Find the value of the a and b

**zorro** · December 3rd 2009, 03:40 PM

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. $

**CaptainBlack** · December 3rd 2009, 10:12 PM

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

**zorro** · December 4th 2009, 02:19 PM

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 ....

**CaptainBlack** · December 5th 2009, 01:24 AM

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

**zorro** · December 7th 2009, 12:24 AM

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

**mr fantastic** · December 7th 2009, 02:32 AM

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.

**zorro** · December 7th 2009, 02:57 PM

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?

**CaptainBlack** · December 7th 2009, 08:05 PM

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

**mr fantastic** · December 8th 2009, 12:56 AM

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 ....

**zorro** · December 8th 2009, 01:47 AM

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 ????

**mr fantastic** · December 8th 2009, 01:52 AM

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 ????

Does your answer satisfy

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 $b$ .

**zorro** · December 22nd 2009, 07:51 PM

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 ............

**CaptainBlack** · December 22nd 2009, 11:03 PM

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

**zorro** · December 23rd 2009, 01:01 AM

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 ???

**CaptainBlack** · December 23rd 2009, 01:56 AM

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

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Find the value of the a and b

Could u please explain me this a little more

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

Is this correct

Is the answer correct ?

Trying the L'Hospital rule

I did get u

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