Manipulating Equivalent Expressions -4/(4x+1)^(1/2) to (1/4+x)^(-1/2)

In summary, the conversation discusses a problem with manipulating an expression and finding the Taylor series of a function. The individual attempts to use integration and index laws but realizes they are not equivalent. They also receive help from others in finding the correct form of the expression and finding the constant needed to solve the problem.
  • #1
cosmictide
37
0

Homework Statement



Hi guys, I'm getting confused trying to manipulate an expression. Any help here would be hugely appreciated.

How do I manipulate the expression -4 / (4x + 1)1/2 to become (1/4 + x)-1/2

The Attempt at a Solution



I've tried dividing by -4 to get and used an index law to get (1/2)*(x+1)-1/2 but that's clearly not the expression I want. I'd really appreciate if someone could point me in the right direction.

Thanks in advance. :smile:
 
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  • #2
cosmictide said:

Homework Statement



Hi guys, I'm getting confused trying to manipulate an expression. Any help here would be hugely appreciated.

How do I manipulate the expression -4 / (4x + 1)1/2 to become (1/4 + x)-1/2

The Attempt at a Solution



I've tried dividing by -4 to get and used an index law to get (1/2)*(x+1)-1/2 but that's clearly not the expression I want. I'd really appreciate if someone could point me in the right direction.

Thanks in advance. :smile:
The two expressions you wrote are actually not equal. You can easily check by plugging a value of x, for example x=0.
 
  • #3
nrqed said:
The two expressions you wrote are actually not equal. You can easily check by plugging a value of x, for example x=0.

Hi, thanks for your reply. :smile:

I know they're not equivalent, what I'm trying to do is manipulating the first expression to become the second expression if that makes sense.
 
  • #4
cosmictide said:
Hi, thanks for your reply. :smile:

I know they're not equivalent, what I'm trying to do is manipulating the first expression to become the second expression if that makes sense.

Since they are not equivalent it is impossible to manipulate one to bring it in the form of the other.
 
  • #5
nrqed said:
Since they are not equivalent it is impossible to manipulate one to bring it in the form of the other.

Okay so I must be missing something else. I'm trying to find the Taylor series about 0 for the function g(x) = (1/4 + x)-1/2 from the series of the function f(x) =

(1/4 + x)-3/2 = 8 - 48x + 240x2 - 1120x3 by using integration.

I've done the integration to get

-4 / (4x + 1)1/2 = c + 8x - 24x2 + 80x3 - ...

Taking x=0 I calculated the constant c = -4 leaving me with

-4 / (4x + 1)1/2 = -4 + 8x - 24x2 + 80x3 - ...

This is where I'm struggling. If I am asked to find the Taylor series about 0 for the function g(x) = (1/4 + x)-1/2 then shouldn't the left-hand side of the equation above be equivalent to this or am I missing something else?

Thanks. :smile:
 
  • #6
Since the problem involves Taylor series, I moved it to the Calculus section.
 
  • #7
cosmictide said:
Okay so I must be missing something else. I'm trying to find the Taylor series about 0 for the function g(x) = (1/4 + x)-1/2 from the series of the function f(x) =

(1/4 + x)-3/2 = 8 - 48x + 240x2 - 1120x3 by using integration.

I've done the integration to get

-4 / (4x + 1)1/2 = c + 8x - 24x2 + 80x3 - ...

Taking x=0 I calculated the constant c = -4 leaving me with

-4 / (4x + 1)1/2 = -4 + 8x - 24x2 + 80x3 - ...

This is where I'm struggling. If I am asked to find the Taylor series about 0 for the function g(x) = (1/4 + x)-1/2 then shouldn't the left-hand side of the equation above be equivalent to this or am I missing something else?

Thanks. :smile:

You had a similar problem in the Calculus and Beyond Forum, and made the same error there. The fact is that you cannot get the series for ##(1/4 + x)^{-1/2}## directly by integrating the series for ##(1/4+x)^{-3/2}##. You ought to be able to see the reason: if ##F'(x)
\neq G(x)## the series for ##G(x)## is not gotten from that of ##F(x)## by differentiating and vice-versa (integrating instead of differentiating). However, if ##F'(x) = c G(x)## for some known constant ##c## then it is not difficult to pass from one to the other. Do you see how?

Note: typo edited out now.
 
Last edited:
  • #8
cosmictide said:
Okay so I must be missing something else. I'm trying to find the Taylor series about 0 for the function g(x) = (1/4 + x)-1/2 from the series of the function f(x) =

(1/4 + x)-3/2 = 8 - 48x + 240x2 - 1120x3 by using integration.

I've done the integration to get

-4 / (4x + 1)1/2 = c + 8x - 24x2 + 80x3 - ...

Taking x=0 I calculated the constant c = -4 leaving me with

-4 / (4x + 1)1/2 = -4 + 8x - 24x2 + 80x3 - ...

This is where I'm struggling. If I am asked to find the Taylor series about 0 for the function g(x) = (1/4 + x)-1/2 then shouldn't the left-hand side of the equation above be equivalent to this or am I missing something else?

Thanks. :smile:

I did not check all the numbers but it does look good to me, so far.
Ok, now you do need to rewrite the left side in the form

[itex] \frac{C}{(1/4 +x)^{1/2}} [/itex]

where C is a constant.
Once you have that, just divide both sides of your result by the constant and you will have your final answer.

So what you need to do now is to find the constant required so that

[itex] \frac{C}{(1/4 +x)^{1/2}} = - \frac{4}{(4x +1)^{1/2}} [/itex]
 
  • #9
Thanks guys, much appreciated.
 

What is an Equivalent Expression?

Equivalent expression refers to two mathematical expressions that have the same value, even though they may look different. In other words, they have the same solution when evaluated.

How do you determine if two expressions are equivalent?

To determine if two expressions are equivalent, you can simplify both expressions and see if they result in the same value. Another way is to plug in different values for the variables and see if the two expressions have the same output.

What are some examples of equivalent expressions?

Some examples of equivalent expressions include: 2x and x + x, 3y - 5 and 2y + y - 5, and 4x^2 and 2x * 2x.

Why is it important to understand equivalent expressions?

Understanding equivalent expressions is important because it helps in simplifying complex expressions, solving equations, and making connections between different mathematical concepts.

How can I practice identifying equivalent expressions?

You can practice identifying equivalent expressions by solving math problems, creating your own expressions and simplifying them, and using online resources such as practice worksheets and quizzes.

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