Why is the taylor series for ln(x+1) so similar to the tailor series sin(x) or cos(x), yet the functions themselves are not at all?
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I’m a bit confused, they aren’t the same at all. The taylor series terms for ln(x+1) are x^n / n alternating signs.
The taylor series for sin(x) has terms x^2n+1 / (2n+1)! (alternating signs).
The taylor series for sin(x) has terms x^2n / (2n)! (alternating signs).
These are very different.
I haven’t looked at the details in years, but the equation:
e^(ix) = cos(x) + i sin(x) , where i is the imaginary number sqrt(-1)
should show a relationship between the functions.
What does “so similar” mean?
The whole point of Taylor series is that, for certain convenient functions (analytic functions), the value of the function at values “near” a point of interest can be approximated with increasing accuracy as you increase the number of polynomial terms. That is, if you know the value of sin(x) at x = 1 and you want to figure out what sin(x) equals at x = 1.1, you don’t need to know that value explicitly; you only need to know what sin'(x)|x=1 and sin”(x)|=1 and so on are, and the more terms you add, the better you can approximate sin(1.1) by only knowing sin(x)|x=1 and its derivatives.
The thing is, this means literally all Taylor series will look “similar” in the sense that they all look like polynomials.
To be perfectly honest with you, the Taylor series are really not that similar. The coefficients are very different and sin(x)’s expansion only has odd terms (and cosine’s has only even terms) whereas ln(x)’s Taylor series has even and odd terms.
All Taylor series are constructed using the same general form. The similarities that you recognise for the series of different functions are the “bones” of this generic shape. The differences in how these forms translate back to their function comes from whether different terms in the series cancel one another.
OP, I think you might need to give us some clarification.
There isn’t much similarity at all
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