hcju/proofwikiqa huggingface dataset:
Proof by induction: For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: : $\dfrac {\d^n} {\d x^n} \left({f \left({a x + b}\right)}\right) = a^n \dfrac {\d^n} {\d z^n} \left({f \left({z}\right)}\right)$ where $z = a x + b$. === Basis for the Induction === $P(1)$ is the case:. : $\dfrac \d {\d x} \left({f \left({a x + b}\right)}\right) = a \dfrac \d {\d z} \left({f \left({z}\right)}\right)$ where $z = a x + b$. This is proved in Derivative of Function of Constant Multiple: Corollary. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: : $\dfrac {\d^k} {\d x^k} \left({f \left({a x + b}\right)}\right) = a^k \dfrac {\d^n} {\d z^k} \left({f \left({z}\right)}\right)$ where $z = a x + b$. Then we need to show: : $\dfrac {\d^{k + 1} } {\d x^{k + 1} } \left({f \left({a x + b}\right)}\right) = a^{k + 1} \dfrac {\d^{k + 1} } {\d z^{k + 1} } \left({f \left({z}\right)}\right)$ where $z = a x + b$. === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \dfrac {\d^{k+1} } {\d x^{k+1} } \left({f \left({a x + b}\right)}\right)       | r = \dfrac \d {\d x} \left({\dfrac {\d^k} {\d x^k} \left({f \left({a x + b}\right)}\right)}\right)       | c = Definition of Higher Derivatives }} {{eqn | r = \dfrac \d {\d x} \left({a^k \dfrac {\d^k} {\d z^k} \left({f \left({z}\right)}\right)}\right)       | c = Induction Hypothesis }} {{eqn | r = a^k \dfrac \d {\d x} \left({\dfrac {\d^k} {\d z^k} \left({f \left({z}\right)}\right)}\right)       | c = Derivative of Constant Multiple }} {{eqn | r = a^k \cdot a \dfrac \d {\d z} \left({\dfrac {\d^k} {\d z^k} \left({f \left({z}\right)}\right)}\right)       | c = Basis of the Induction }} {{eqn | r = a^{k+1} \dfrac \d {\d z} \left({\dfrac {\d^k} {\d z^k} \left({f \left({z}\right)}\right)}\right)       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: : $\dfrac {\d^n} {\d x^n} \left({f \left({a x + b}\right)}\right) = a^n \dfrac {\d^n} {\d z^n} \left({f \left({z}\right)}\right)$ where $z = a x + b$. {{qed}} Category:Differential Calculus Category:Derivatives hqcxbd1rneyggvge6y19oy59zpiav1a
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$ === Basis for the Induction === $P(1)$ is true, as this is the case: {{begin-eqn}} {{eqn | l = \frac {\d} {\d x} \frac 1 {x^m}       | r = \frac {\d} {\d x} x^{-m}       | c =  }} {{eqn | r = \paren {-m} x^{-m - 1}       | c = Power Rule for Derivatives }} {{eqn | r = \frac {-m} {x^{m+1} }       | c =  }} {{end-eqn}} which matches the proposition as $m^{\overline 1} = m$ from the definition of rising factorial. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\dfrac {\d^k} {\d x^k} \dfrac 1 {x^m} = \dfrac {\paren {-1}^k m^{\overline k} } {x^{m + k} }$ Then we need to show: :$\dfrac {\d^{k + 1} } {\d x^{k + 1} } \dfrac 1 {x^m} = \dfrac {\paren {-1}^{k + 1} m^{\overline {k + 1} } } {x^{m + k + 1} }$ === Induction Step === This is our induction step: First, let $k < m$. Then we have: {{begin-eqn}} {{eqn | l = \frac {\d^{k + 1} } {\d x^{k + 1} } \frac 1 {x^m}       | r = \map {\frac \d {\d x} } {\frac {\d^k} {\d x^k} \frac 1 {x^m} }       | c =  }} {{eqn | r = \frac \d {\d x} \frac {\paren {-1}^k m^{\overline k} } {x^{m + k} }       | c = Induction Hypothesis }} {{eqn | r = \paren {-1}^k m^{\overline k} \frac \d {\d x} \frac 1 {x^{m + k} }       | c = Derivative of Constant Multiple }} {{eqn | r = \paren {-1}^k m^{\overline k} \paren {\frac {-\paren {m + k} } {x^{m + k + 1} } }       | c = Basis for the Induction }} {{eqn | r = \frac {\paren {-1}^{k + 1} m^{\overline {k + 1} } } {x^{m + k + 1} }       | c = {{Defof|Rising Factorial}} }} {{end-eqn}} {{qed}} Category:Derivatives Category:Reciprocals 28ff3vh7wh9tscj8moc7wtefd1hbwbw
The Probability Generating Function of Geometric Distribution is: :$\map {\Pi_X} s = \dfrac q {1 - p s}$ where $q = 1 - p$. From Derivatives of Function of $a x + b$, we have that: :$\map {\dfrac {\d^n} {\d s^n} } {\map f {1 - p s} } = \paren {-p}^n \map {\dfrac {\d^n} {\d z^n} } {\map f z}$ where $z = 1 - ps$. Here we have that $\map f z = \dfrac q z$. From Nth Derivative of Reciprocal of Mth Power: :$\dfrac {\d^n} {\d z^n} \dfrac q z = q \dfrac {\d^n} {\d z^n} \dfrac 1 z = q \dfrac {\paren {-1}^n n!} {z^{n + 1} }$ where $n!$ denotes $n$ factorial. So putting it together: :$\dfrac {\d^n} {\d s^n} \map {\Pi_X} s = q \paren {-p}^n \dfrac {\paren {-1}^n n!} {\paren {1 - p s}^{n + 1} }$ whence (after algebra): :$\dfrac {\d^n} {\d s^n} \map {\Pi_X} s = \dfrac {q p^n n!} {\paren {1 - p s}^{n + 1} }$ {{qed}} Category:Geometric Distribution Category:Derivatives of PGFs sju647ktnbvavrs3k0l214xz8lffn0p
The Probability Generating Function of Shifted Geometric Distribution is: :$\Pi_X \left({s}\right) = \dfrac {p s} {1 - q s}$ where $q = 1 - p$. First we need to obtain the first derivative: {{begin-eqn}} {{eqn | l = \Pi'_X \left({s}\right)       | r = \frac {\mathrm d} {\mathrm d s} \left({\frac {p s} {1 - q s} }\right)       | c =  }} {{eqn | r = p s \frac {\mathrm d} {\mathrm d s} \left({\frac 1 {1 - q s} }\right) + \frac 1 {1 - q s} \frac {\mathrm d} {\mathrm d s} \left({p s}\right)       | c = Sum Rule for Derivatives }} {{eqn | r = \frac {p q s} {\left({1 - q s}\right)^2} + \frac p {1 - q s}       | c = LHS is first derivative of PGF of Geometric Distribution }} {{eqn | r = \frac {p q s + p \left({1 - q s}\right)} {\left({1 - q s}\right)^2}       | c =  }} {{eqn | r = \frac p {\left({1 - q s}\right)^2}       | c = after some algebra }} {{end-eqn}} From Derivatives of Function of $a x + b$: :$\dfrac {\mathrm d^n} {\mathrm d s^n} \left({f \left({1 - q s}\right)}\right) = \left({-q}\right)^n \dfrac {\mathrm d^n} {\mathrm d z^n} \left({f \left({z}\right)}\right)$ where $z = 1 - q s$. Here we have that: :$f \left({z}\right) = p \dfrac 1 {z^2}$ From Nth Derivative of Reciprocal of Mth Power: :$\dfrac {\mathrm d^{n - 1} } {\mathrm d z^{n - 1} } \dfrac 1 {z^2} = \dfrac {\left({-1}\right)^{n - 1} 2^{\overline {n - 1} } } {z^{\left({n - 1}\right) + 2} }$ where $\overline {n-1}$ denotes the rising factorial. Note that we consider the $n-1$th derivative because we have already taken the first one. Also note that $2^{\overline {n - 1}} = 1^{\overline {n - 1} } = \left({n - 1}\right)!$ So putting it together: :$\dfrac {\mathrm d^n} {\mathrm d s^n} \Pi_X \left({s}\right) = p \left({-q}\right)^{n - 1} \dfrac {\left({-1}\right)^{n - 1} \left({n - 1}\right)!} {\left({1 - q s}\right)^{n + 1} }$ whence (after algebra): :$\dfrac {\mathrm d^n} {\mathrm d s^n} \Pi_X \left({s}\right) = \dfrac {p q^{n - 1} \left({n - 1}\right)!} {\left({1 - q s}\right)^{n + 1} }$ {{qed}} {{proofread}} Category:Geometric Distribution Category:Derivatives of PGFs 2i2gg831vegh2umfaarsg8wufcum0gp
The Probability Generating Function of Discrete Uniform Distribution is: :$\map {\Pi_X} s = \dfrac {s \paren {1 - s^n} } {n \paren {1 - s} } = \dfrac 1 n \displaystyle \sum_{k \mathop = 1}^n s^k$ From Nth Derivative of Mth Power: :$\dfrac {\d^k} {\d s^k} s^n = \begin {cases} n^{\underline k} s^{n - k} & : k \le n \\ 0 & : k > n \end {cases}$ The result follows. {{qed}} Category:Discrete Uniform Distribution Category:Derivatives of PGFs sejl92b6043shst00y11squf7yjwlrw
As $\operatorname{Im} \left({X}\right)$ and $\operatorname{Im} \left({Y}\right)$ are countable, then so is $\operatorname{Im} \left({g \left({X, Y}\right)}\right)$. Now consider $g^{-1} \left({Z}\right)$. We have that: * $\forall x \in \R: X^{-1} \left({x}\right) \in \Sigma$. * $\forall y \in \R: Y^{-1} \left({x}\right) \in \Sigma$. We also have that: : $\displaystyle \forall z \in \R: g^{-1} \left({z}\right) = \bigcup_{\left({x, y}\right): g \left({x, y}\right) = z} \left\{{\left({x, y}\right)}\right\}$ But $\Sigma$ is a sigma-algebra and therefore closed for unions. Thus $\forall z \in \R: g^{-1} \left({z}\right) \in \Sigma^2$. Now consider $\left({v, w}\right) \in g^{-1} \left({Z}\right)$. As $\Sigma$ is a sigma-algebra it follows directly that: *$\forall x \in v: X^{-1} \left({x}\right) \in \Sigma$ *$\forall y \in w: Y^{-1} \left({y}\right) \in \Sigma$ Hence the result. {{qed}} === Proof of Corollary === Follows straightforwardly by induction from the main result. {{qed}} Category:Probability Theory 4n7jjhws4qt0p3jwk32xuo2q5klf416
Let $\Omega_X = \operatorname{Im} \left({X}\right) = I_X$ and $\Omega_Y = \operatorname{Im} \left({Y}\right) = I_Y$. Let $Z = g \left({X, Y}\right)$. Thus $\Omega_Z = \operatorname{Im} \left({Z}\right) = g \left({I_X, I_Y}\right)$. So: {{begin-eqn}} {{eqn | l = E \left({Z}\right)       | r = \sum_{z \mathop \in g \left({I_X, I_Y}\right)} z \Pr \left({Z = z}\right)       | c =  }} {{eqn | r = \sum_{z \mathop \in g \left({I_X, I_Y}\right)} z \sum_{ {x \mathop \in I_X, y \mathop \in I_Y} \atop {g \left({x, y}\right) \mathop = z} } \Pr \left({X = x, Y = y}\right)       | c = Probability Mass Function of Function of Discrete Random Variable }} {{eqn | r = \sum_{x \mathop \in I_X} \sum_{y \mathop \in I_Y} g \left({x, y}\right) \Pr \left({X = x, Y = y}\right)       | c =  }} {{eqn | r = \sum_{x \mathop \in I_X} \sum_{y \mathop \in I_Y} g \left({x, y}\right) p_{X, Y} \left({x, y}\right)       | c =  }} {{end-eqn}} From the definition of expectation, this last sum applies only when the last sum is absolutely convergent. {{qed}}
=== Discrete Random Variable === {{:Linearity of Expectation Function/Discrete}}  === Continuous Random Variable === {{:Linearity of Expectation Function/Continuous}}
We have by definition of joint mass function that: :$x \notin \Omega_X \implies \map {p_{X, Y} } {x, y} = 0$ :$y \notin \Omega_Y \implies \map {p_{X, Y} } {x, y} = 0$ Hence we only need to worry about values of $x$ and $y$ in their appropriate $\Omega$ spaces. === Necessary Condition === Suppose there exist functions $f, g: \R \to \R$ such that: :$\forall x, y \in \R: \map {p_{X, Y} } {x, y} = \map f x \map g y$ Then by definition of marginal probability mass function: :$\displaystyle \map {p_X} x = \map f x \sum_y \map g y$ :$\displaystyle \map {p_Y} y = \map g y \sum_x \map f x$ Hence: {{begin-eqn}} {{eqn | l = 1       | r = \sum_{x, y} \map {p_{X, Y} } {x, y}       | c = {{Defof|Joint Probability Mass Function}} }} {{eqn | r = \sum_{x, y} \map f x \map g y       | c = by hypothesis }} {{eqn | r = \sum_x \map f x \sum_y \map g y       | c =  }} {{end-eqn}} So it follows that: {{begin-eqn}} {{eqn | l = \map {p_{X, Y} } {x, y}       | r = \map f x \map g y       | c = by hypothesis }} {{eqn | r = \map f x \map g y \sum_x \map f x \sum_y \map g y       | c = from above }} {{eqn | r = \paren {\map f x \sum_y \map g y} \paren {\map g y \sum_x \map f x}       | c =  }} {{eqn | r = \map {p_X} x \map {p_Y} y       | c = from above }} {{end-eqn}} Hence the result from the definition of independent random variables. {{qed}} === Sufficient Condition === Suppose that $X$ and $Y$ are independent. Then we can take the variables: :$\map f x = \map {p_X} x$ :$\map g y = \map {p_Y} y$ and the result follows by definition of independence. {{Qed}}
=== Sufficient Condition === Suppose that $X$ and $Y$ were ''not'' independent. That is: :$\map \Pr {X = a, Y = b} \ne \map \Pr {X = a} \map \Pr {Y = b}$ for some $a, b \in \R$. Now, suppose that :$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$ for all functions $g, h: \R \to \R$ for which the latter two expectations exist. Let us define $g$ and $h$ as examples of such functions as follows: :$(1): \quad \map g x := \begin{cases} 1 & : x = a \\ 0 & : x \ne a \end{cases}$ :$(2): \quad \map h y := \begin{cases} 1 & : y = b \\ 0 & : y \ne b \end{cases}$ where $a \in \R$ and $b \in \R$ are arbitrary real numbers. Then: {{begin-eqn}} {{eqn | l = \expect {\map g x \map h y}       | r = \map \Pr {X = a, Y = b}       | c = }} {{eqn | l = \expect {\map g X} \expect {\map h Y}       | r = \map \Pr {X = a} \map \Pr {Y = b}       | c = }} {{end-eqn}} So by hypothesis: :$\map \Pr {X = a, Y = b} = \map \Pr {X = a} \map \Pr {Y = b}$ But also by hypothesis: :$\map \Pr {X = a, Y = b} \ne \map \Pr {X = a} \map \Pr {Y = b}$ This contradicts our supposition that: :$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$ for all functions $g, h: \R \to \R$ for which the latter two expectations exist. So, if the above supposition holds, then $X$ and $Y$ have to be independent. {{qed|lemma}} === Necessary Condition === Suppose $X$ and $Y$ are independent. Let $g, h: \R \to \R$ be any real functions such that $\expect {\map g x}$ and $\expect {\map h y}$ exist. Then: {{begin-eqn}} {{eqn | l = \expect {\map g x \map h y}       | r = \sum_{x, y} \map g x \map h y \map \Pr {X = x, Y = y}       | c = }} {{eqn | r = \sum_{x, y} \map g x \map h y \map \Pr {X = x} \map \Pr {Y = y}       | c = by definition of independent random variables }} {{eqn | r = \sum_x \map g x \map \Pr {X = x} \sum_y \map h y \map \Pr {Y = y}       | c = }} {{eqn | r = \expect {\map g x} \expect {\map h y}       | c = }} {{end-eqn}} thus proving that :$\expect {\map g x \map h y} = \expect {\map g x} \expect {\map h y}$ whatever $g$ and $h$ are. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\Pi_Z} s       | r = \expect {s^Z}       | c = {{Defof|Probability Generating Function}} }} {{eqn | r = \expect {s^{X + Y} }       | c = Definition of $Z$ (see above) }} {{eqn | r = \expect {s^X s^Y}       | c = Exponential of Sum }} {{eqn | r = \expect {s^X} \expect {s^Y}       | c = Condition for Independence from Product of Expectations }} {{eqn | r = \map {\Pi_X} s \, \map {\Pi_Y} s       | c = {{Defof|Probability Generating Function}} }} {{end-eqn}} {{qed}}
:450px === Sufficient Condition === Suppose that $EF$ is perpendicular to $BC$. We need to prove that $AF = FD$. Thus: {{begin-eqn}} {{eqn | l = \angle FAM       | r = \angle CAD       | c = producing $AF$ to $D$ and $AM$ to $C$ }} {{eqn | r = \angle CBD       | c = Angles in Same Segment of Circle are Equal: both subtend $CD$ }} {{eqn | r = \angle CBM       | c = producing $BM$ to $D$ }} {{eqn | r = \angle CME       | c = both are complements of $\angle BCM$ }} {{eqn | r = \angle FMA       | c = Vertical Angle Theorem }} {{end-eqn}} Then by Triangle with Two Equal Angles is Isosceles, it follows that $AF = FM$. Similarly: {{begin-eqn}} {{eqn | l = \angle FDM       | r = \angle ADB       | c = producing $DF$ to $A$ and $DM$ to $B$ }} {{eqn | r = \angle ACB       | c = Angles in Same Segment of Circle are Equal: both subtend $AB$ }} {{eqn | r = \angle BCM       | c = producing $BM$ to $A$ }} {{eqn | r = \angle BME       | c = both are complements to $\angle CBM$ }} {{eqn | r = \angle DMF       | c = Vertical Angle Theorem }} {{end-eqn}} Then by Triangle with Two Equal Angles is Isosceles, it follows that $FD = FM$. So $AF = FD$, as we needed to show. {{qed|lemma}} === Necessary Condition === Now suppose that $AF = FD$. We now need to show that $EF$ is perpendicular to $BC$. From Thales' Theorem (indirectly) we have that $AF = FM = FD$. So: {{begin-eqn}} {{eqn | l = \angle EBM       | r = \angle CBD       | c = producing $EB$ to $C$ and $BM$ to $D$ }} {{eqn | r = \angle CAD       | c = Angles in Same Segment of Circle are Equal: both subtend $CD$ }} {{eqn | r = \angle FAM       | c = producing $AM$ to $C$ and $FA$ to $D$ }} {{eqn | r = \angle AMF       | c = Isosceles Triangle has Two Equal Angles, and $AF = FM$ }} {{eqn | r = \angle EMC       | c = Vertical Angle Theorem }} {{end-eqn}} We note the result Sum of Angles of Triangle equals Two Right Angles. We have that $\angle EBM$ and $\angle ECM$ are complementary, as both are angles in $\triangle CBM$, which is a right triangle. So $\angle EMC$ and $\angle ECM$ are complementary, which means that $\angle CEM$ must be a right angle. Hence by definition $EF$ is perpendicular to $BC$, as we were to show. {{qed}} {{Namedfor|Brahmagupta|cat = Brahmagupta}} Category:Cyclic Quadrilaterals abd75qwx5riomyj946xa36dod8wtrqw
:380px {{AimForCont}} $BC$ does not meet ray $\overrightarrow {AD}$. Either $BC$ meets line $\overleftrightarrow {AD}$ or it does not. If it meets $\overleftrightarrow {AD}$, by the Line Separation Property it must meet the ray opposite to $\overrightarrow {AD}$ at a point $E \ne A$. According to Proposition 3.8 (b), $E$ is not in the interior of $\angle CAB$. Point $B$ does not lie on $\overleftrightarrow {AD}$; this is because $D \in \operatorname {int} \angle CAB$, so $D$ and $C$ lie on the same side of $\overleftrightarrow {AB}$, so $D \notin \set {\overleftrightarrow {AB} }$, so $B \notin \set {\overleftrightarrow {AB} }$. Thus, since $E \in \set {\overleftrightarrow {AD} }$, we have $E \ne B$. By the same reasoning with $C$ and $B$ interchanged, we have $E \ne C$. Since $E \in BC$ and $E$ is not an endpoint, we have $B ∗ E ∗ C$. Thus by Proposition 3.7, $E$ is in the interior of $\angle CAB$, which is a contradiction. Thus $\overleftrightarrow {AD}$ does not meet $BC$ at all; that is, $B$ and $C$ are on the same side of $\overleftrightarrow {AD}$. By B-2, we have a point $E$ such that $C ∗ A ∗ E$. By Lemma 3.2.2, $C$ and $E$ are on opposite sides of $\overleftrightarrow {AD}$. Thus, by B-4(iii), E and B are on opposite sides of $\overleftrightarrow {AD}$. But by Proposition 3.8(c), $B$ is on the interior of $\angle DAE$, so $E$ and $B$ are on the same side of $\overleftrightarrow {AD}$ This is a contradiction. Thus $\overrightarrow {AD}$ meets $BC$. {{MissingLinks}} Category:Triangles Category:Named Theorems 0y4vo2pufqg5j3elrgp06jh2f2rekym
:380px Let $ABC$ be a circle, let $\angle BEC$ be an angle at its center, and let $\angle BAC$ be an angle at the circumference. Let these angles have the same arc $BC$ at their base. Let $AE$ be joined and drawn through to $F$. Since $EA = EB$, then from Isosceles Triangle has Two Equal Angles we have that $\angle EBA = \angle EAB$. So $\angle EBA + \angle EAB = 2 \angle EAB$. But from Sum of Angles of Triangle equals Two Right Angles we have that $\angle BEF = \angle EBA + \angle EAB$. That is, $\angle BEF = 2 \angle EAB$. For the same reason, $\angle FEC = 2 \angle EAC$. So adding them together, we see that $\angle BEC = 2 \angle BAC$. Now consider the point $D$, from which we have another angle $\angle BDC$. Let $DE$ be joined and produced to $G$. Similarly, we prove that $\angle GEC = 2 \angle EDC$. Then $\angle GEB = 2 \angle EDB$. Therefore $\angle BEC$ which remains is equal to $2 \angle BDC$. Hence the result. {{Qed}}
Consider the simplest case that occurs when $AC$ is a diameter of the circle: :300px Because all lines radiating from $D$ to the circumference are radii and thus equal: :$AD = BD = CD$ Hence the triangles $\triangle ADB$ and $\triangle BDC$ are isosceles. Therefore from Isosceles Triangle has Two Equal Angles: :$\angle DBC = \angle DCB$. From Sum of Angles of Triangle equals Two Right Angles: :$\angle BDC$ is a supplement of $\angle DBC + \angle DCB = 2 \angle DCB$. From Thales' Theorem, $\angle ABC$ is right. By similar reasoning $\angle DAB$ is the complement of $\angle DCB$. If $\angle BDC$ is the supplement of twice the complement of $\angle DAB$, then $\angle BDC = 2 \angle DAB$. That proves the theorem for this case. {{qed|lemma}} The general case is illustrated below. :300px A diameter is drawn from $A$ through the center $D$ to $E$.  By the previous logic: :$\angle BDE = 2 \angle BAE$ :$\angle CDE = 2 \angle CAE$ Subtracting the latter from the former equation obtains the general result.  {{qed}}
{{ProofWanted|There should be a proof that the net of the polyhedron is a planar graph. The result then follows from Euler's Theorem for Planar Graphs.}}
Suppose the opposite, and let circle $ABC$ cut the circle $DEF$ at more points than two, that is at $B, G, F, H$. :400px Clearly, in the diagram, circle $DEF$ is not actually a circle. For the sake of demonstrating a contradiction, we suppose that it is. Join $BH$ and $BG$, and bisect them at $K$ and $L$. From $K$ and $L$, draw perpendiculars to $BH$ and $BG$, through to $A$ and $E$. Since $AC$ cuts $BH$ into two equal parts at right angles, from the porism to Finding Center of Circle, the center of circle $ABC$ is on $AC$. For the same reason, the center of circle $ABC$ is also on $NO$. But as the only point on both $AC$ and $NO$ is $P$, it follows that $P$ is the center of circle $ABC$. Similarly, we can show that $P$ is also the center of circle $DEF$. So we have two circles that intersect which have the same center. From Intersecting Circles have Different Centers, this statement is contradictory. Hence such a pair of circles, i.e. those that intersect at more points than two, can not exist. {{qed}} {{Euclid Note|10|III}}
Let the circles $ABC$ and $ADE$ touch internally at $A$. Let $F$ be the center of $ABC$ and let $G$ be the center of $ADE$. We are to show that the straight line joining $F$ to $G$ passes through $A$. :400px Suppose, as in the diagram above, that it does not. Let $FG$ fall on $H$ instead. It will also pass through $D$ on its way, which lies on circle $ADE$. Join $AF$ and $AG$. From Sum of Two Sides of Triangle Greater than Third Side $AG + GF$ is greater than $AF$. Therefore $AG + GF$ is greater than $FH$ as $F$ is the center and both $AF$ and $FH$ are radii. So, subtract $FG$ from both $AG + GF$ and $FH$. It follows that $AG$ is greater than $GH$, and hence also greater than $GD$. But as $G$ is the center of $ADE$, it follows that $AG$ is supposed to be equal to $GD$. So the $AG + GF$ can not be greater than $AF$ and so must be coincident with it. Hence the result. {{Qed}} {{Euclid Note|11|III}}
{{begin-eqn}} {{eqn | l=\Pi_Z \left({s}\right)       | r=E \left({s^{X_1 + X_2 + \cdots + X_N} }\right)       | c=by definition of probability generating function }} {{eqn | r=\sum_{n \mathop \ge 0} E \left({s^{X_1 + X_2 + \cdots + X_N} \mid N = n}\right) \Pr \left({N = n}\right)       | c=Total Expectation Theorem }} {{eqn | r=\sum_{n \mathop \ge 0} E \left({s^{X_1 + X_2 + \cdots + X_n} }\right) \Pr \left({N = n}\right)       | c= }} {{eqn | r=\sum_{n \mathop \ge 0} \Pi_X \left({s}\right)^n \Pr \left({N = n}\right)       | c=PGF of Sum of Independent Discrete Random Variables }} {{eqn | r=\Pi_{N} \left({\Pi_X \left({s}\right)}\right)       | c=by definition of $\Pi_N \left({s}\right)$ }} {{end-eqn}} {{qed}}
We assume the two hypotheses of the theorem. We have that: {{begin-eqn}} {{eqn | l = \map {\frac \d {\d t} } {\map \Phi {t + T} }       | r = \map {\Phi'} {t + T}       | c =  }} {{eqn | r = \map {\mathbf A} {t + T} \map \Phi {t + T}       | c =  }} {{eqn | r = \map {\mathbf A} t \map \Phi {t + T}       | c =  }} {{end-eqn}} So the first implication of the theorem holds, that is: that $\map \Phi {t + T}$ is a fundamental matrix. Because $\map \Phi t$ and $\map \Phi {t + T}$ are both fundamental matrices, there must exist some matrix $\mathbf C$ such that: :$\map \Phi {t + T} = \map \Phi t \mathbf C$ Hence by the existence of the matrix logarithm, there exists a matrix $\mathbf B$ such that: :$\mathbf C = e^{\mathbf BT}$ Defining $\map {\mathbf P} t = \map \Phi t e^{-\mathbf B t}$, it follows that: {{begin-eqn}} {{eqn | l = \map {\mathbf P} {t + T}       | r = \map \Phi {t + T} e^{-\mathbf B t - \mathbf B T}       | c =  }} {{eqn | r = \map \Phi t C e^{-\mathbf B T} e^{-\mathbf B t}       | c =  }} {{eqn | r = \map \Phi t e^{-\mathbf B t}       | c =  }} {{eqn | r = \map {\mathbf P} t       | c =  }} {{end-eqn}} and hence $\map {\mathbf P} t$ is periodic with period $T$. As $\map \Phi t = \map {\mathbf P} t e^{\mathbf B t}$, the second implication also holds. {{qed}}