- #1

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- Homework Statement
- look at the image

- Relevant Equations
- Sommation

taylor series

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Besides convergence of the Taylor series, you're trying to calculate a finite sum whereas the Taylor series is an infinite sum.$$\sum_{k=4}^{N+4} \frac{x^k}{k} = \sum_{k=4}^{N+4} \int x^{k-1}\,dx$$$$\sum_{k=4}^{N+4} \frac{x^k}{k} = \sum_{k=4}^{N+4} \int x^{k-1}\,dx$$The Taylor series for ##x## is:$$\displaystyle \sum_{n=0}^N \ {2^nf

- #1

- 310

- 54

- Homework Statement
- look at the image

- Relevant Equations
- Sommation

taylor series

- #2

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That is derailing spectacularly ! Write down a few terms of the one and the other and see that the two are very different !I started by removing the 4

Also: renaming an index (##n## here) works better if you use a different symbol (e.g. use ##k = n+4##) and that changes the summation limits to:

##n=0,\quad k = n +4 \Rightarrow ## lower limit is ##k = 4##

##n=N,\quad k = n+4 \Rightarrow ## upper limit is ##k = N+4##

And in the numerator you have ##n=k-4## so the exponent is not ##k## but ##k-4##

##\ ##

- #3

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you are right thank youHomework Statement::look at the image

Relevant Equations::Sommation

taylor series

Greetings!

I want to caluculate the summation of this following serie

View attachment 297263

I started by removing the 4 by

View attachment 297264

and then

View attachment 297265

and I thought of the taylor expansion of

Log(1-x)=-∑x^{n}/n but as the 2 is not inside (-1,1) I couldn´t use it

any hint?

thank you!

Best !

and then? the main problem remain unsolved?

- #4

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[tex] \int_{-\infty}^1 2^{mt} dt = [\frac{2^{mt}}{m \ln 2}]_{-\infty}^1=\frac{1}{\ln 2}\frac{2^m}{m}[/tex]

for easy summation.

- #5

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thank you!

[tex] \int_{-\infty}^1 2^{mt} dt = [\frac{2^{mt}}{m \ln 2}]_{-\infty}^1=\frac{1}{\ln 2}\frac{2^m}{m}[/tex]

for easy summation.

- #6

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Haha, if you're starting in the wrong direction it won't be solved anywaythe main problem remain unsolved

I don't have all the answers but a modest beginning might come about by writing down a bunch of terms for a low value of ##N## ...

So the actual homework statement isHomework Statement::look at the image

Relevant Equations::Sommation

taylor series

I want to caluculate

[edit] Ah, we have another contributor !

##\ ##

- #7

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I don't have all the answers but a modest beginning might come about by writing down a bunch of terms for a low value of ##N## ...

Indeed, that was a huge contribution! I can do this kind of deadly mistakes during exams, thanks a million for point it out!

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- #9

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Indeed, my instructor said he can asked it in the exam, so I tried to solved it, :)

- #10

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Besides convergence of the Taylor series, you're trying to calculate a finite sum whereas the Taylor series is an infinite sum.I thought of the taylor expansion of

Log(1-x)=-∑x^{n}/n but as the 2 is not inside (-1,1) I couldn´t use it

Here's a hint:

$$\sum_{k=4}^{N+4} \frac{x^k}{k} = \sum_{k=4}^{N+4} \int x^{k-1}\,dx$$ Remember that you're evaluating a finite sum, so the problem actually isn't too bad.

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