Shas72
Badges: 10
Rep:
?
#1
Report Thread starter 10 months ago
#1
given that A/(x+a) +B/(x+b)+C/(x+c)=f(x)/[(x+a)(x+b)(x+c)], prove that if f(x) is linear then A+B+C=0
0
reply
ghostwalker
  • Study Helper
Badges: 17
#2
Report 10 months ago
#2
(Original post by Shas72)
given that A/(x+a) +B/(x+b)+C/(x+c)=f(x)/[(x+a)(x+b)(x+c)], prove that if f(x) is linear then A+B+C=0
Start with the lefthand side. Put it over a common denominator. What do you get?
0
reply
Shas72
Badges: 10
Rep:
?
#3
Report Thread starter 10 months ago
#3
(Original post by ghostwalker)
Start with the lefthand side. Put it over a common denominator. What do you get?
So it will be A(x+b)(x+c)+B(x+a)(x+c)+C(x+a)(x+b)= f(x)
0
reply
Shas72
Badges: 10
Rep:
?
#4
Report Thread starter 10 months ago
#4
(Original post by ghostwalker)
Start with the lefthand side. Put it over a common denominator. What do you get?
Do I also take x=-a, x=-b and x=-c
0
reply
ghostwalker
  • Study Helper
Badges: 17
#5
Report 10 months ago
#5
(Original post by Shas72)
So it will be A(x+b)(x+c)+B(x+a)(x+c)+C(x+a)(x+b)= f(x)
Yes. So, expand and group terms in x^2, x, and constants.

(Original post by Shas72)
Do I also take x=-a, x=-b and x=-c
No need.
0
reply
Shas72
Badges: 10
Rep:
?
#6
Report Thread starter 10 months ago
#6
(Original post by ghostwalker)
Yes. So, expand and group terms in x^2, x,
Ok let me try
0
reply
Shas72
Badges: 10
Rep:
?
#7
Report Thread starter 10 months ago
#7
(Original post by ghostwalker)
Yes. So, expand and group terms in x^2, x, and constants.



No need.
So I get
Ax^2+Bx^2+Cx^2+[A(b+c)+B(a+c)+C(a+b)]x + Abc +Bac +Cab
0
reply
ghostwalker
  • Study Helper
Badges: 17
#8
Report 10 months ago
#8
(Original post by Shas72)
So I get
Ax^2+Bx^2+Cx^2+[A(b+c)+B(a+c)+C(a+b)]x + Abc +Bac +Cab
Not checked the whole thing, so I assume it's correct.

=(A+B+C)x^2+[A(b+c)+B(a+c)+C(a+b)]x + (Abc +Bac +Cab)

OK. So, you're told it's a linear function. What's the coefficient of x^2 going to be then?
0
reply
Shas72
Badges: 10
Rep:
?
#9
Report Thread starter 10 months ago
#9
(Original post by ghostwalker)
Not checked the whole thing, so I assume it's correct.

=(A+B+C)x^2+[A(b+c)+B(a+c)+C(a+b)]x + (Abc +Bac +Cab)

OK. So, you're told it's a linear function. What's the coefficient of x^2 going to be then?
A+B+C
0
reply
ghostwalker
  • Study Helper
Badges: 17
#10
Report 10 months ago
#10
(Original post by Shas72)
A+B+C
Yes, well, that's what it is in the function. But if the function is linear (i.e. not quadratic), what must that equal.
0
reply
Shas72
Badges: 10
Rep:
?
#11
Report Thread starter 10 months ago
#11
(Original post by ghostwalker)
Not checked the whole thing, so I assume it's correct.

=(A+B+C)x^2+[A(b+c)+B(a+c)+C(a+b)]x + (Abc +Bac +Cab)

OK. So, you're told it's a linear function. What's the coefficient of x^2 going to be then?
So linear function is a straight line and is not quadratic. So x^2 doesnt exist. So A+B+C=0
0
reply
ghostwalker
  • Study Helper
Badges: 17
#12
Report 10 months ago
#12
(Original post by Shas72)
So linear function is a straight line and is not quadratic. So x^2 doesnt exist. So A+B+C=0
Yep. Though we would say, the coefficient of x^2 must be zero, and therefore A+B+C=0,
0
reply
Shas72
Badges: 10
Rep:
?
#13
Report Thread starter 10 months ago
#13
(Original post by ghostwalker)
Not checked the whole thing, so I assume it's correct.

=(A+B+C)x^2+[A(b+c)+B(a+c)+C(a+b)]x + (Abc +Bac +Cab)

OK. So, you're told it's a linear function. What's the coefficient of x^2 going to be then?
Thank you so so much!
Can you also pls explain the reasoning whether it's possible to find the binomial expansion for sqrt(x-1)
0
reply
ghostwalker
  • Study Helper
Badges: 17
#14
Report 10 months ago
#14
(Original post by Shas72)
Thank you so so much!
Can you also pls explain the reasoning whether it's possible to find the binomial expansion for sqrt(x-1)
Suppose f(x)=\sqrt{x-1}


and further suppose you could write it as a power series. a_0 +a_1x +a_2x^2 + \cdots

Then a_0=f(0) =\sqrt{-1} :eek:

Fine if you're working with complex numbers, but otherwise not.
0
reply
Shas72
Badges: 10
Rep:
?
#15
Report Thread starter 10 months ago
#15
(Original post by ghostwalker)
Suppose f(x)=\sqrt{x-1}


and further suppose you could write it as a power series. a_0 +a_1x +a_2x^2 + \cdots

Then a_0=f(0) =\sqrt{-1} :eek:

Fine if you're working with complex numbers, but otherwise not.
Thank you very very much!!
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

If you haven't confirmed your firm and insurance choices yet, why is that?

I don't want to decide until I've received all my offers (39)
40.63%
I am waiting until the deadline in case anything in my life changes (21)
21.88%
I am waiting until the deadline in case something in the world changes (ie. pandemic-related) (5)
5.21%
I am waiting until I can see the unis in person (9)
9.38%
I still have more questions before I made my decision (5)
5.21%
No reason, just haven't entered it yet (6)
6.25%
Something else (let us know in the thread!) (11)
11.46%

Watched Threads

View All
Latest
My Feed