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These are questions asked by one pound

Its difficult for me to explain what I'm looking for so I will try with an example:

Example 1

Is there anyway any function that will say for example the sum of squares of integers for a given n say n=4, 1^2 + 2^2 + 3^2 + 4^2 = 30 that will give the equation ((n^3)/3) + ((n^2)/2) + (n/6) that gives the sum for a given n?

It would be determined from the series 1^2 + 2^2 + 3^2 + 4^2 + ... or the equation for the elements 1^2, 2^2, 3^2, 4^2 the equation that gives the sum; in this case the well known ((n^3)/3) + ((n^2)/2) + (n/6) .

 

So far I've got FindSequenceFunction[{1, 4, 9, 16}, n] but this only finds the individual elements not an equation for the sum of them. Of course I already know the equation for producing the individual elements.

1) I do not understand,

Ei(1, -ln(a-2)) for real numbers > +2

how to I convert it into this series format are there are 2 parameters ? (from https://en.wikipedia.org/wiki/Exponential_integral)

I'm aware of https://www.maplesoft.com/support/help/Maple/view.aspx?path=Ei&term=Ei

but as you can deduce I'm still having difficulties. 

2) Also In mary boas's book she only deals with Ei(x) no where can I figure out how 0.2193839344 is calculated from Ei(1,1.)?

Ei(1,1.)

0.2193839344

 

I'm  trying to converting an infinite series from several terms to sigma form.

The series in question is:

x-x^2/(t*Log[t])+(2/3)*x^3/(t^2*Log[t]^2)+(-1/(2*t^3*Log[t]^3)-1/(4*t^3*Log[t]^2)-1/(6*t^3*Log[t]))*x^4+(2/(5*t^4*Log[t]^4)+2/(5*t^4*Log[t]^3)+4/(15*t^4*Log[t]^2))*x^5+(-1/(3*t^5*Log[t]^5)-1/(2*t^5*Log[t]^4)-5/(12*t^5*Log[t]^3)-5/(36*t^5*Log[t]^2)-1/(15*t^5*Log[t]))*x^6+(2/(7*t^6*Log[t]^6)+4/(7*t^6*Log[t]^5)+25/(42*t^6*Log[t]^4)+1/(3*t^6*Log[t]^3)+46/(315*t^6*Log[t]^2))*x^7+(-1/(4*t^7*Log[t]^7)-5/(8*t^7*Log[t]^6)-19/(24*t^7*Log[t]^5)-19/(32*t^7*Log[t]^4)-19/(60*t^7*Log[t]^3)-7/(80*t^7*Log[t]^2)-1/(28*t^7*Log[t]))*x^8+(2/(9*t^8*Log[t]^8)+2/(3*t^8*Log[t]^7)+1/(t^8*Log[t]^6)+25/(27*t^8*Log[t]^5)+163/(270*t^8*Log[t]^4)+106/(405*t^8*Log[t]^3)+88/(945*t^8*Log[t]^2))*x^9+(-1/(5*t^9*Log[t]^9)-7/(10*t^9*Log[t]^8)-73/(60*t^9*Log[t]^7)-4/(3*t^9*Log[t]^6)-149/(144*t^9*Log[t]^5)-691/(1200*t^9*Log[t]^4)-18569/(75600*t^9*Log[t]^3)-761/(12600*t^9*Log[t]^2)-1/(45*t^9*Log[t]))*x^10I've attached the worksheet 1.mw.

The example below works but not for the above I just get:

Error, invalid arguments to coeffs

Not sure where I'm going wrong any idea how to do this?

EXAMPLE that works from https://math.stackexchange.com/questions/2786296/strategy-or-software-for-representing-infinite-series-in-sigma-form

I've also attached the worksheet for this 2.mw

=>

S:= 1-x+4/3*(x^2)-2*x^3+16/5*(x^4)-16/3*(x^5)+64/7*(x^6)-16*x^7+256/9*(x^8)-256/5*(x^9)+1024/11*(x^10);

Coeffs:= [seq(coeff(S, x, j), j=0..10)];

g:= gfun:-guessgf(Coeffs,x);

convert(g[1], FPS, x);

results in:

Sum((-1)^k*2^k*x^k/(k+1), k = 0 .. infinity)

or 

rec:= gfun:-listtorec(Coeffs, a(n));

aa:= rsolve(rec[1],a(n));

Sum(aa*x^n, n=0..infinity);

results in the same desired result.

 

Hi, I'd like to use something to compare expressions, so I thought evalb or verify.

I'd like to evaluate things like the identity,

logb(y) = y ∙ logb(x)

However using either I get:

> verify(2*log(x), log(x^2));

                             false
> evalb(2*log(x) = log(x^2));

                             false
now I thought maybe conditions might be different but as none are specified they should be the same in any domain. 

Not sure what is wrong or which comand is more suitable.

 

 

Supposing as a nice simple example I use the power series command,

series(ex, x=0,8)

to get,

1+x+12x+ 16x+ 124x+ 1120x+ 1720x+ 15040x7

Is there automated anyway to get this as a Sigma representation? 

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