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come up with a formula for the function f(x) whose expansion in powers of x is ((x^2)/2! )+((x^3)/3! )+((x^4)/4! )+((x^5)/5! )+...+((x^n)/n! )
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looks a lot like \(e^x\) except you are missing the first two terms
clear or no?
i would say \(f(x)=e^x-1-x\)
it's not e^x maybe i think it's second one thankyou
oh it is not \(e^x\) because you are missing the first two terms of the expansion
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\(e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\) so \[e^x-1-x=\frac{x^2}{2!}+\frac{x^2}{3!}+...\]
can you help me on another question?
sure go ahead and post, i will help if i can
it's two parametric function to get the volume
probably not, but post in a new question and i bet you will get lots of replies i will try it too
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i hate integrating, it is a pain
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