Felix' Math Place (Posts about Taylor expansion.)
https://math.fontein.de/tag/taylor-expansion.atom
2019-11-17T10:38:21Z
Felix Fontein
Nikola
The Hasse derivative.
https://math.fontein.de/2009/08/12/the-hasse-derivative/
2009-08-12T05:43:38+02:00
2009-08-12T05:43:38+02:00
Felix Fontein
<div>
<p>
In real and complex analysis, one has a powerful tool, namely the <a href="https://en.wikipedia.org/wiki/Taylor_expansion">Taylor expansion</a>, which expands an analytic function into a power series. In algebra, one can define the derivative of a polynomial aswell; for <span class="inline-formula"><img class="img-inline-formula img-formula" width="166" height="20" src="https://math.fontein.de/formulae/Y4grbVouSCvO.cptaDQIItA0wHwkLHEhrR8TCg.svgz" alt="f = \sum_{i=0}^n a_i x^i \in R[x]" title="f = \sum_{i=0}^n a_i x^i \in R[x]"></span>, <span class="inline-formula"><img class="img-inline-formula img-formula" width="14" height="12" src="https://math.fontein.de/formulae/vYu2wcShMlDKJ.IrmXbb2no6ZOHI_2bIn_7ZWQ.svgz" alt="R" title="R"></span> being a unitary ring, define <span class="inline-formula"><img class="img-inline-formula img-formula" width="182" height="20" src="https://math.fontein.de/formulae/I5O5SVdlWfucmQQ_wnt.xbLsjSzssdW9xA305A.svgz" alt="f' := \sum_{i=1}^n i a_i x^i \in R[x]" title="f' := \sum_{i=1}^n i a_i x^i \in R[x]"></span>. This satisfies the same rules as the usual derivative, for example <span class="inline-formula"><img class="img-inline-formula img-formula" width="16" height="12" src="https://math.fontein.de/formulae/08bW5Zvy2ST6Ewwt6yOyAbfn7ZY0nrbV5GNE.Q.svgz" alt="K" title="K"></span>-linearity and the product rule <span class="inline-formula"><img class="img-inline-formula img-formula" width="133" height="18" src="https://math.fontein.de/formulae/xP43qOuCbbfQ_L2a0UPcOYBZ6n6yvIBxBNZ9_g.svgz" alt="(f g)' = f' g + f g'" title="(f g)' = f' g + f g'"></span>. One can also define <span class="inline-formula"><img class="img-inline-formula img-formula" width="30" height="19" src="https://math.fontein.de/formulae/tW3uScd44Le5oTCJEC3WUGQtBaAtTRv1LSdNjA.svgz" alt="f^{(k)}" title="f^{(k)}"></span> recursively by <span class="inline-formula"><img class="img-inline-formula img-formula" width="64" height="19" src="https://math.fontein.de/formulae/NXHOW4wyoFkakg7zm_1bY_v_JKcDNpP9uky6Fg.svgz" alt="f^{(0)} = f" title="f^{(0)} = f"></span>, <span class="inline-formula"><img class="img-inline-formula img-formula" width="121" height="20" src="https://math.fontein.de/formulae/LGZVjMi95II9tofORhVyvv7yayrKzvfi5u0HAw.svgz" alt="f^{(k + 1)} = (f^{(k)})'" title="f^{(k + 1)} = (f^{(k)})'"></span> for <span class="inline-formula"><img class="img-inline-formula img-formula" width="44" height="13" src="https://math.fontein.de/formulae/G5w7Ev5Uv.kUdRWdTjhMpFCIr1OwBxoxggSxyw.svgz" alt="k \in \N" title="k \in \N"></span>. Unfortunately, one looses certain properties; for example, if <span class="inline-formula"><img class="img-inline-formula img-formula" width="14" height="12" src="https://math.fontein.de/formulae/vYu2wcShMlDKJ.IrmXbb2no6ZOHI_2bIn_7ZWQ.svgz" alt="R" title="R"></span> is of finite characteristic <span class="inline-formula"><img class="img-inline-formula img-formula" width="48" height="12" src="https://math.fontein.de/formulae/aaLQN98v4mp9Md5sD1RKJiZj53kVRieQFE4A1A.svgz" alt="m > 0" title="m > 0"></span>, the polynomial <span class="inline-formula"><img class="img-inline-formula img-formula" width="113" height="18" src="https://math.fontein.de/formulae/_7McjNK_CcIhIwbgRr8mCx9ToM6FUp2XZDzC5w.svgz" alt="f = x^m \in R[x]" title="f = x^m \in R[x]"></span> satisfies <span class="inline-formula"><img class="img-inline-formula img-formula" width="48" height="17" src="https://math.fontein.de/formulae/wMSk4p4azl8RkqVWsdeQ3P6b51fMvFJ_2GvsmA.svgz" alt="f' = 0" title="f' = 0"></span>, but is not constant as <span class="inline-formula"><img class="img-inline-formula img-formula" width="155" height="18" src="https://math.fontein.de/formulae/JhZpmYRxCCtwAh9uUs_VS8SNTFtoQWCgTq8K2Q.svgz" alt="f(0) = 0 \neq 1 = f(1)" title="f(0) = 0 \neq 1 = f(1)"></span> (assuming <span class="inline-formula"><img class="img-inline-formula img-formula" width="14" height="12" src="https://math.fontein.de/formulae/vYu2wcShMlDKJ.IrmXbb2no6ZOHI_2bIn_7ZWQ.svgz" alt="R" title="R"></span> is not the zero ring). In particular, the <a href="https://en.wikipedia.org/wiki/Identity_theorem">Identity Theorem</a> does not work. This example also shows one problem with a possible Taylor expansion: for that, one needs to compute <span class="inline-formula"><img class="img-inline-formula img-formula" width="47" height="27" src="https://math.fontein.de/formulae/4XeRfJkjcCK0C6vuaX8tm7ux6YWml8ZZ1WbNmA.svgz" alt="\frac{f^{(i)}(a)}{i!}" title="\frac{f^{(i)}(a)}{i!}"></span> for <span class="inline-formula"><img class="img-inline-formula img-formula" width="118" height="16" src="https://math.fontein.de/formulae/0LOSJq_DYv.rK25rHL3id.apovUVzhK5Qp2O8Q.svgz" alt="i = 0, \dots, \deg f" title="i = 0, \dots, \deg f"></span>; but <span class="inline-formula"><img class="img-inline-formula img-formula" width="48" height="11" src="https://math.fontein.de/formulae/.ZeZmgAL2YB1ovs0Kr1YPbnVER10gPU6Gh07gA.svgz" alt="m = 0" title="m = 0"></span> in <span class="inline-formula"><img class="img-inline-formula img-formula" width="14" height="12" src="https://math.fontein.de/formulae/vYu2wcShMlDKJ.IrmXbb2no6ZOHI_2bIn_7ZWQ.svgz" alt="R" title="R"></span>, whence <span class="inline-formula"><img class="img-inline-formula img-formula" width="20" height="12" src="https://math.fontein.de/formulae/wdtEPZUq_XQtFrP1AdijaTpexkW6L84DvKH50A.svgz" alt="m!" title="m!"></span> has no inverse in <span class="inline-formula"><img class="img-inline-formula img-formula" width="14" height="12" src="https://math.fontein.de/formulae/vYu2wcShMlDKJ.IrmXbb2no6ZOHI_2bIn_7ZWQ.svgz" alt="R" title="R"></span>! Hence, a Taylor expansion in the classical sense cannot be defined. A “fix” for this problem is offered by <em>Hasse derivatives</em>: they are defined to make both the Identity Theorem and Taylor expansions work again.
</p>
<div class="theorem-environment theorem-definition-environment">
<div class="theorem-header theorem-definition-header">
Definition.
</div>
<div class="theorem-content theorem-definition-content">
<p>
Let <span class="inline-formula"><img class="img-inline-formula img-formula" width="166" height="20" src="https://math.fontein.de/formulae/Y4grbVouSCvO.cptaDQIItA0wHwkLHEhrR8TCg.svgz" alt="f = \sum_{i=0}^n a_i x^i \in R[x]" title="f = \sum_{i=0}^n a_i x^i \in R[x]"></span> and <span class="inline-formula"><img class="img-inline-formula img-formula" width="44" height="13" src="https://math.fontein.de/formulae/G5w7Ev5Uv.kUdRWdTjhMpFCIr1OwBxoxggSxyw.svgz" alt="k \in \N" title="k \in \N"></span>. Define
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="249" height="52" src="https://math.fontein.de/formulae/xCfdhAcLnM5hnaUf5TFkqPAfylm5Q8G2Sc_aRg.svgz" alt="D^{(k)} f := \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} \in R[x]." title="D^{(k)} f := \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} \in R[x].">
</div>
<p>
The function <span class="inline-formula"><img class="img-inline-formula img-formula" width="144" height="20" src="https://math.fontein.de/formulae/HfPe1kAb0bWjM_jVkumA89ncExQHIVip6MreSg.svgz" alt="D^{(k)} : R[x] \to R[x]" title="D^{(k)} : R[x] \to R[x]"></span> is called the <em><span class="inline-formula"><img class="img-inline-formula img-formula" width="10" height="12" src="https://math.fontein.de/formulae/KIbZSMqheEMj2LBRLlu5emEXwvNaAcODJL5YWQ.svgz" alt="k" title="k"></span>-th Hasse derivative</em>.
</p>
</div>
</div>
<p>
The Hasse derivative shares several properties with the usual derivative, but not all of them; for example, <span class="inline-formula"><img class="img-inline-formula img-formula" width="143" height="19" src="https://math.fontein.de/formulae/mJ0ah0ZOjnUMiO.isAF0JBlisqhHM4jWCPHQLA.svgz" alt="D^{(k)} D^{(\ell)} \neq D^{(k+\ell)}" title="D^{(k)} D^{(\ell)} \neq D^{(k+\ell)}"></span> in general. But we have the following properties:
</p>
<div class="theorem-environment theorem-theorem-environment">
<div class="theorem-header theorem-theorem-header">
Theorem.
</div>
<div class="theorem-content theorem-theorem-content">
<p>
Let <span class="inline-formula"><img class="img-inline-formula img-formula" width="82" height="18" src="https://math.fontein.de/formulae/ULS.mqaHyKw6YoRAYgA1FcrASVSb_9SzsRtvxA.svgz" alt="f, g \in R[x]" title="f, g \in R[x]"></span> and <span class="inline-formula"><img class="img-inline-formula img-formula" width="46" height="13" src="https://math.fontein.de/formulae/B6vI2KC8PMuA.i2MVLjtLEZ7bEEHtcSAh43E4A.svgz" alt="\lambda \in R" title="\lambda \in R"></span>, and <span class="inline-formula"><img class="img-inline-formula img-formula" width="59" height="16" src="https://math.fontein.de/formulae/hD88YLgyJraNF7MoMtlhbk8lBYZovbzQsMHNWA.svgz" alt="k, \ell \in \N" title="k, \ell \in \N"></span>.
</p>
<ol class="enum-level-1">
<li>We have that <span class="inline-formula"><img class="img-inline-formula img-formula" width="35" height="16" src="https://math.fontein.de/formulae/n2APuXqE3xuOtTZ7hzwulQ6hgi0.hxRyFQqBpA.svgz" alt="D^{(k)}" title="D^{(k)}"></span> is <span class="inline-formula"><img class="img-inline-formula img-formula" width="14" height="12" src="https://math.fontein.de/formulae/vYu2wcShMlDKJ.IrmXbb2no6ZOHI_2bIn_7ZWQ.svgz" alt="R" title="R"></span>-linear, i.e. <span class="inline-formula"><img class="img-inline-formula img-formula" width="225" height="20" src="https://math.fontein.de/formulae/QIFcOJxOdyQqj5zOcKTwyLcXc6Z4WYNbba3S3g.svgz" alt="D^{(k)} (f + g) = D^{(k)} f + D^{(k)} g" title="D^{(k)} (f + g) = D^{(k)} f + D^{(k)} g"></span> and <span class="inline-formula"><img class="img-inline-formula img-formula" width="149" height="20" src="https://math.fontein.de/formulae/3GcrQ1_Flt6AcxkRs9J6vrKTMrgGjiab.aDyxA.svgz" alt="D^{(k)}(\lambda f) = \lambda D^{(k)} f" title="D^{(k)}(\lambda f) = \lambda D^{(k)} f"></span>.</li>
<li>We have <span class="inline-formula"><img class="img-inline-formula img-formula" width="127" height="19" src="https://math.fontein.de/formulae/M9fDO6rcIa7U867DGJlZp3ne_CCZ1JAlb7HQdg.svgz" alt="k! \cdot D^{(k)} f = f^{(k)}" title="k! \cdot D^{(k)} f = f^{(k)}"></span>; in particular, <span class="inline-formula"><img class="img-inline-formula img-formula" width="84" height="19" src="https://math.fontein.de/formulae/MzY68svXYUiIB6gTMlMZHqKmML.3eTnDsqEo6g.svgz" alt="D^{(1)} f = f'" title="D^{(1)} f = f'"></span>.</li>
<li>We have <span class="inline-formula"><img class="img-inline-formula img-formula" width="205" height="23" src="https://math.fontein.de/formulae/RCWBN6iv5J4GGyoUN1.k0V4aN851bXY.otMIKg.svgz" alt="D^{(k)} D^{(\ell)} f = \binom{k + \ell}{\ell} D^{(k+\ell)} f" title="D^{(k)} D^{(\ell)} f = \binom{k + \ell}{\ell} D^{(k+\ell)} f"></span>.</li>
<li>
<p>
(<a href="https://en.wikipedia.org/wiki/Leibniz_rule_%28generalized_product_rule%29">Leibniz Rule</a>) We have
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="241" height="55" src="https://math.fontein.de/formulae/qc7qk.WeKpCw17aMf3HizoHqcfXW45f4zTyuMQ.svgz" alt="D^{(k)}(f g) = \sum_{i=0}^k D^{(i)} f \cdot D^{(k-i)} g;" title="D^{(k)}(f g) = \sum_{i=0}^k D^{(i)} f \cdot D^{(k-i)} g;">
</div>
<p>
more generally, for <span class="inline-formula"><img class="img-inline-formula img-formula" width="126" height="18" src="https://math.fontein.de/formulae/dllHV.9ISQsvpkvB.BFfvZJDg.ckSe7_sTK69Q.svgz" alt="f_1, \dots, f_t \in R[x]" title="f_1, \dots, f_t \in R[x]"></span>, we have
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="287" height="56" src="https://math.fontein.de/formulae/w.LhgLB8LOujfyIiKdXh6FUprwUj3jeONV2QCg.svgz" alt="D^{(k)} \prod_{i=1}^t f_i = \sum_{m_1 + \dots + m_t = k} \prod_{i=1}^t D^{(m_i)} f_i," title="D^{(k)} \prod_{i=1}^t f_i = \sum_{m_1 + \dots + m_t = k} \prod_{i=1}^t D^{(m_i)} f_i,">
</div>
<p>
where the sum goes over all such tuples <span class="inline-formula"><img class="img-inline-formula img-formula" width="139" height="18" src="https://math.fontein.de/formulae/Ts.OAnmHKeELUV2YmWvLlc624ISFRBxfVQzipQ.svgz" alt="(m_1, \dots, m_t) \in \N^t" title="(m_1, \dots, m_t) \in \N^t"></span>.
</p>
</li>
<li>
<p>
(<a href="https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula">Faà di Bruno's Formula</a>) We have
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="511" height="58" src="https://math.fontein.de/formulae/IKMWR1cBce.4w_k80xNFORlomoq_.eQ7IrBeeg.svgz" alt="D^{(k)} (f \circ g) = \sum \binom{c_1 + \dots + c_k}{c_1, \dots, c_k} (D^{(c_1 + \dots + c_k)} f) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}," title="D^{(k)} (f \circ g) = \sum \binom{c_1 + \dots + c_k}{c_1, \dots, c_k} (D^{(c_1 + \dots + c_k)} f) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j},">
</div>
<p>
where the sum goes over all tuples <span class="inline-formula"><img class="img-inline-formula img-formula" width="128" height="19" src="https://math.fontein.de/formulae/nceALFyr8i9xxuAX6XFDR9Qm2Mnok4CjzNtxbw.svgz" alt="(c_1, \dots, c_k) \in \N^k" title="(c_1, \dots, c_k) \in \N^k"></span> with <span class="inline-formula"><img class="img-inline-formula img-formula" width="99" height="23" src="https://math.fontein.de/formulae/oBe7ZHbW7wG4UXmyZwwLZyeiAtCWfId1DctvcA.svgz" alt="\sum_{i=1}^k i c_i = k" title="\sum_{i=1}^k i c_i = k"></span>; here, <span class="inline-formula"><img class="img-inline-formula img-formula" width="78" height="24" src="https://math.fontein.de/formulae/nfkp.14Mj8OEkQmbtdCmR8h8s3lCKKyZ7_oUHQ.svgz" alt="\binom{c_1 + \dots + c_k}{c_1, \dots, c_k}" title="\binom{c_1 + \dots + c_k}{c_1, \dots, c_k}"></span> is a <a href="https://en.wikipedia.org/wiki/Multinomial_coefficient">multinomial coefficient</a> having the value
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="124" height="40" src="https://math.fontein.de/formulae/C.fDoAQulCS1mgzv2FpXyJx0zuOVUkrOQ4DsDQ.svgz" alt="\frac{(c_1 + \dots + c_k)!}{c_1! \cdots c_k!}." title="\frac{(c_1 + \dots + c_k)!}{c_1! \cdots c_k!}.">
</div>
</li>
<li>
<p>
(Taylor Formula) We have
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="214" height="56" src="https://math.fontein.de/formulae/._W2rklXpma1v_TGDikoEfloC7KCqzBuLrwVUw.svgz" alt="f = \sum_{i=0}^{\deg f} (D^{(i)} f)(\lambda) (x - \lambda)^i." title="f = \sum_{i=0}^{\deg f} (D^{(i)} f)(\lambda) (x - \lambda)^i.">
</div>
</li>
<li>(Identity Theorem) If we have <span class="inline-formula"><img class="img-inline-formula img-formula" width="113" height="20" src="https://math.fontein.de/formulae/bR38a9R2809_6l6VPBiJdfm6aMLGuajTCVp3CQ.svgz" alt="(D^{(i)} f)(\lambda) = 0" title="(D^{(i)} f)(\lambda) = 0"></span> for all <span class="inline-formula"><img class="img-inline-formula img-formula" width="39" height="14" src="https://math.fontein.de/formulae/7bOS.gW7YEc_Vl5o5rMonnTQYAeECKSB0p0pwQ.svgz" alt="i \ge 0" title="i \ge 0"></span>, then <span class="inline-formula"><img class="img-inline-formula img-formula" width="43" height="16" src="https://math.fontein.de/formulae/DjePK8KCxuUnedU8zeqe207OqfO2FVL6xAtBSg.svgz" alt="f = 0" title="f = 0"></span>.</li>
</ol>
</div>
</div>
<p>
For that reason, one can define <span class="inline-formula"><img class="img-inline-formula img-formula" width="104" height="26" src="https://math.fontein.de/formulae/S4HMNfwb0tYNWsCDkMXJM6YysBKuRoIgPoX4YQ.svgz" alt="\frac{f^{(k)}}{k!} := D^{(k)} f" title="\frac{f^{(k)}}{k!} := D^{(k)} f"></span>, so that we can write Taylor's formula in a more tempting form as
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="185" height="52" src="https://math.fontein.de/formulae/D_pCR6P19LtjkToALkcybYJKuVlzIumVmWZ3IA.svgz" alt="f = \sum_{i=0}^n \frac{f^{(i)}}{i!}(\lambda) (x - \lambda)^i," title="f = \sum_{i=0}^n \frac{f^{(i)}}{i!}(\lambda) (x - \lambda)^i,">
</div>
<p>
which almost equals the classical form.
</p>
<p>
Note that the Leibniz rule, Faà di Bruno's formula and Taylor's formula take simpler forms than their classical counterparts; this is due to the fact that the additional factorial terms or binomial coefficients already hide in the definition of the Hasse derivative.
</p>
<div class="theorem-environment theorem-proof-environment qed">
<div class="theorem-header theorem-proof-header">
Proof.
</div>
<div class="theorem-content theorem-proof-content">
<ol class="enum-level-1">
<li>This follows from the definition of <span class="inline-formula"><img class="img-inline-formula img-formula" width="35" height="16" src="https://math.fontein.de/formulae/n2APuXqE3xuOtTZ7hzwulQ6hgi0.hxRyFQqBpA.svgz" alt="D^{(k)}" title="D^{(k)}"></span>.</li>
<li>
<p>
Write <span class="inline-formula"><img class="img-inline-formula img-formula" width="111" height="20" src="https://math.fontein.de/formulae/eVyidLoEydKuaeHkZB27H0clhkz9KArZkAHXtQ.svgz" alt="f = \sum_{i=0}^n a_i x^i" title="f = \sum_{i=0}^n a_i x^i"></span>; then
</p>
<div class="align-formula">
<img class="img-align-formula img-formula" width="405" height="168" src="https://math.fontein.de/formulae/g_9sLMysJo5XmMocTt2JmXEeJQ0nQ5PU4WPWjA.svgz" alt="k! \cdot D^{(k)} f ={} & k! \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} = \sum_{i=k}^n k! \binom{i}{k} a_i x^{i-k} \\
{}={} & \sum_{i=k}^n a_i \cdot i (i - 1) (i - 2) \cdots (i - k + 1) x^{i - k} \\
{}={} & \sum_{i=k}^n a_i (x^i)^{(k)} = f^{(k)}." title="k! \cdot D^{(k)} f ={} & k! \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} = \sum_{i=k}^n k! \binom{i}{k} a_i x^{i-k} \\
{}={} & \sum_{i=k}^n a_i \cdot i (i - 1) (i - 2) \cdots (i - k + 1) x^{i - k} \\
{}={} & \sum_{i=k}^n a_i (x^i)^{(k)} = f^{(k)}.">
</div>
</li>
<li>
<p>
By 1., it is suffices to show this for <span class="inline-formula"><img class="img-inline-formula img-formula" width="50" height="18" src="https://math.fontein.de/formulae/8oHllh8M3TPErCsW_Ly5cefHOv7Wajm2p4LTmA.svgz" alt="f = x^i" title="f = x^i"></span>, <span class="inline-formula"><img class="img-inline-formula img-formula" width="69" height="15" src="https://math.fontein.de/formulae/mdErj84mnLslMupmXrTMTwM1rp5H1dCS0J6mEg.svgz" alt="i \ge k + \ell" title="i \ge k + \ell"></span> (for smaller <span class="inline-formula"><img class="img-inline-formula img-formula" width="6" height="12" src="https://math.fontein.de/formulae/S43oPTrFqmoVC.yqOcgzvrroaMU3pS7pa40ROQ.svgz" alt="i" title="i"></span>, both sides will be zero). We have
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="374" height="43" src="https://math.fontein.de/formulae/WLUSpuxJmO3HqVYXsakam1mm5fyBfYFNAUmatg.svgz" alt="D^{(k)} D^{(\ell)} f = D^{(k)} \binom{i}{\ell} x^{i - \ell} = \binom{i - \ell}{k} \binom{i}{\ell} x^{i - k - \ell}" title="D^{(k)} D^{(\ell)} f = D^{(k)} \binom{i}{\ell} x^{i - \ell} = \binom{i - \ell}{k} \binom{i}{\ell} x^{i - k - \ell}">
</div>
<p>
and
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="208" height="43" src="https://math.fontein.de/formulae/Vmh1l1tiqhUE9U7VIsT6pgwsPkSSS7D.Kzx58g.svgz" alt="D^{(k + \ell)} f = \binom{i}{k + \ell} x^{i - k - \ell}." title="D^{(k + \ell)} f = \binom{i}{k + \ell} x^{i - k - \ell}.">
</div>
<p>
But since
</p>
<div class="align-formula">
<img class="img-align-formula img-formula" width="451" height="140" src="https://math.fontein.de/formulae/uHjkpC1nGYau9.4ja7O_IxKOghXcIEWrP8cyOQ.svgz" alt="\binom{k + \ell}{\ell} \binom{i}{k + \ell} ={} & \frac{(k + \ell)!}{\ell! k!} \frac{i!}{(k + \ell)! (i - k - \ell)!} \\
{}={} & \frac{i!}{\ell! k! (i - k - \ell!)} \\
{}={} & \frac{(i - \ell)!}{k! (i - k - \ell)!} \frac{i!}{\ell! (i - \ell)!} = \binom{i - \ell}{k} \binom{i}{\ell}," title="\binom{k + \ell}{\ell} \binom{i}{k + \ell} ={} & \frac{(k + \ell)!}{\ell! k!} \frac{i!}{(k + \ell)! (i - k - \ell)!} \\
{}={} & \frac{i!}{\ell! k! (i - k - \ell!)} \\
{}={} & \frac{(i - \ell)!}{k! (i - k - \ell)!} \frac{i!}{\ell! (i - \ell)!} = \binom{i - \ell}{k} \binom{i}{\ell},">
</div>
<p>
these terms are equal.
</p>
</li>
<li>
<p>
Note that if we fix <span class="inline-formula"><img class="img-inline-formula img-formula" width="11" height="16" src="https://math.fontein.de/formulae/M.ji_f0zsVnTLyaKegBkRJLOeqrrt6brNnapOQ.svgz" alt="f" title="f"></span>, we get an <span class="inline-formula"><img class="img-inline-formula img-formula" width="14" height="12" src="https://math.fontein.de/formulae/vYu2wcShMlDKJ.IrmXbb2no6ZOHI_2bIn_7ZWQ.svgz" alt="R" title="R"></span>-linear function <span class="inline-formula"><img class="img-inline-formula img-formula" width="95" height="18" src="https://math.fontein.de/formulae/cn7i7Zn3nmF.9.2s0Ata6rpHjlzeja0vnymbRw.svgz" alt="R[x] \to R[x]" title="R[x] \to R[x]"></span>, <span class="inline-formula"><img class="img-inline-formula img-formula" width="105" height="20" src="https://math.fontein.de/formulae/3Z1p9hvOOF8VBEwnCUa.kAbt7ENsBv2u2PheHw.svgz" alt="g \mapsto D^{(k)} (f g)" title="g \mapsto D^{(k)} (f g)"></span>. Hence, it suffices to show this for arbitrary <span class="inline-formula"><img class="img-inline-formula img-formula" width="66" height="18" src="https://math.fontein.de/formulae/rKBraqOZ.NbqH1d2x4aJDWVb4j1QTkoqEzjIww.svgz" alt="f \in R[x]" title="f \in R[x]"></span> and <span class="inline-formula"><img class="img-inline-formula img-formula" width="56" height="15" src="https://math.fontein.de/formulae/be8wb8TF0v9F_IYIwtFCasig9CoZo2LYlBUTXg.svgz" alt="g = x^m" title="g = x^m"></span>, <span class="inline-formula"><img class="img-inline-formula img-formula" width="50" height="13" src="https://math.fontein.de/formulae/lZq6w7FC4RcOgrJ2Nfd2.bCqYfEwb5lFwKGBOg.svgz" alt="m \in \N" title="m \in \N"></span>. By the same argument, for <span class="inline-formula"><img class="img-inline-formula img-formula" width="56" height="15" src="https://math.fontein.de/formulae/be8wb8TF0v9F_IYIwtFCasig9CoZo2LYlBUTXg.svgz" alt="g = x^m" title="g = x^m"></span>, we get an <span class="inline-formula"><img class="img-inline-formula img-formula" width="14" height="12" src="https://math.fontein.de/formulae/vYu2wcShMlDKJ.IrmXbb2no6ZOHI_2bIn_7ZWQ.svgz" alt="R" title="R"></span>-linbear function <span class="inline-formula"><img class="img-inline-formula img-formula" width="95" height="18" src="https://math.fontein.de/formulae/cn7i7Zn3nmF.9.2s0Ata6rpHjlzeja0vnymbRw.svgz" alt="R[x] \to R[x]" title="R[x] \to R[x]"></span>, <span class="inline-formula"><img class="img-inline-formula img-formula" width="121" height="20" src="https://math.fontein.de/formulae/Ne69UgNlQtWCcD7VbPxuDDY9UviO27XGgD1Euw.svgz" alt="f \mapsto D^{(k)} (f x^m)" title="f \mapsto D^{(k)} (f x^m)"></span>; therefore, it suffices to consider <span class="inline-formula"><img class="img-inline-formula img-formula" width="54" height="16" src="https://math.fontein.de/formulae/B9VYHSYPG0rfTwZEUyrzgMMYEywa5Jt04x4acA.svgz" alt="f = x^n" title="f = x^n"></span>, <span class="inline-formula"><img class="img-inline-formula img-formula" width="45" height="13" src="https://math.fontein.de/formulae/UeChwhh6h7Bw5vi122OTLvfxRX.k6iU0MpIILg.svgz" alt="n \in \N" title="n \in \N"></span>. But now,
</p>
<div class="align-formula">
<img class="img-align-formula img-formula" width="446" height="142" src="https://math.fontein.de/formulae/m5bOr_WWF3qknj98FQe4zEnc9hiImYurlwL.aw.svgz" alt="D^{(k)} (x^n x^m) ={} & D^{(k)} x^{n + m} = \binom{n + m}{k} x^{n + m - k} \\
\text{and} \quad D^{(i)} x^n \cdot D^{(k-i)} x^m ={} & \binom{n}{i} x^{n - i} \binom{m}{k - i} x^{m - (k - i)} \\
{}={} & \binom{n}{i} \binom{m}{k - i} x^{n + m - k}." title="D^{(k)} (x^n x^m) ={} & D^{(k)} x^{n + m} = \binom{n + m}{k} x^{n + m - k} \\
\text{and} \quad D^{(i)} x^n \cdot D^{(k-i)} x^m ={} & \binom{n}{i} x^{n - i} \binom{m}{k - i} x^{m - (k - i)} \\
{}={} & \binom{n}{i} \binom{m}{k - i} x^{n + m - k}.">
</div>
<p>
Hence, it suffices to show <span class="inline-formula"><img class="img-inline-formula img-formula" width="183" height="25" src="https://math.fontein.de/formulae/H72mRULl.OKelCDKcnStS1BlaLlAbJx3P1xcKg.svgz" alt="\sum_{i=0}^k \binom{n}{i} \binom{m}{k - i} = \binom{n + m}{k}" title="\sum_{i=0}^k \binom{n}{i} \binom{m}{k - i} = \binom{n + m}{k}"></span>. By reorganizing the binomial coefficients, one transforms this into the equality
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="275" height="55" src="https://math.fontein.de/formulae/9jqqwbd5gDRlN_UMmJkaKGUUfq9RWTOYywj1bA.svgz" alt="\sum_{i=0}^k \binom{k}{i} \binom{n + m - k}{n - i} = \binom{n + m}{n}," title="\sum_{i=0}^k \binom{k}{i} \binom{n + m - k}{n - i} = \binom{n + m}{n},">
</div>
<p>
which is <a href="https://en.wikipedia.org/wiki/Vandermonde%27s_identity">Vandermonde's Identity</a> and, hence, true.
</p>
<p>
The more general equation is shown by induction on <span class="inline-formula"><img class="img-inline-formula img-formula" width="6" height="11" src="https://math.fontein.de/formulae/Fb4QjIncBFMP2NDS0yCZ5Fvi7l8GB681giwdQQ.svgz" alt="t" title="t"></span>. For <span class="inline-formula"><img class="img-inline-formula img-formula" width="39" height="11" src="https://math.fontein.de/formulae/pcuDtqzBhs5mZOQwquMXekvxdsfb3JlRTmmFRw.svgz" alt="t = 1" title="t = 1"></span>, we have <span class="inline-formula"><img class="img-inline-formula img-formula" width="292" height="24" src="https://math.fontein.de/formulae/tEgJxTwnV9tybupBLnFJSptgZIhKCub.7QCR_A.svgz" alt="\sum_{m_1 + \dots + m_t = k} \prod_{i=1}^t D^{(m_i)} f_i = D^{(k)} f_1" title="\sum_{m_1 + \dots + m_t = k} \prod_{i=1}^t D^{(m_i)} f_i = D^{(k)} f_1"></span>. Now, assume that the equation is true for all <span class="inline-formula"><img class="img-inline-formula img-formula" width="10" height="12" src="https://math.fontein.de/formulae/KIbZSMqheEMj2LBRLlu5emEXwvNaAcODJL5YWQ.svgz" alt="k" title="k"></span> for one <span class="inline-formula"><img class="img-inline-formula img-formula" width="39" height="14" src="https://math.fontein.de/formulae/fOmKsfRNmJwngAAt_aHR0Dbtj0RUgJNcuLhy5A.svgz" alt="t \ge 1" title="t \ge 1"></span>. Then, for any <span class="inline-formula"><img class="img-inline-formula img-formula" width="10" height="12" src="https://math.fontein.de/formulae/KIbZSMqheEMj2LBRLlu5emEXwvNaAcODJL5YWQ.svgz" alt="k" title="k"></span>,
</p>
<div class="align-formula">
<img class="img-align-formula img-formula" width="406" height="182" src="https://math.fontein.de/formulae/8jxbzG3DbL6IxoUh07BjLd8ffUu1QSNvXU64ow.svgz" alt="& D^{(k)} \prod_{i=1}^{t+1} f_i = D^{(k)} \biggl( f_{t+1} \cdot \prod_{i=1}^t f_i \biggr) \\
{}={} & \sum_{m_{t+1} = 0}^k D^{(m_{t+1})} f_{t+1} \cdot D^{(k-m_{t+1})} \biggl( \prod_{i=1}^t f_i \biggr) \\
{}={} & \sum_{m_{t+1} = 0}^k D^{(m_{t+1})} f_{t+1} \cdot \sum_{m_1 + \dots + m_t = k - m_{t+1}} \prod_{i=1}^t D^{(m_i)} f_i" title="& D^{(k)} \prod_{i=1}^{t+1} f_i = D^{(k)} \biggl( f_{t+1} \cdot \prod_{i=1}^t f_i \biggr) \\
{}={} & \sum_{m_{t+1} = 0}^k D^{(m_{t+1})} f_{t+1} \cdot D^{(k-m_{t+1})} \biggl( \prod_{i=1}^t f_i \biggr) \\
{}={} & \sum_{m_{t+1} = 0}^k D^{(m_{t+1})} f_{t+1} \cdot \sum_{m_1 + \dots + m_t = k - m_{t+1}} \prod_{i=1}^t D^{(m_i)} f_i">
</div>
<p>
by the Leibniz rule and by the induction hypothesis; here, the second sum goes over all such tuples <span class="inline-formula"><img class="img-inline-formula img-formula" width="139" height="18" src="https://math.fontein.de/formulae/Ts.OAnmHKeELUV2YmWvLlc624ISFRBxfVQzipQ.svgz" alt="(m_1, \dots, m_t) \in \N^t" title="(m_1, \dots, m_t) \in \N^t"></span>. But this equals <span class="inline-formula"><img class="img-inline-formula img-formula" width="262" height="25" src="https://math.fontein.de/formulae/xkuYRdfSIxXmx1BOGRm8yBPKQ_icWtXGuiUIqQ.svgz" alt="\sum_{m_1+\dots+m_t+m_{t+1}=k} \prod_{i=1}^{t+1} D^{(m_i)} f_i" title="\sum_{m_1+\dots+m_t+m_{t+1}=k} \prod_{i=1}^{t+1} D^{(m_i)} f_i"></span>, what we had to show.
</p>
</li>
<li>
<p>
Again, by 1., it suffices to show this for <span class="inline-formula"><img class="img-inline-formula img-formula" width="54" height="16" src="https://math.fontein.de/formulae/B9VYHSYPG0rfTwZEUyrzgMMYEywa5Jt04x4acA.svgz" alt="f = x^n" title="f = x^n"></span>, <span class="inline-formula"><img class="img-inline-formula img-formula" width="45" height="13" src="https://math.fontein.de/formulae/UeChwhh6h7Bw5vi122OTLvfxRX.k6iU0MpIILg.svgz" alt="n \in \N" title="n \in \N"></span>. Now, by the second part of 4.,
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="390" height="54" src="https://math.fontein.de/formulae/_mS3OunxtLyX1s_l2_hEaXgtiUkeljrIl7uESw.svgz" alt="D^{(k)}(g^n) = D^{(k)}(g \cdots g) = \sum_{m_1 + \dots + m_n = k} \prod_{i=1}^n D^{(m_i)} g," title="D^{(k)}(g^n) = D^{(k)}(g \cdots g) = \sum_{m_1 + \dots + m_n = k} \prod_{i=1}^n D^{(m_i)} g,">
</div>
<p>
where the sum goes over all such <span class="inline-formula"><img class="img-inline-formula img-formula" width="146" height="18" src="https://math.fontein.de/formulae/i2.8GvdFhFWLyb3.bMNLkJdp22Y_jNxIMxibvg.svgz" alt="(m_1, \dots, m_n) \in \N^n" title="(m_1, \dots, m_n) \in \N^n"></span>. The formula we want is now obtained by sorting the summands by the different powers of <span class="inline-formula"><img class="img-inline-formula img-formula" width="41" height="19" src="https://math.fontein.de/formulae/lTmZ6Js0m0.FViqgP0zsJ552qoUOgyjRUQwvgQ.svgz" alt="D^{(i)} g" title="D^{(i)} g"></span> appearing, <span class="inline-formula"><img class="img-inline-formula img-formula" width="72" height="15" src="https://math.fontein.de/formulae/7zAsSCyNIpg6oj3DtJNy71ZEC.kwUnXUtzgknA.svgz" alt="0 \le i \le k" title="0 \le i \le k"></span>.
</p>
<p>
Consider the map
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="459" height="20" src="https://math.fontein.de/formulae/Z_OYhNMCRvXjd3NYhcaOcHt8KIJz4cUUD3CLEA.svgz" alt="\varphi : \N^n \to \N^{k+1}, \quad m = (m_1, \dots, m_n) \mapsto (c_0(m), \dots, c_k(m))," title="\varphi : \N^n \to \N^{k+1}, \quad m = (m_1, \dots, m_n) \mapsto (c_0(m), \dots, c_k(m)),">
</div>
<p>
there <span class="inline-formula"><img class="img-inline-formula img-formula" width="273" height="18" src="https://math.fontein.de/formulae/7YhYsSlixv1DK6fegNGyVddrEsluBS1Nl85n8g.svgz" alt="c_i(m) := \abs{\{ j \in \{ 1, \dots, n \} \mid m_j = i \}}" title="c_i(m) := \abs{\{ j \in \{ 1, \dots, n \} \mid m_j = i \}}"></span>, <span class="inline-formula"><img class="img-inline-formula img-formula" width="72" height="15" src="https://math.fontein.de/formulae/7zAsSCyNIpg6oj3DtJNy71ZEC.kwUnXUtzgknA.svgz" alt="0 \le i \le k" title="0 \le i \le k"></span>. Now, if <span class="inline-formula"><img class="img-inline-formula img-formula" width="185" height="18" src="https://math.fontein.de/formulae/3ChFma9wYXWckAteQkIVqPORyOCpupt4ytozoQ.svgz" alt="m = (m_1, \dots, m_n) \in \N^n" title="m = (m_1, \dots, m_n) \in \N^n"></span> satisfies <span class="inline-formula"><img class="img-inline-formula img-formula" width="100" height="20" src="https://math.fontein.de/formulae/DtYZsrxQEWDOsPsqbw8o5SeF8ti6rwZs23pGDQ.svgz" alt="\sum_{i=1}^n m_i = k" title="\sum_{i=1}^n m_i = k"></span>, then <span class="inline-formula"><img class="img-inline-formula img-formula" width="133" height="25" src="https://math.fontein.de/formulae/tgQYR3VCWtjWSL2jebybcHj2lR4_zKhwcqU9Og.svgz" alt="\sum_{j=0}^k j c_j(m) = k" title="\sum_{j=0}^k j c_j(m) = k"></span> and <span class="inline-formula"><img class="img-inline-formula img-formula" width="126" height="25" src="https://math.fontein.de/formulae/8RqIITv3k2eF1a.fmhgavpYsIDU.CYqWWuaThQ.svgz" alt="\sum_{j=0}^k c_j(m) = n" title="\sum_{j=0}^k c_j(m) = n"></span>. Now, for a fixed <span class="inline-formula"><img class="img-inline-formula img-formula" width="146" height="19" src="https://math.fontein.de/formulae/TOGd5PV9DTTgBhON_uWFL1x8g3rSR2i6VytTDg.svgz" alt="(c_0, \dots, c_k) \in \N^{k+1}" title="(c_0, \dots, c_k) \in \N^{k+1}"></span> with <span class="inline-formula"><img class="img-inline-formula img-formula" width="93" height="23" src="https://math.fontein.de/formulae/O9KnNjUK0Ws.Cy.JDcYnjw.kcKGtU2EY1TUVNA.svgz" alt="\sum_{i=0}^k c_i = n" title="\sum_{i=0}^k c_i = n"></span> and <span class="inline-formula"><img class="img-inline-formula img-formula" width="99" height="23" src="https://math.fontein.de/formulae/CsXru8qUj7fEtxtHjjQVJ8eIl7zbIPaJsgGVxg.svgz" alt="\sum_{i=0}^k i c_i = k" title="\sum_{i=0}^k i c_i = k"></span>, the number <span class="inline-formula"><img class="img-inline-formula img-formula" width="128" height="21" src="https://math.fontein.de/formulae/_3gAl1g3lApMVyb4Xwbma2qQK3y6sDDae533SA.svgz" alt="\abs{\varphi^{-1}(c_0, \dots, c_k)}" title="\abs{\varphi^{-1}(c_0, \dots, c_k)}"></span> equals the <a href="https://en.wikipedia.org/wiki/Multinomial_coefficient#Number_of_unique_permutations_of_words">multinomial coefficient</a>
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="256" height="43" src="https://math.fontein.de/formulae/doohH95PjRCc8SkgVu_CS5sFcDU_hdUc.ZMyjQ.svgz" alt="\binom{n}{c_0, c_1, \dots, c_k} = \frac{n!}{c_0! \cdot c_1! \cdots c_n!}," title="\binom{n}{c_0, c_1, \dots, c_k} = \frac{n!}{c_0! \cdot c_1! \cdots c_n!},">
</div>
<p>
whence we get that the above formula for <span class="inline-formula"><img class="img-inline-formula img-formula" width="67" height="20" src="https://math.fontein.de/formulae/m6mK3fLm6np3I_geIOb6Byl5ZKa1aaOdqhKTmg.svgz" alt="D^{(k)}(g^n)" title="D^{(k)}(g^n)"></span> equals
</p>
<div class="display-formula">
<img class="img-display-formula img-formula" width="288" height="58" src="https://math.fontein.de/formulae/NbHl4ieiB99702PNSocKVfMs.m1L7ZRlt99cUg.svgz" alt="\sum \binom{n}{c_0, c_1, \dots, c_k} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}," title="\sum \binom{n}{c_0, c_1, \dots, c_k} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j},">
</div>
<p>
where the sum goes over all tuples <span class="inline-formula"><img class="img-inline-formula img-formula" width="152" height="19" src="https://math.fontein.de/formulae/wdGAGkF6Q5NCeQsB.qaUq4jcvMw5Wss.FKo5Xg.svgz" alt="(c_0, c_1, \dots, c_k) \in \N^k" title="(c_0, c_1, \dots, c_k) \in \N^k"></span> with <span class="inline-formula"><img class="img-inline-formula img-formula" width="99" height="23" src="https://math.fontein.de/formulae/CsXru8qUj7fEtxtHjjQVJ8eIl7zbIPaJsgGVxg.svgz" alt="\sum_{i=0}^k i c_i = k" title="\sum_{i=0}^k i c_i = k"></span> and <span class="inline-formula"><img class="img-inline-formula img-formula" width="93" height="23" src="https://math.fontein.de/formulae/O9KnNjUK0Ws.Cy.JDcYnjw.kcKGtU2EY1TUVNA.svgz" alt="\sum_{i=0}^k c_i = n" title="\sum_{i=0}^k c_i = n"></span>.
</p>
<p>
Now note that <span class="inline-formula"><img class="img-inline-formula img-formula" width="339" height="24" src="https://math.fontein.de/formulae/lMLCcfbMrHJWHqYYqiej3dHHeOre8n03_JXKQQ.svgz" alt="\binom{n}{n - c_0} g^{c_0} = \binom{n}{n - c_0} x^{c_0} \circ g = D^{(n - c_0)}(x^n) \circ g" title="\binom{n}{n - c_0} g^{c_0} = \binom{n}{n - c_0} x^{c_0} \circ g = D^{(n - c_0)}(x^n) \circ g"></span> and <span class="inline-formula"><img class="img-inline-formula img-formula" width="143" height="25" src="https://math.fontein.de/formulae/T53264CThfVgH5RMrmGkgwYzE9rdhbxbdyW0jw.svgz" alt="\binom{n}{n - c_0} = \frac{n!}{c_0! (n - c_0)!}" title="\binom{n}{n - c_0} = \frac{n!}{c_0! (n - c_0)!}"></span>. Therefore, we get
</p>
<div class="align-formula">
<img class="img-align-formula img-formula" width="357" height="248" src="https://math.fontein.de/formulae/IEZfnJ6Umz97KR5pdqHqLHn0YDECb0lC4zpEMA.svgz" alt="& \sum \binom{n}{c_0, c_1, \dots, c_k} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j} \\
{}={} & \sum \frac{n!}{c_0! \cdot c_1! \cdots c_n!} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j} \\
{}={} & \sum \frac{n!}{c_0! \cdot (n - c_0)!} \frac{(n - c_0)!}{c_1! \cdots c_n!} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j} \\
{}={} & \sum \frac{(n - c_0)!}{c_1! \cdots c_n!} D^{(n - c_0)}(x^n) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}." title="& \sum \binom{n}{c_0, c_1, \dots, c_k} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j} \\
{}={} & \sum \frac{n!}{c_0! \cdot c_1! \cdots c_n!} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j} \\
{}={} & \sum \frac{n!}{c_0! \cdot (n - c_0)!} \frac{(n - c_0)!}{c_1! \cdots c_n!} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j} \\
{}={} & \sum \frac{(n - c_0)!}{c_1! \cdots c_n!} D^{(n - c_0)}(x^n) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}.">
</div>
<p>
If one now replaces <span class="inline-formula"><img class="img-inline-formula img-formula" width="48" height="13" src="https://math.fontein.de/formulae/ztOtm_ramZEUwLRt2SeC.1U0svkiHbZkjpYE3Q.svgz" alt="n - c_0" title="n - c_0"></span> by <span class="inline-formula"><img class="img-inline-formula img-formula" width="97" height="13" src="https://math.fontein.de/formulae/A9QyWfSpWvjYEyifV8E6GY7tUFqTYJ5lAKWLbQ.svgz" alt="c_1 + \dots + c_n" title="c_1 + \dots + c_n"></span>, we obtain the formula from the statement.
</p>
</li>
<li>
<p>
By 1., it suffices to show this for <span class="inline-formula"><img class="img-inline-formula img-formula" width="54" height="16" src="https://math.fontein.de/formulae/B9VYHSYPG0rfTwZEUyrzgMMYEywa5Jt04x4acA.svgz" alt="f = x^n" title="f = x^n"></span>, <span class="inline-formula"><img class="img-inline-formula img-formula" width="45" height="13" src="https://math.fontein.de/formulae/UeChwhh6h7Bw5vi122OTLvfxRX.k6iU0MpIILg.svgz" alt="n \in \N" title="n \in \N"></span>. Now
</p>
<div class="align-formula">
<img class="img-align-formula img-formula" width="413" height="140" src="https://math.fontein.de/formulae/TjmCquP97jxKXkHcdjd13NoFN5ehvZhlIUgkmw.svgz" alt="\sum_{i=0}^{\deg f} (D^{(i)} f)(\lambda) (x - \lambda)^k ={} & \sum_{i=0}^n \biggl(\binom{n}{i} x^{n - i}\biggr)(\lambda) (x - \lambda)^i \\
{}={} & \sum_{i=0}^n \binom{n}{i} \lambda^{n-i} (x - \lambda)^i \\
{}={} & ((x - \lambda) + \lambda)^n = x^n" title="\sum_{i=0}^{\deg f} (D^{(i)} f)(\lambda) (x - \lambda)^k ={} & \sum_{i=0}^n \biggl(\binom{n}{i} x^{n - i}\biggr)(\lambda) (x - \lambda)^i \\
{}={} & \sum_{i=0}^n \binom{n}{i} \lambda^{n-i} (x - \lambda)^i \\
{}={} & ((x - \lambda) + \lambda)^n = x^n">
</div>
<p>
by the <a href="https://en.wikipedia.org/wiki/Binomial_theorem">Binomial Theorem</a>, what we had to show.
</p>
</li>
<li>This follows directly from 6.</li>
</ol>
</div>
<div class="qed-block"><span class="qed-sign"></span></div>
</div>
</div>