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In IB Calculus Option you are introduced to number of different tests. They are Divergence Test, Comparison Test (simple and limit forms), Absolute Convergence Test, Alternating Series Test, Ratio Test an...
a) f(x)=ln(x), so f'(x)=1/x. By MVT, f'(c) = (f(b)-f(a))/(b-a) = (ln b - ln a)/(b-a) = ln(b/a)/(b-a), where c lies between a and b. Now, since 1/x is a decreasing function, and a < c < b, we get 1/...
There are many methods to check it, but the the most useful is: By comparison: if we find a series whose terms are greater than the given ones which is known to be convergent (an armonic series for exampl...
Here we are examining the imaginary number ‘i’, defined as the square root (sqrt) of ‘-1’.We begin by using Euler’s identity:e^(iπ)+1=0e^(iπ)=-1Since sqrt(x) is the same as x^(1/2):(e^(iπ))^(...
Here, we must remember the definition of a bijection. To be bijective, a function must be both injective and surjective. For a function to be injective, we must have that f(a) = f(b) implies that a = b...
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