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@ -368,10 +368,11 @@ The area is invariant under shifts left or right.
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Any partition $a =x_0 < x_1 < \cdots < x_n=b$ is related to a partition of $[a-c, b-c]$ through $a-c < x_0-c < x_1-c < \cdots < x_n - c = b-c$. Let $d_i=c_i-c$ denote this partition, then we have:
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$$
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f(c_1 -c) \cdot (x_1 - x_0) + f(c_2 -c) \cdot (x_2 - x_1) + \cdots + f(c_n -c) \cdot (x_n - x_{n-1}) =
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f(d_1) \cdot(x_1-c - (x_0-c)) + f(d_2) \cdot(x_2-c - (x_1-c)) + \cdots + f(d_n) \cdot(x_n-c - (x_{n-1}-c)).
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$$
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\begin{align*}
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f(c_1 -c) \cdot (x_1 - x_0) &+ f(c_2 -c) \cdot (x_2 - x_1) + \cdots + f(c_n -c) \cdot (x_n - x_{n-1}) = \\
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& f(d_1) \cdot(x_1-c - (x_0-c)) + f(d_2) \cdot(x_2-c - (x_1-c)) + \cdots + f(d_n) \cdot(x_n-c - (x_{n-1}-c)).
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\end{align*}
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The left side will have a limit of $\int_a^b f(x-c) dx$ the right would have a "limit" of $\int_{a-c}^{b-c}f(x)dx$.
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@ -498,7 +499,7 @@ Cauchy showed this using a *geometric series* for the partition, not the arithme
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\begin{align*}
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S &= a^{k+1}\alpha u^0 + a^{k+1}\alpha u^1 + \cdots + a^{k+1}\alpha u^{n-1}
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S &= a^{k+1}\alpha u^0 + a^{k+1}\alpha u^1 + \cdots + a^{k+1}\alpha u^{n-1}\\
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&= a^{k+1} \cdot \alpha \cdot (u^0 + u^1 + \cdot u^{n-1}) \\
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&= a^{k+1} \cdot \alpha \cdot \frac{u^n - 1}{u - 1}\\
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&= (b^{k+1} - a^{k+1}) \cdot \frac{\alpha}{(1+\alpha)^{k+1} - 1} \\
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