How Do You Apply Taylor Series to Find Terms for f(x) = ln(3+x)?

morbello
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ive got a question to ask I am working on taylor series and want to know

f(x)=In(3+x) and g(x)=In (1+x)

by writing

In(3+x)=In3+In(1+1/3x)
im asked to use substitution in one off the standard taylor series given in the course.to find about 0 for f

explicitly all terms up to term in x^3

im not sure were to start.

I have

x+1/3 x^2+1*2/3 x^3+1*2*3/3 x^4+...
 
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I think you forgot to provide the function you're trying to find its Taylor's polynomial.
 
the function is ln(3+x) but you are told to write it as ln3 + ln(1 + x/3)

I believe you just need the standard series of ln (1+x) but sub in x/3 for the x.

So you would end up with

ln3 + x/3 - (x^2)/18 + (x^3)/81
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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