# Taylor series with using geometric series

• Pietervv
In summary: Taking the third derivative of both sides is one of those manipulations. I'm glad you were able to solve it. In summary, the key to determining the Taylor series of f(x) at x=c(≠B) using a geometric series is to first determine the geometric series for g(x)=A/(x-B) and then take the third derivative of both sides of the equation obtained from this geometric series.
Pietervv
Member warned about not using the homework template
The question is:

Determine the Taylor series of f(x) at x=c(≠B) using geometric series

f(x)=A/(x-B)4

My attempt to the solution is:

4√f(x) = 4√A/((x-c)-B = (4√A/B) * 1/(((x-c)/B)-1) = (4√A/-B) * 1/(1-((x-c)/B))

using geometric series : 4√f(x) = (4√A/-B) Σ((x-c)/B)n

f(x)= A/B4 * Σ((x-c)/B)4n with abs((x-c)/B)<1

But i am not very sure if this is true.

f(x)= A/(x- B)4 is NOT the same as A/(x- c- B)^4 so $\sqrt[4]{f(x)}$ is NOT equal to $\sqrt[4]{A/((x- c)- B)^4}$. To put the "c" in you need
$f(x)= A/((x- c)- (B- c))^4$.

Last edited by a moderator:
Pietervv said:
The question is:

Determine the Taylor series of f(x) at x=c(≠B) using geometric series

f(x)=A/(x-B)4

My attempt to the solution is:

4√f(x) = 4√A/((x-c)-B = (4√A/B) * 1/(((x-c)/B)-1) = (4√A/-B) * 1/(1-((x-c)/B))

using geometric series : 4√f(x) = (4√A/-B) Σ((x-c)/B)n

f(x)= A/B4 * Σ((x-c)/B)4n with abs((x-c)/B)<1

But i am not very sure if this is true.

As HallofIvy has pointed out, the manipulations you have done above are incorrect. After correcting them, you will be left with an an expression that involves, in part, an expansion of the form
$$\frac{1}{(1-u)^4} = c_0 + c_1 u + c_2 u^2 + \cdots .$$
That is a not a "geometric" series, because you have the 4th power in the denominator. I think the title of the problem is incorrect---you cannot do it using a geometric series (at least, not directly). However, you can start with a geometric series and do some calculus manipulations on it to obtain the series above.

Last edited:
HallsofIvy said:
f(x)= A/(x- B)4 is NOT the same as A/(x- c- B)^4 so $\sqrt[4]{f(x)}$ is NOT equal to $\sqrt[4]{A/((x- c)- B)^4}$. To put the "c" in you need
$f(x)= A/((x- c)- (B- c))^4$.

Then the c disappears in the nominator and my f(x) is again f(x)= A/(x- B)4 after put the c in??

And is my tactic for solving this problem, with taking the 4√ the good tactic, or is it totaly the wrong way to solve it?

Pietervv said:
Then the c disappears in the nominator and my f(x) is again f(x)= A/(x- B)4 after put the c in??

And is my tactic for solving this problem, with taking the 4√ the good tactic, or is it totaly the wrong way to solve it?

No, it is totally wrong. In simplified notation, here is what you said:
$$\left(\sum_n a_n y^n\right)^4 = \sum_n a_n^4 \, y^{4n} \; \Longleftarrow \; \text{False!}$$

For the people who are still interested, after watched the solution, the key was to first determine the series of g(x)=A/(x-B) using the geometric series (1) and then take the third derivative of both sides of the equation obtained by step 1

Pietervv said:
For the people who are still interested, after watched the solution, the key was to first determine the series of g(x)=A/(x-B) using the geometric series (1) and then take the third derivative of both sides of the equation obtained by step 1

Yes, that is exactly what I meant in Post #3 when I said to start with a geometric series and perform calculus manipulations on it.

## What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms. It is used to approximate a function by using a polynomial with an infinite number of terms.

## What is the relationship between Taylor series and geometric series?

A geometric series is a special case of a Taylor series where the terms have a common ratio. In other words, the Taylor series can be written as a geometric series by factoring out a common ratio from the terms.

## How do you find the coefficients in a Taylor series using geometric series?

The coefficients in a Taylor series can be found by using the formula cn = f(n)(a)/n!, where cn is the coefficient of the nth term, f(n)(a) is the nth derivative of the function at the point a, and n! is the factorial of n. This formula is based on the fact that the coefficients in a geometric series can be found by using the formula an = a1rn-1, where a1 is the first term and r is the common ratio.

## What are the applications of Taylor series with using geometric series?

Taylor series with using geometric series have various applications in mathematics and physics. They are commonly used to approximate complex functions and solve differential equations. They also have applications in areas such as signal processing, control systems, and numerical analysis.

## What are the limitations of using Taylor series with geometric series?

One limitation of using Taylor series with geometric series is that they only provide an approximation of a function and may not accurately represent the exact behavior of the function. Additionally, they can only be used for functions that are infinitely differentiable, which may not be the case for all functions. Furthermore, the convergence of a Taylor series may be limited to a specific interval, making it unsuitable for representing the entire function.

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