# how do Taylor series adjustments affect original equation?

• Sep 3rd 2009, 02:26 PM
rainer
how do Taylor series adjustments affect original equation?
If I make adjustments to a function's Taylor series (for example raise all of the powers by 1 or change the signs), how do these changes alter the original function? Or rather what is the process for finding out how the original function has been altered by such changes? Does such a process even exist?
• Sep 3rd 2009, 02:44 PM
Calculus26
Usually it is done the other way around given the Taylor Series of a function what does a change in the function do to the series.

For Eg e^x = sum (x^k/(k!))

e^(2x) =sum [(2x)^k/k!] = sum(2^k) (x^k)/k!)

Or 1/(1-x) = sum (x^k)

1/(1+x) = sum ((-x)^k) = sum ((-1)^k x^k)
• Sep 4th 2009, 08:21 AM
rainer
Thanks Calculus 26.

The example you provide is fairly straightforward. Any chance you could help me out with a more complicated series? In particular I want to take the series for arcosh(x), square the first part of all of the terms, subtract 1 from the denominator of the second part of all the terms, subtract the "ln2x", add 1 to the entire series, and multiply the entire series by 2pi. How do these adjustments alter the original arcosh(x) function?

$arcosh(x)= \ln 2x - \sum_{n=1}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right) \frac {x^{-2n}} {(2n)} , \qquad x > 1$

i.e.

$arcosh(x)= \ln 2x - \left( \left( \frac {1} {2} \right) \frac {x^{-2}} {2} + \left( \frac {1 \cdot 3} {2 \cdot 4} \right) \frac {x^{-4}} {4} + \left( \frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} \right) \frac {x^{-6}} {6} +\cdots \right)$

The adjustments I want to make:

$2\pi \left( \ln 2x - \ln 2x + 1 - \sum_{n=1}^\infty \left( \frac {(2n)!} {2^{2n}(n!)^2} \right)^2 \frac {x^{-2n}} {(2n-1)} \right) , \qquad x > 1$

Quote:

Originally Posted by Calculus26
Usually it is done the other way around given the Taylor Series of a function what does a change in the function do to the series.

For Eg e^x = sum (x^k/(k!))

e^(2x) =sum [(2x)^k/k!] = sum(2^k) (x^k)/k!)

Or 1/(1-x) = sum (x^k)

1/(1+x) = sum ((-x)^k) = sum ((-1)^k x^k)