Further MAth doubt!! please help asap

[ima_shah2002]

hey guys i came across this induction question ? can u please solve it for me...
MJ04 the Question is 4... in both the parts my answers are similar but the ms doesnt have the constants of B - is B removed ? .Please look it up... m sure a proficient mathematician will do it in no time

 

[ahmed t]

is this mechanics 2

 

[ima_shah2002]

no it is Further Math

 

[ahmed t]

is it cambridge

 

[ima_shah2002]

yes CIE

 

[OakMoon!]

I am sure you were done with part one as it is simple differentiation with the quotient rule. We don't really need to find the value of b. We just need the 3 values of a that are -1,2 and -6. And we have to establish a formula by observing this and then prove it.
As we can see the three values are successive factorials with alternating negative signs. The conclusion we can come to is the an=(-1)^n*n!
Now use the three inductive hypothesis. Prove for n=1. a1=-1^1*z=-1 which is true. And for a2=(-1)^2*2!=2. For a3=(-1)^3*3!=-6
Move on to the second hypothesis. Let it be true for n=k so ak=(-1)^k*k
Now move on the 3rd inductive hypothesis and prove for n=k+1.. Before doing this look at a1, a2 and a3, What do you do to a1 to get a2 i.e.a1+1 you multiply the answer by -(n+1) or -(1+1) so a2=a1*-(n+1)..
So to find a(k+1)=ak*-1(k+1)
=(-1)^k*k!*-1(k+1)
(k+1)*k! is (k+1)!.. And (-1)^k*(-1)= -1^(k+1)
So a(k+1)=(-1)^(k+1)*(k+1)!.... PROVEN as you can check by directly substituting n by k+1 and you'll get this.
So the formula is proved and established. i.e. an=(-1)^n*!

 

[ima_shah2002]

hey... thanx... i got everythin except one part... after differntiation my eqns are a1 = (1-b1)/(lnx) -1 , a2 = (-3-b1)/(lnx) +2 and a3=(-7-b3)/lnx - 6 , now i do see the resemblance however do i just omit the B part of the eqn?? is it mathematically right??

 

[OakMoon!]

Where are you finding a1 like this. Use the identities method.
Find the first differential of (lnx/x) ofr a1 and then the 2nd differential for a2 and third differential for a3. Why are you finding these equations?
d/dx(lnx/x)=-lnx+1/x^2 over here we put in n as 1. So a1 will be -1 as it is the coeffecient of lnx is -1. Next find out d2y/dx2 for a2 which is (-3+2lnx)/x^3 make an identity with anlnx+bn/x^(n+1). an is the coeffecient of lnx so for a2 the coeffecient of lnx is 2. Do the same for a3.

I hope you understand.
You are welcome.

 

[ima_shah2002]

Ohh ofcourse... how did I overlook that... makes it so simple.... m super happy now... the thing is i have barely a month to go and i have to prepare further maths... thanks a million... i will try and thank you several times if possible ... on da top rite

 

[ima_shah2002]

hey man... one more doubt... please please if u dont mind.. help.. its regarding series.... 0n04 of the series file i have attached.. in da first part .. SN = n=1(sigma)N , (-1)^n-1 n^3 this has been simplified for 2N to S2N= n=1(sigma)2N ,n^3 - k,n=1(sigma)N, n^3 ... how is this posssible..

 

[OakMoon!]

Remember one thing. In series when you can't come to a conclusion by just looking at the formula, then you should put in the values of n and observe the sequence formed.
Look at the picture below. Tell me if you don't understand anything.

 

[ima_shah2002]

shit man ur bloody awesome!!! took me hours and couldnt figure this out... nice ... btw u a tutor or something... ahhhh and the next part is pretty easy... no problem in that(for once )

 

[OakMoon!]

lol! I am just a student. Its just that this question came in a class test and I was totally blank and then the teacher explained it to us. So I remember the soution to this question quite well because I made a mistake in it. You can ask any problem. I'll be glad to help.

 

[ima_shah2002]

lol! I am just a student. Its just that this question came in a class test and I was totally blank and then the teacher explained it to us. So I remember the soution to this question quite well because I made a mistake in it. You can ask any problem. I'll be glad to help.

well thats good that ur glad to help... pheww... cos doubts keep pourin... i dont have a teacher for further math .. neways in vector spaces when ur trying to reduce to echlon wat if in the leadin diagonals u get zeros how do u continue? came across this in this question MJ03 Q8.. uploaded here

 

[OakMoon!]

Sorry man, I won't be able to help you here. I am still set to do vector spaces in my class. So I am not aware of this topic.

 

[ima_shah2002]

no problem... thanx... btw... btw if u can get an answer from a tutor... it would be grateful...

 

[OakMoon!]

You are welcome.
Even if I do get an answer, I won't be able to explain it to you as it'll just pass over my head until I get the concept. Most probably, I'll be done with this topic in 2 weeks. I can give it a try then.

 

[usman]

salam
i hav recently started studying linear algebra (matrix algebra and vector spaces), so i may nt b a gr8 help.
Well, I culdn't get ur question clearly @ "in vector spaces when ur trying to reduce to echlon wat if in the leadin diagonals u get zeros how do u continue."

Anyways, since the reduced echelon form of a given matrix is unique [in CONTRAST of the echelon (NOT reduced) form, since any multiple of a matrix in echelon form is also in echelon form], i guess my attempt to get the reduced echelon form may help u or atleast provide u an alternative way. So here it is:- (This is the sequence of operations i applied to get to the reduced echelon form)
1. Add -2 times row 1 to row 2.
2. Add -3 times row 1 to row 3.
3. Add -4 times row 1 to row 4.
4. Add -1 times row 2 to row 3.
5. Add -1 times row 2 to row 4. (After step 5 u will reach the echelon (NOT reduced) form of the given matrix)
6. Add 1 times row 2 to row 1. (After step 6 u will finally reach the reduced echelon form of the given matrix)


I hop this helps....;p

 

[ima_shah2002]

salam
i hav recently started studying linear algebra (matrix algebra and vector spaces), so i may nt b a gr8 help.
Well, I culdn't get ur question clearly @ "in vector spaces when ur trying to reduce to echlon wat if in the leadin diagonals u get zeros how do u continue."

Anyways, since the reduced echelon form of a given matrix is unique [in CONTRAST of the echelon (NOT reduced) form, since any multiple of a matrix in echelon form is also in echelon form], i guess my attempt to get the reduced echelon form may help u or atleast provide u an alternative way. So here it is:- (This is the sequence of operations i applied to get to the reduced echelon form)
1. Add -2 times row 1 to row 2.
2. Add -3 times row 1 to row 3.
3. Add -4 times row 1 to row 4.
4. Add -1 times row 2 to row 3.
5. Add -1 times row 2 to row 4. (After step 5 u will reach the echelon (NOT reduced) form of the given matrix)
6. Add 1 times row 2 to row 1. (After step 6 u will finally reach the reduced echelon form of the given matrix)


I hop this helps....;p

THANKS!!!!!

 

[ima_shah2002]

I hope u guys are good at this one... complex numbers. MJ05 q10 part 2 after z+2/z how do u further reduce it... uploade it chek it out