Am in

(1)what is the real application of mathematics in our live?

(2) prove that:
a. Sin^3 x = 1/4(3sinx - sin3x)
b. Find 2nd derivative of:
y = cos^3 x

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

'trial n error' method. U try sha.

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.
logx + logy = 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.

.........hmmmmmmmm.........
.........Quite dubious........

How is logx+logy=log5
How is logx^x+logy^y=log31
Why would you approximate 6.2 to 6.......can't I as well use 6.2......

brilliand dudes in here

Richiez:

well funny as the name sounds, the pigeonhole principle is a very important principle in mathematics. it states that if there are x<y and there are x pigeonholes and y pigeons, then at least one pigeonhole must contain more than one pigeon.

this principle is used in solving basic mathematical arguments, for example; if there are only 4 colours of shirts and five students are required to choose any colour of shirt that they prefer, then atleast two students must choose same colour.

wow.....maths is really interesting!

thanks bro,u just increased ma knowledge... I look up into dis.

1stknight:

Bro pls I don 't understand how you got the fifth equation.

The fifth equation follows the principle/laws of logarithm.
E.g
1)logx +logy =logxy
2)alogx +blogy=ablogxy.
3)logx-logy=logx/logy

solomon111: The fifth equation follows the principle/laws of logarithm.
E.g
1)logx +logy =logxy
2)alogx +blogy=ablogxy.
3)logx-logy=logx/logy

.... the three above equations are correct......

But x+y=5
Cannot be said to be equal to logx+logy=log5.....
and same to the remaining equations

Hey Richiez, you guys should [come back and] prolong this 'thing'; I'm so having fun.

Gboliwe:
Maths is frightening with equations like this

Hehehe, i find this interesting. But what i ask myself is. What am i really doing with all my A's and B's in school

A man needs to locate his talent and convert intelligence into Money(£, €, $, =N=). Otherwise nah die be that.

olasesi:

.........hmmmmmmmm.........
.........Quite dubious........

How is logx+logy=log5
How is logx^x+logy^y=log31
Why would you approximate 6.2 to 6.......can't I as well use 6.2......

To solve the above problem,you have to apply the laws of logarithm,since it involves exponential values.
If logarithm is applied in the first equation,then it would also be applied in the second equation to create uniformity.
X+Y=5,becomes
logx+logy=5.
x^x+y^y=31,becomes
xlogx+ylogy=log31.

mathematics is a column of numbers performing increasingly confusing things with themselves.

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

Following

x^x^x=log3
x^y^y=log8.
Find x and y.

solomon111: The fifth equation follows the principle/laws of logarithm.
E.g
1)logx +logy =logxy
2)alogx +blogy=ablogxy.
3)logx-logy=logx/logy

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

Real gurus here... Wonder where dem statisticians at? I wanna get practical.

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

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

That only applies when (x+y) is acting as a single variable,hence the bracket.
In the question,x and y are different variables.

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.

Ara ga agba ndi Ara taa ta...

i never knew we had many maths guru here o...i dey envy u guys o bt na biological science na im sure pass sha..

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

u try sha bt u r nt a gud teacher, in your presentations you did not consider the level of your audience.
you don't use this method to a 'lower level' audience

@RICHEZ. Concerning d pigeonhole example, does it imply dat at most 6 pairs will write on d same topic and how do u interprete d extra 2 pairs with regard to d topic of discussion? By d way, i love ur confidence.

Guy u are good I have a brother who is also good in mathematics I will love for u guyz to meet. Plz post ur bb pin, e-mail add or twitter handler. Thanks

olasesi:

.... the three above equations are correct......

But x+y=5
Cannot be said to be equal to logx+logy=log5.....
and same to the remaining equations

Is correct, all he did is take log of both sides.

paranorman: Real gurus here... Wonder where dem statisticians at? I wanna get practical.

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

His target is the maths gurus. You and i seem like the lower level audience so we just have to watch. Too bad

galadee:

u try sha bt u r nt a gud teacher, in your presentations you did not consider the level of your audience.
you don't use this method to a 'lower level' audience

solomon111: That only applies when (x+y) is acting as a single variable,hence the bracket.
In the question,x and y are different variables.

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

U guys make me remember ma Secondary School Maths Teacher... The Youngest Mathematician In The Rivers State Chapter Of Mathematicians.....

All i pray is to be Mathematically Inclined Like Him... Bt I wasted My Time With Rubbish...

Pray To Recover Soon...

majortele: Am in

welcome

Actually i just got admission to study mathematics & statistics in Uniport, hmm so to al dose gurus in dis room, wat do u advice i do to bcom a guru like dis.. I admire ur confidence in handlin som questions here