# Thread: Find the value of the a and b

1. ## Find the value of the a and b

Question :
Find the value of a and b in order that

$
\lim_{x \to 0} \frac{x(1 + a cosx) - b sinx}{x^3} = 1.
$

2. Originally Posted by zorro
Question :
Find the value of a and b in order that

$
\lim_{x \to 0} \frac{x(1 + a cosx) - b sinx}{x^3} = 1.
$
The numerator behaves like $x(1+a+b)-x^3(a/2 + b/6)$ for small $x$ and we need this to be $\sim x^3$ so $a=-5/2$ and $b=3/2$.

Note: you need to expand $x(1+ a \cos(x) )$ and $b \sin(x)$ up to and including the terms in $x^3$ then the coefficient of $x^3$ in the numerator should be $1$ and the coefficients for all lower power terms should be $0$ .

CB

3. ## Could u please explain me this a little more

Originally Posted by CaptainBlack
The numerator behaves like $x(1+a+b)-x^3(a/2 + b/6)$ for small $x$ and we need this to be $\sim x^3$ so $a=-5/2$ and $b=3/2$.
CB

I am not that brilliant in this and need a little more explanation ....

4. Originally Posted by zorro
I am not that brilliant in this and need a little more explanation ....
Replace sin and cos with their series expansions up to and including the third and second degree terms respectivly and simplify. Alternatively try L'Hopital's rule.

CB

5. ## Sir i am not that brilliant to replace sin and cos with series expansion

Originally Posted by CaptainBlack
Replace sin and cos with their series expansions up to and including the third and second degree terms respectivly and simplify. Alternatively try L'Hopital's rule.

CB

Sir i am not that brilliant to replace sin and cos with series expansion...If u could just guide me through the first few step i would be able to do it .....sorry for the language but i am very frustrated right now with this question

6. Originally Posted by zorro
Sir i am not that brilliant to replace sin and cos with series expansion...If u could just guide me through the first few step i would be able to do it .....sorry for the language but i am very frustrated right now with this question
You will find the required series for sin x and cos x here: Taylor series - Wikipedia, the free encyclopedia

Now apply the advice given in post #4.

7. ## Is this correct

Originally Posted by mr fantastic
You will find the required series for sin x and cos x here: Taylor series - Wikipedia, the free encyclopedia

Now apply the advice given in post #4.

$\lim_{x \to 0} \frac{x(1 - a cosx) - bsinx}{x^3} = 1$

$sinx = \sum_{n = 0}^{ \infty } \frac{(-1)^n}{(2n + 1)} = x - \frac{x^3}{3!}$

$cosx = \sum_{n = 0}^{ \infty} \frac{(-1)^n}{2n} x^{2n} = 1 - \frac{x^2}{2!}$

Substituting in the equation we get

$\lim_{x \to 0} \frac{x \left[ 1 + a \left(1 - \frac{x^2}{2!} \right) \right] - b \left(x - \frac{x^3}{3!} \right)}{x^3}$

$\lim_{x \to 0} \frac{x \left[ 1 + a \left( \frac{2 - x^2}{2} \right) \right] - b \left(\frac{6x - x^3}{6} \right)}{x^3}$

$\lim_{x \to 0} \frac{f(x)}{f(y)} = \lim_{x \to 0} \frac{f'(x)}{f'(y)}$

$\lim_{x \to 0} \frac{f'(x)}{f'(y)}= \lim_{x \to 0} \frac{ \frac{6x + 6ax - 3ax^3 - 6bx +bx^3}{6}}{3x^2}$

$\lim_{x \to 0} \frac{(1 + a - b) - \frac{1}{2} x^2(3a - b)}{3x^2}$

I am stuck here !!!!
What i have done is it correct or no?

8. Originally Posted by zorro
$\lim_{x \to 0} \frac{x(1 - a cosx) - bsinx}{x^3} = 1$

$sinx = \sum_{n = 0}^{ \infty } \frac{(-1)^n}{(2n + 1)} = x - \frac{x^3}{3!}$

$cosx = \sum_{n = 0}^{ \infty} \frac{(-1)^n}{2n} x^{2n} = 1 - \frac{x^2}{2!}$

Substituting in the equation we get

$\lim_{x \to 0} \frac{x \left[ 1 + a \left(1 - \frac{x^2}{2!} \right) \right] - b \left(x - \frac{x^3}{3!} \right)}{x^3}$
The original limit is equal to (since we are ignoring a term or order $x$ in our truncation of the series):

$\lim_{x \to 0} \frac{x \left[ 1 + a \left( \frac{2 - x^2}{2} \right) \right] - b \left(\frac{6x - x^3}{6} \right)}{x^3} =$ $
\lim_{x \to 0} \frac{x(1+a-b)-x^3(a/2 +b/6)}{x^3}=$
$
\lim_{x \to 0} \frac{(1+a-b)}{x^2}- (a/2 +b/6)$

Now if this last limit is equal to $1$, we must have $1+a-b=0$ (as otherwise the limit would be infinite), and $a/2+b/6=-1$

CB

9. I'll just add that the equation $a/2+b/6=-1$ can be re-written as $3a + b = -6$, which may be easier to work with ....

10. ## Is the answer correct ?

Originally Posted by mr fantastic
I'll just add that the equation $a/2+b/6=-1$ can be re-written as $3a + b = -6$, which may be easier to work with ....

since $(1 + a - b) = 0$

therefore, $a = b - 1$

Substituting $a = b - 1$ in $3a + b = -6$

$3(b - 1) + b = -6$

$3b - 3 + b = -6$

$4b = -6 + 3$

$4b = -3$

$b$ = - $\frac{3}{4}$

Substiting b in $3a + b = -6$ weget

$3a + \left( - \frac{3}{4} \right)$ = $-6$

$3a$ = $-6 + \frac{3}{4}$

$3a$ = $\frac{-24 + 3}{4}$

$3a$ = $\frac{-21}{4}$

$a$ = $3 \times \frac{-21}{4}$

a = - $\frac{63}{4}$

Is this correct ????

11. Originally Posted by zorro
since $(1 + a - b) = 0$

therefore, $a = b - 1$

Substituting $a = b - 1$ in $3a + b = -6$

$3(b - 1) + b = -6$

$3b - 3 + b = -6$

$4b = -6 + 3$

$4b = -3$

$b$ = - $\frac{3}{4}$

Substiting b in $3a + b = -6$ weget

$3a + \left( - \frac{3}{4} \right)$ = $-6$

$3a$ = $-6 + \frac{3}{4}$

$3a$ = $\frac{-24 + 3}{4}$

$3a$ = $\frac{-21}{4}$

$a$ = $3 \times \frac{-21}{4}$

a = - $\frac{63}{4}$

Is this correct ????

1 + a - b = 0 .... (1)

3a + b = -6 .... (2)

You should know how to solve simultaneous equations at the level you're apparently studying at. Personally, I'd just add equations (1) and (2) and solve for $a$. Then I'd substitute my value for $a$ into (1) and solve for $b$.

12. ## Trying the L'Hospital rule

Originally Posted by CaptainBlack
Replace sin and cos with their series expansions up to and including the third and second degree terms respectivly and simplify. Alternatively try L'Hopital's rule.

CB

I am trying to solve the same question by using the L'Hospital rule , but i have got stuck at the following step

$\lim_{x \to 0} \ \frac{x(1+acosx)-bsinx}{x^3} = 1$

$f'(x) \ = \ x(1+acosx) - bsinx$

$= 1- ax (sinx) + (cosx) - b(cosx)$

and $g'(x) = 3 x^2$

$\lim_{x \to 0} \frac{ 1- ax (sinx) + (cosx) - b(cosx)}{3 x^2}$

$\lim_{x \to 0} \frac{1}{3x^2} - \frac{a}{3} \frac{sinx}{x} + (a-b). \frac{cosx}{3x^2}$

$\lim_{x \to 0} \frac{1}{3x^2} - \frac{a}{3} (1) + \lim_{x \to 0} (a-b). \frac{cosx}{3x^2}$
I am stuck here ............

13. Originally Posted by zorro
I am trying to solve the same question by using the L'Hospital rule , but i have got stuck at the following step

$\lim_{x \to 0} \ \frac{x(1+acosx)-bsinx}{x^3} = 1$

${\color{red}{f(x)}} \ = \ x(1+acosx) - bsinx$

$\color{red}f'(x)=(1+a \cos(x))+x(-a\sin(x))+b \cos(x)$

....... $\color{red}= 1- ax \sin(x) + a\cos(x) - b\cos(x)$

and $g'(x) = 3 x^2$

$\lim_{x \to 0} \frac{{\color{red} 1- ax \sin(x) + a\cos(x) - b\cos(x) }}{3 x^2}$
Stop here. If this limit is finite the numerator must be equal to zero at $x=0$. Since otherwise the limit does not exist. So:

$1+a-b=0$

or:

$b=1+a$

use this to substitute for $b$ in the last limit. Now another stage of L'Hopitals rule is needed

CB

14. ## I did get u

Originally Posted by CaptainBlack
Stop here. If this limit is finite the numerator must be equal to zero at $x=0$. Since otherwise the limit does not exist. So:

$1+a-b=0$

or:

$b=1+a$

use this to substitute for $b$ in the last limit. Now another stage of L'Hopitals rule is needed

CB

Where should i substitute the value of b ???

15. Originally Posted by zorro
Where should i substitute the value of b ???
Here:

$

\lim_{x \to 0} \frac{{ 1- ax \sin(x) + a\cos(x) - b\cos(x) }}{3 x^2}
$
$
=\lim_{x \to 0}\frac{ 1- ax \sin(x) + a\cos(x) - (a+1)\cos(x)}{3x^2}$
$
=\lim_{x \to 0}\frac{ 1- ax \sin(x) - \cos(x)}{3x^2}$

CB