Series expansion of logarithmic function

AI Thread Summary
To find the first three non-zero terms in the series expansion of ln(5+p) for small values of p, it is suggested to rewrite the function as ln(5[1+(p/5)]). This allows the use of logarithmic properties to separate the terms. The series expansion can then be applied to the term ln(1+(p/5)), which is suitable for small p. The accuracy of the expansion diminishes as p moves away from -4, but the focus remains on deriving the first three non-zero terms. The approach emphasizes the importance of correctly applying logarithmic identities for effective expansion.
seboastien
Messages
52
Reaction score
0

Homework Statement


Find first three non zero terms in series expansion where the argument of funstion is small

ln(5+p)

Homework Equations





The Attempt at a Solution



The only way I could think how to do this is by saying ln(5+p) = ln(1+(4+p)) and expanding to

(4+p)- 1/2(4+p)^2 + 1/3(4+p)^3 - ... however, I imagine that this would only work if p was approx -4.
 
Physics news on Phys.org
Come on guys! I really need to know how to do this!
 
seboastien said:

Homework Statement


Find first three non zero terms in series expansion where the argument of funstion is small

ln(5+p)

Homework Equations





The Attempt at a Solution



The only way I could think how to do this is by saying ln(5+p) = ln(1+(4+p)) and expanding to

(4+p)- 1/2(4+p)^2 + 1/3(4+p)^3 - ... however, I imagine that this would only work if p was approx -4.
Well, the accuracy, for any finite polynomial expansion, deteriorates as p gets farther from -4 but is the accuracy really relevant? You are only asked to "Find first three non zero terms".
 
How do I find the first three non-zero terms of ln(5+p), I'm pretty sure that my answer is wrong.
 
You'll want to write this as ln( 5 [ 1+(p/5) ] ) ; then use the properties of logarithms to write it as two terms, one of which is the term you would do the series expansion for.
 
Thank you
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Back
Top