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Answers for “如何逐项展开求和式 $\sum_{k=0}^{\infty}\frac{(-1)^k x^{2k}}{(2k)!}$?”

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To expand the summation, substitute successive integer values for the index kk into the general term. For k=0k=0, the term simplifies to 11.

Conditions: The summation index kk starts at 00 and increases by 11 for each term.; The general term is (−1)kx2k(2k)!\frac{(-1)^k x^{2k}}{(2k)!}.

Meet the concept

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The trailing ellipsis indicates that the series continues indefinitely following the exact same generating rule established by the first few terms. It means you keep substituting successive integers for kk into the formula (−1)kx2k(2k)!\frac{(-1)^k x^{2k}}{(2k)!} to produce the next terms, alternating signs and increasing the even powers and factorials.

Conditions: The first few terms have been explicitly calculated to establish a pattern.; The summation is an infinite series.

Understand why

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The k=0k=0 term equals 11 because substituting k=0k=0 into the general term yields (−1)0x00!\frac{(-1)^0 x^0}{0!}. The video uses the standard conventions that (−1)0=1(-1)^0 = 1, x0=1x^0 = 1, and 0!=10! = 1.

Conditions: The index value is k=0k=0.; The standard mathematical conventions x0=1x^0=1 and 0!=10!=1 are assumed.

Understand why

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The video states that a function can be defined by a series to counter the misconception that functions must only be given by simple closed-form algebraic or geometric formulas. By setting f(x)f(x) equal to an infinite sum, the series itself serves as the rule that determines the value of the function for any given xx.

Conditions: The expression is an infinite series.; The equality f(x)=seriesf(x) = \text{series} is taken as a definition.