Difference between revisions of "Euler-Mascheroni Constant"
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− | The <b>Euler-Mascheroni constant</b> <math>\gamma</math> is defined by <cmath>\gamma = \lim_{n \rightarrow \infty} \left( \left( \sum_{k=1}^n \frac{1}{k} \right) - \ln(n) \right).</cmath> Its value is approximately <cmath>\gamma = 0.577215 \dots .</cmath> | + | The <b>Euler-Mascheroni constant</b> <math>\gamma</math> is a [[constant]] defined by the [[limit]] <cmath>\gamma = \lim_{n \rightarrow \infty} \left( \left( \sum_{k=1}^n \frac{1}{k} \right) - \ln(n) \right).</cmath> Its value is approximately <cmath>\gamma = 0.577215 \dots .</cmath> |
− | Whether <math>\gamma</math> is rational or irrational and (if irrational) algebraic or transcendental is an open question. | + | Whether <math>\gamma</math> is [[Rational number|rational]] or [[Irrational number|irrational]] and (if irrational) [[Algebraic number|algebraic]] or [[Transcendental number|transcendental]] is an open question. |
==Proof of existence== | ==Proof of existence== | ||
===Alternate formulation of the limit=== | ===Alternate formulation of the limit=== | ||
− | The tangent-line approximation (first-degree Taylor polynomial) of <math>\ln(k + 1)</math> about <math>x = k</math> is <cmath>\ln(k + 1) = \ln(k) + ((k + 1) - k)\ln'(k) + E_k</cmath> for some error term <math>E_k</math>. Using <math>\ln'(x) = \frac{1}{x}</math> and simplifying, <cmath>\ln(k + 1) = \ln(k) + \frac{1}{k} + E_k.</cmath> | + | The tangent-line approximation (first-degree [[Taylor polynomial]]) of <math>\ln(k + 1)</math> about <math>x = k</math> is <cmath>\ln(k + 1) = \ln(k) + ((k + 1) - k)\ln'(k) + E_k</cmath> for some error term <math>E_k</math>. Using <math>\ln'(x) = \frac{1}{x}</math> and simplifying, <cmath>\ln(k + 1) = \ln(k) + \frac{1}{k} + E_k.</cmath> Applying the tangent-line formula [[Recursion|recursively]] for all <math>k</math> descending from <math>n</math> to <math>1</math>, |
<cmath>\begin{align*} \ln(n) &= \ln(n-1) + \frac{1}{n-1} + E_{n-1} \\ &= \left( \ln (n-2) + \frac{1}{n-2} + E_{n-2} \right) + \frac{1}{n - 1} + E_{n-1} \\ &= \dots \\ &= \ln(1) + \left( \sum_{k=1}^{n-1} \frac{1}{k} \right) + \left( \sum_{k=1}^{n-1} E_k \right) . \end{align*}</cmath> | <cmath>\begin{align*} \ln(n) &= \ln(n-1) + \frac{1}{n-1} + E_{n-1} \\ &= \left( \ln (n-2) + \frac{1}{n-2} + E_{n-2} \right) + \frac{1}{n - 1} + E_{n-1} \\ &= \dots \\ &= \ln(1) + \left( \sum_{k=1}^{n-1} \frac{1}{k} \right) + \left( \sum_{k=1}^{n-1} E_k \right) . \end{align*}</cmath> | ||
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===Convergence of the sum of error terms=== | ===Convergence of the sum of error terms=== | ||
− | We have <math>\ln''(x) = \left(\frac{1}{x} \right)' = -\frac{1}{x^2}</math>. For <math>k \geq 1</math>, the maximum absolute value of <math>-\frac{1}{x^2}</math> for <math>x \in [k, k+1]</math> is <math>\frac{1}{k^2}</math>. Therefore, by the Lagrange Error Bound, <cmath>|E_k| \leq \left| \frac{1^2 \left( \frac{1}{k^2} \right) }{2!} \right| = \frac{1}{2k^2}.</cmath> | + | We have <math>\ln''(x) = \left(\frac{1}{x} \right)' = -\frac{1}{x^2}</math>. For <math>k \geq 1</math>, the maximum [[absolute value]] of <math>-\frac{1}{x^2}</math> for <math>x \in [k, k+1]</math> is <math>\frac{1}{k^2}</math>. Therefore, by the [[Taylor polynomial#Error bound|Lagrange Error Bound]], <cmath>|E_k| \leq \left| \frac{1^2 \left( \frac{1}{k^2} \right) }{2!} \right| = \frac{1}{2k^2}.</cmath> |
− | The series <math>\sum_{k=1}^{\infty} \frac{1}{k^2}</math> famously converges to <math>\frac{\pi^2}{6}</math> by the Basel problem, so <math>\sum_{k=1}^{\infty} -\frac{1}{2k^2}</math> converges to <math>-\frac{\pi^2}{12}</math> and <math>\sum_{k=1}^{\infty} \frac{1}{2k^2}</math> converges to <math>\frac{\pi^2}{12}</math>. | + | The series <math>\sum_{k=1}^{\infty} \frac{1}{k^2}</math> famously converges to <math>\frac{\pi^2}{6}</math> by the [[Riemann zeta function|Basel problem]], so <math>\sum_{k=1}^{\infty} -\frac{1}{2k^2}</math> converges to <math>-\frac{\pi^2}{12}</math> and <math>\sum_{k=1}^{\infty} \frac{1}{2k^2}</math> converges to <math>\frac{\pi^2}{12}</math>. |
− | Because <math>E_k \in \left[ -\frac{1}{2k^2}, \frac{1}{2k^2} \right]</math> for all <math>k</math>, the Series Comparison Test gives that <math>\sum_{k=1}^{\infty} E_k</math> must converge to a value in <math>\left[-\frac{\pi^2}{12}, \frac{\pi^2}{12} \right]</math>. | + | Because <math>E_k \in \left[ -\frac{1}{2k^2}, \frac{1}{2k^2} \right]</math> for all <math>k</math>, the [[Series Comparison Test]] gives that <math>\sum_{k=1}^{\infty} E_k</math> must converge to a value in <math>\left[-\frac{\pi^2}{12}, \frac{\pi^2}{12} \right]</math>. |
Hence, <math>\gamma = - \sum_{k=1}^{\infty} E_k</math> is a defined constant. | Hence, <math>\gamma = - \sum_{k=1}^{\infty} E_k</math> is a defined constant. | ||
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+ | [[Category: Constants]] |
Revision as of 13:15, 9 March 2022
The Euler-Mascheroni constant is a constant defined by the limit
Its value is approximately
Whether is rational or irrational and (if irrational) algebraic or transcendental is an open question.
Proof of existence
Alternate formulation of the limit
The tangent-line approximation (first-degree Taylor polynomial) of about
is
for some error term
. Using
and simplifying,
Applying the tangent-line formula recursively for all
descending from
to
,
Because , we may rearrange to
Adding
to both sides yields
Taking the limit as
goes to infinity of both sides,
Thus, .
Convergence of the sum of error terms
We have . For
, the maximum absolute value of
for
is
. Therefore, by the Lagrange Error Bound,
The series famously converges to
by the Basel problem, so
converges to
and
converges to
.
Because for all
, the Series Comparison Test gives that
must converge to a value in
.
Hence, is a defined constant.