Mfcttrf

How do I, as a DM, handle a party that decides to set up an ambush in a dungeon?

Is it normal for gliders not to have attitude indicators?

Clean sphere separation with a sphere created from a cube

Is there a proof that the set of real numbers can exactly represent distances?

Meaning of the (idiomatic?) expression "seghe mentali"

How to calculate rate of axial precession?

Krull dimension of the ring of global sections

Is there a word that describes the unjustified use of a more complex word?

How to remap repeating commands i.e. <number><command>?

How do I allocate more memory to an app on Sheepshaver running Mac OS 9?

Has the United States ever had a non-Christian President?

Why did the Apollo 13 crew extend the LM landing gear?

Why didn't this character get a funeral at the end of Avengers: Endgame?

Understanding ties

Constitutional limitation of criminalizing behavior in US law?

My first C++ game (snake console game)

The Adventures of a Chocolate Cookie

Is there a word for food that's gone 'bad', but is still edible?

Sci-fi/fantasy book - ships on steel runners skating across ice sheets

Speed up this NIntegrate

What to do when scriptures go against conscience?

Is there precedent or are there procedures for a US president refusing to concede to an electoral defeat?

In linear regression why does regularisation penalise the parameter values as well?

All superlinear runtime algorithms are asymptotically equivalent to convex function?

Clarification of algebra in moment generating functions

I'm having trouble seeing how they go from step 1 to step 2Simple convolution problemprobability density of the maximum of samples from a normalized uniform distributionCharacterization of point process, given the number of pointsDiscrete random vector and their sumEquivalent condition for independence of discrete random variables (check my work)Simplification step in derivation of posterior pdf using Bayes' ruleFraction simplification numerator unclearFinding the probability that an inner product is positiveFinding a constant in probability function

2

$begingroup$

Suppose $X$ has a range $1,2,dots n $ and $p_X(j)=1/n$ for $1leq j leq n$ (uniform distribution). Then beginalign*
g(t)&=sum_j=1^nfrac1ne^tj\
&=frac1n(e^t+e^2t+cdots+e^nt)\
&=frace^t(e^nt-1)n(e^t-1) endalign*

I don't understand how the algebra goes from step 2 to step 3 here. I understand factoring out an $e^t$, but how does the denominator come about. Is this polynomial division?

probability algebra-precalculus

share|cite|improve this question

asked 31 mins ago

PeetriusPeetrius

350113

$endgroup$

add a comment | 

2

$begingroup$

Suppose $X$ has a range $1,2,dots n $ and $p_X(j)=1/n$ for $1leq j leq n$ (uniform distribution). Then beginalign*
g(t)&=sum_j=1^nfrac1ne^tj\
&=frac1n(e^t+e^2t+cdots+e^nt)\
&=frace^t(e^nt-1)n(e^t-1) endalign*

I don't understand how the algebra goes from step 2 to step 3 here. I understand factoring out an $e^t$, but how does the denominator come about. Is this polynomial division?

probability algebra-precalculus

share|cite|improve this question

asked 31 mins ago

PeetriusPeetrius

350113

$endgroup$

add a comment | 

$begingroup$

Suppose $X$ has a range $1,2,dots n $ and $p_X(j)=1/n$ for $1leq j leq n$ (uniform distribution). Then beginalign*
g(t)&=sum_j=1^nfrac1ne^tj\
&=frac1n(e^t+e^2t+cdots+e^nt)\
&=frace^t(e^nt-1)n(e^t-1) endalign*

I don't understand how the algebra goes from step 2 to step 3 here. I understand factoring out an $e^t$, but how does the denominator come about. Is this polynomial division?

probability algebra-precalculus

share|cite|improve this question

asked 31 mins ago

PeetriusPeetrius

350113

$endgroup$

share|cite|improve this question

asked 31 mins ago

PeetriusPeetrius

350113

share|cite|improve this question

asked 31 mins ago

PeetriusPeetrius

350113

asked 31 mins ago

PeetriusPeetrius

350113

asked 31 mins ago

PeetriusPeetrius

350113

2 Answers
2

active

oldest

votes

3

$begingroup$

They are applying the formula for a finite geometric series,

$$ sum_k=0^n x^k = fracx^n+1-1x-1.$$

This formula can be derived in various ways, including some which involve polynomial division.

Here is one approach :

Let,

$$ S_n(x) = sum_k=0^n x^k ,$$

and note that,

$$ x S_n(x) = sum_k=0^n x^k+1 ,$$

$$ x S_n(x) = sum_k=1^n+1 x^k ,$$

$$ x S_n(x) = sum_k=1^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = 1+sum_k=1^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = sum_k=0^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = S_n(x) + x^n+1 ,$$

$$ x S_n(x) = S_n(x) + x^n+1-1 ,$$

$$ x S_n(x) - S_n(x) = x^n+1-1 ,$$

$$ (x-1) S_n(x) = x^n+1-1 ,$$

$$ S_n(x) = fracx^n+1-1x-1 ,$$

share|cite|improve this answer

answered 22 mins ago

SpencerSpencer

9,06722357

$endgroup$

add a comment | 

1

$begingroup$

It's a geometric series: $1 + (e^t) + (e^t)^2+...+(e^t)^n-1$ with the sum given as a part of your expression.

share|cite|improve this answer

answered 23 mins ago

dnqxtdnqxt

93025

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3213868%2fclarification-of-algebra-in-moment-generating-functions%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

They are applying the formula for a finite geometric series,

$$ sum_k=0^n x^k = fracx^n+1-1x-1.$$

This formula can be derived in various ways, including some which involve polynomial division.

Here is one approach :

Let,

$$ S_n(x) = sum_k=0^n x^k ,$$

and note that,

$$ x S_n(x) = sum_k=0^n x^k+1 ,$$

$$ x S_n(x) = sum_k=1^n+1 x^k ,$$

$$ x S_n(x) = sum_k=1^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = 1+sum_k=1^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = sum_k=0^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = S_n(x) + x^n+1 ,$$

$$ x S_n(x) = S_n(x) + x^n+1-1 ,$$

$$ x S_n(x) - S_n(x) = x^n+1-1 ,$$

$$ (x-1) S_n(x) = x^n+1-1 ,$$

$$ S_n(x) = fracx^n+1-1x-1 ,$$

share|cite|improve this answer

answered 22 mins ago

SpencerSpencer

9,06722357

$endgroup$

add a comment | 

3

$begingroup$

They are applying the formula for a finite geometric series,

$$ sum_k=0^n x^k = fracx^n+1-1x-1.$$

This formula can be derived in various ways, including some which involve polynomial division.

Here is one approach :

Let,

$$ S_n(x) = sum_k=0^n x^k ,$$

and note that,

$$ x S_n(x) = sum_k=0^n x^k+1 ,$$

$$ x S_n(x) = sum_k=1^n+1 x^k ,$$

$$ x S_n(x) = sum_k=1^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = 1+sum_k=1^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = sum_k=0^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = S_n(x) + x^n+1 ,$$

$$ x S_n(x) = S_n(x) + x^n+1-1 ,$$

$$ x S_n(x) - S_n(x) = x^n+1-1 ,$$

$$ (x-1) S_n(x) = x^n+1-1 ,$$

$$ S_n(x) = fracx^n+1-1x-1 ,$$

share|cite|improve this answer

answered 22 mins ago

SpencerSpencer

9,06722357

$endgroup$

add a comment | 

$begingroup$

They are applying the formula for a finite geometric series,

$$ sum_k=0^n x^k = fracx^n+1-1x-1.$$

This formula can be derived in various ways, including some which involve polynomial division.

Here is one approach :

Let,

$$ S_n(x) = sum_k=0^n x^k ,$$

and note that,

$$ x S_n(x) = sum_k=0^n x^k+1 ,$$

$$ x S_n(x) = sum_k=1^n+1 x^k ,$$

$$ x S_n(x) = sum_k=1^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = 1+sum_k=1^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = sum_k=0^n x^k + x^n+1 ,$$

$$ 1+ x S_n(x) = S_n(x) + x^n+1 ,$$

$$ x S_n(x) = S_n(x) + x^n+1-1 ,$$

$$ x S_n(x) - S_n(x) = x^n+1-1 ,$$

$$ (x-1) S_n(x) = x^n+1-1 ,$$

$$ S_n(x) = fracx^n+1-1x-1 ,$$

share|cite|improve this answer

answered 22 mins ago

SpencerSpencer

9,06722357

$endgroup$

share|cite|improve this answer

answered 22 mins ago

SpencerSpencer

9,06722357

answered 22 mins ago

SpencerSpencer

9,06722357

answered 22 mins ago

SpencerSpencer

9,06722357

1

$begingroup$

It's a geometric series: $1 + (e^t) + (e^t)^2+...+(e^t)^n-1$ with the sum given as a part of your expression.

share|cite|improve this answer

answered 23 mins ago

dnqxtdnqxt

93025

$endgroup$

add a comment | 

1

$begingroup$

It's a geometric series: $1 + (e^t) + (e^t)^2+...+(e^t)^n-1$ with the sum given as a part of your expression.

share|cite|improve this answer

answered 23 mins ago

dnqxtdnqxt

93025

$endgroup$

add a comment | 

$begingroup$

It's a geometric series: $1 + (e^t) + (e^t)^2+...+(e^t)^n-1$ with the sum given as a part of your expression.

share|cite|improve this answer

answered 23 mins ago

dnqxtdnqxt

93025

$endgroup$

share|cite|improve this answer

answered 23 mins ago

dnqxtdnqxt

93025

answered 23 mins ago

dnqxtdnqxt

93025

answered 23 mins ago

dnqxtdnqxt

93025