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... Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t. Then (d) lim (sn /tn) = s/t, provided that tn 0 for all n and t 0. Proof: Since sn /tn = sn(1/tn), it suffices from part (c) to show that lim (1/tn) = 1/t. That is, given > 0, we must make t tn ...
... Suppose that (sn) and (tn) are convergent sequences with lim sn = s and lim tn = t. Then (d) lim (sn /tn) = s/t, provided that tn 0 for all n and t 0. Proof: Since sn /tn = sn(1/tn), it suffices from part (c) to show that lim (1/tn) = 1/t. That is, given > 0, we must make t tn ...
B Veitch Calculus 2 Study Guide This study guide is in
... (c) Find f 0 (x) i. Find the critical values, all x-values where f 0 (x) = 0 or when f 0 (x) does not exist. ii. Plot the critical values on a number line. iii. Find increasing / decreasing intervals using number line iv. Use The First Derivative Test to find local maximums / minimums (if any exist) ...
... (c) Find f 0 (x) i. Find the critical values, all x-values where f 0 (x) = 0 or when f 0 (x) does not exist. ii. Plot the critical values on a number line. iii. Find increasing / decreasing intervals using number line iv. Use The First Derivative Test to find local maximums / minimums (if any exist) ...
Notes on space complexity of integration of computable real
... It is known [1] that real function g(x) = 0 f (t)dt is polynomial-time computable real function on interval [0, 1] iff FP = #P wherein f is a polynomial-time computable real function on interval [0, 1]. It means integration of polynomial-time computable real functions is as hard as string functions ...
... It is known [1] that real function g(x) = 0 f (t)dt is polynomial-time computable real function on interval [0, 1] iff FP = #P wherein f is a polynomial-time computable real function on interval [0, 1]. It means integration of polynomial-time computable real functions is as hard as string functions ...
Fundamental theorem of calculus
The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function with the concept of the function's integral.The first part of the theorem, sometimes called the first fundamental theorem of calculus, is that the definite integration of a function is related to its antiderivative, and can be reversed by differentiation. This part of the theorem is also important because it guarantees the existence of antiderivatives for continuous functions.The second part of the theorem, sometimes called the second fundamental theorem of calculus, is that the definite integral of a function can be computed by using any one of its infinitely-many antiderivatives. This part of the theorem has key practical applications because it markedly simplifies the computation of definite integrals.