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FP2 Taylor's expansion

If you have a function in the form y=ln[1+cos(x)], do you just expand it as if it were ln(1+x) but replace the x with cos(x)?

Also, could you do this for any other function of x that isn't x, for example
y=ln[1+(x^2)]?

Thanks for any help!

Reply 1

Zacken

Original post by PhyM23

If you have a function in the form y=ln[1+cos(x)], do you just expand it as if it were ln(1+x) but replace the x with cos(x)?

This will get you a series in

\cos x

. Instead, expand

\cos x

first - then you'll get

Unparseable latex formula:

\ln(1 + \text{polynomial})}

and expand this normally.

Also, could you do this for any other function of x that isn't x, for example
y=ln[1+(x^2)]?

Thanks for any help!

In this case, you can just replace

x

with

x^2

everywhere in the expansion and you're good to go.

Reply 2

φM23

Original post by Zacken

This will get you a series in

\cos x

. Instead, expand

\cos x

first - then you'll get

Unparseable latex formula:

\ln(1 + \text{polynomial})}

and expand this normally.

In this case, you can just replace

x

with

x^2

everywhere in the expansion and you're good to go.

PRSOM

Brilliant. Thank you for confirming this

Reply 3

joostan

Original post by PhyM23

Yes, you can, provided

|f(x)|<1

, though this is often not the most convenient way to find the Taylor series, and a little fiddling needs to be done if

|f(x)|>1

Reply 4

φM23

Original post by joostan

Yes, you can, provided

|f(x)|<1

, though this is often not the most convenient way to find the Taylor series, and a little fiddling needs to be done if

|f(x)|>1

Thank you for the reply!

Please may you explain what you mean by this. Why isn't it the most convenient way, and what's the difference between what happens when

|f(x)|<1

as opposed to when

|f(x)|>1

Reply 5

Zacken

Original post by PhyM23

Thank you for the reply!

Please may you explain what you mean by this. Why isn't it the most convenient way, and what's the difference between what happens when

|f(x)|<1

as opposed to when

|f(x)|>1

The series for

\ln (1 + x)

is only valid for

|x| < 1

(otherwise, your series would just diverge to infinity, this condition is given next to the series in your textbook or formula booklet). So if you replace

x

f(x)

your series is now only valid for

|f(x)| < 1

Reply 6

joostan

Original post by PhyM23

Thank you for the reply!

Please may you explain what you mean by this. Why isn't it the most convenient way, and what's the difference between what happens when

|f(x)|<1

as opposed to when

|f(x)|>1

Well normally one is interested in a power series of the form

\displaystyle \sum_{n=0}^{\infty} a_nx^n

, but

\cos(x)

is itself such an expansion, and you end up with:

\displaystyle \sum_{n=0}^{\infty} a_n\cos^n(x)=\sum_{n=0}^{\infty}\left(\sum_{m=0}^{\infty}\dfrac{(-1)^mx^{2m}}{(2m)!}\right)^n

.
As I'm sure you can appreciate this is not very pleasant.

For suitable

f

with

|f(x)|>1

one would need to write

\ln(1+f(x))=\ln(f(x))+\ln\left(1+\dfrac{1}{f(x)}\right)

in order to use the expansion you wanted.

Reply 7

φM23

Original post by Zacken

The series for

\ln (1 + x)

is only valid for

|x| < 1

(otherwise, your series would just diverge to infinity, this condition is given next to the series in your textbook or formula booklet). So if you replace