do that equation quick before i go gaga

solomon111: dynamic and programming mathematical algorithm are for those studying mathematics as a course on it own.
Though,it's very intriguing,and looks superior to the logarithmic method.