expand f(x)=2^x in Maclaurin series. and what is the general term for this power series expansion?
N NJ_84 New member Joined Apr 20, 2012 Messages 7 Apr 22, 2012 #1 expand f(x)=2^x in Maclaurin series. and what is the general term for this power series expansion?
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Apr 22, 2012 #2 You WILL need some derivatives. Let's see a couple.
N NJ_84 New member Joined Apr 20, 2012 Messages 7 Apr 23, 2012 #3 i could probably use the u substitution ? 2=e^ln(2) d/dx(a^x)=(e^(ln a))^x d/dx=e^(ln a /x) 2^x= e^(ln 2 /x) u=ln(2), it gives us e^u/x is that right?
i could probably use the u substitution ? 2=e^ln(2) d/dx(a^x)=(e^(ln a))^x d/dx=e^(ln a /x) 2^x= e^(ln 2 /x) u=ln(2), it gives us e^u/x is that right?
N NJ_84 New member Joined Apr 20, 2012 Messages 7 Apr 23, 2012 #4 NJ_84 said: i could probably use the u substitution ? 2=e^ln(2) y=2^x=(e^ln2)^x=e^((ln2)*x) u=(ln2)*x e^u ? Click to expand...
NJ_84 said: i could probably use the u substitution ? 2=e^ln(2) y=2^x=(e^ln2)^x=e^((ln2)*x) u=(ln2)*x e^u ? Click to expand...
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Apr 23, 2012 #5 I can't tell what it is you are doing. Are you trying to find an antiderivative? \(\displaystyle f(x) = 2^{x}\) \(\displaystyle f'(x) = 2^{x}\cdot log(2)\) \(\displaystyle f"(x) = 2^{x}\cdot [log(2)]^{2}\)
I can't tell what it is you are doing. Are you trying to find an antiderivative? \(\displaystyle f(x) = 2^{x}\) \(\displaystyle f'(x) = 2^{x}\cdot log(2)\) \(\displaystyle f"(x) = 2^{x}\cdot [log(2)]^{2}\)
