Why does the $k=0$ term of the series $\sum_{k=0}^{\infty}\frac{(-1)^k x^{2k}}{(2k)!}$ equal $1$?

Conditions

• The index value is $k=0$.
• The standard mathematical conventions $x^0=1$ and $0!=1$ are assumed.

Reasoning, step by step

1. Substitute $k=0$ into the general term to get $\frac{(-1)^0 x^{2\cdot 0}}{(2\cdot 0)!}$.
2. Evaluate the sign factor: $(-1)^0 = 1$.
3. Evaluate the power of $x$: $x^0 = 1$.
4. Evaluate the factorial in the denominator: $0! = 1$.
5. Combine the results: $\frac{1 \cdot 1}{1} = 1$.

Example

For $k=0$, the speaker says the factor becomes $1$, $x$ to the power $0$ will be $1$, and zero factorial is $1$, so the first term simplifies to $1$.

Common misconceptions

• Assuming $x^0 = 0$ instead of $1$.
• Believing that $0! = 0$ instead of $1$.
• Thinking the first term should contain $x$ because the series is a power series.

Watch the explanation

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