# Find the value of the a and b

• Dec 3rd 2009, 03:40 PM
zorro
Find the value of the a and b
Question :
Find the value of a and b in order that

$\displaystyle \lim_{x \to 0} \frac{x(1 + a cosx) - b sinx}{x^3} = 1.$
• Dec 3rd 2009, 10:12 PM
CaptainBlack
Quote:

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

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

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

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

CB
• Dec 4th 2009, 02:19 PM
zorro
Could u please explain me this a little more
Quote:

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

I am not that brilliant in this and need a little more explanation ....
• Dec 5th 2009, 01:24 AM
CaptainBlack
Quote:

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
• Dec 7th 2009, 12:24 AM
zorro
Sir i am not that brilliant to replace sin and cos with series expansion
Quote:

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(Headbang)
• Dec 7th 2009, 02:32 AM
mr fantastic
Quote:

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(Headbang)

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.
• Dec 7th 2009, 02:57 PM
zorro
Is this correct
Quote:

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.

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

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

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

Substituting in the equation we get

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

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

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

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

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

I am stuck here !!!!
What i have done is it correct or no?
• Dec 7th 2009, 08:05 PM
CaptainBlack
Quote:

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

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

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

Substituting in the equation we get

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

The original limit is equal to (since we are ignoring a term or order $\displaystyle x$ in our truncation of the series):

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

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

CB
• Dec 8th 2009, 12:56 AM
mr fantastic
I'll just add that the equation $\displaystyle a/2+b/6=-1$ can be re-written as $\displaystyle 3a + b = -6$, which may be easier to work with ....
• Dec 8th 2009, 01:47 AM
zorro
Quote:

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

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

therefore, $\displaystyle a = b - 1$

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

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

$\displaystyle 3b - 3 + b = -6$

$\displaystyle 4b = -6 + 3$

$\displaystyle 4b = -3$

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

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

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

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

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

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

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

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

Is this correct ????
• Dec 8th 2009, 01:52 AM
mr fantastic
Quote:

Originally Posted by zorro
since $\displaystyle (1 + a - b) = 0$

therefore, $\displaystyle a = b - 1$

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

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

$\displaystyle 3b - 3 + b = -6$

$\displaystyle 4b = -6 + 3$

$\displaystyle 4b = -3$

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

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

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

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

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

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

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

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

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 $\displaystyle a$. Then I'd substitute my value for $\displaystyle a$ into (1) and solve for $\displaystyle b$.
• Dec 22nd 2009, 07:51 PM
zorro
Trying the L'Hospital rule
Quote:

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

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

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

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

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

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

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

$\displaystyle \lim_{x \to 0} \frac{1}{3x^2} - \frac{a}{3} (1) + \lim_{x \to 0} (a-b). \frac{cosx}{3x^2}$
I am stuck here ............
• Dec 22nd 2009, 11:03 PM
CaptainBlack
Quote:

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

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

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

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

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

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

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

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

$\displaystyle 1+a-b=0$

or:

$\displaystyle b=1+a$

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

CB
• Dec 23rd 2009, 01:01 AM
zorro
I did get u
Quote:

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

$\displaystyle 1+a-b=0$

or:

$\displaystyle b=1+a$

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

CB

Where should i substitute the value of b ???
• Dec 23rd 2009, 01:56 AM
CaptainBlack
Quote:

Originally Posted by zorro
Where should i substitute the value of b ???

Here:

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

CB