Find the sum of a function given a series

[chwala]

Homework Statement

given ##g(x)=\frac {4^x} {4^x+2}##, find the value of ##g(o)+g(\frac {1} {2020})+g(\frac{2} {2020})+....g(1)##

Relevant Equations

series

since the first term is ##g(0)= \frac {1}{3}##
& last term is ##g(1)=\frac {4}{6}##
it follows that the ##\sum_{0}^1 g(x)##= ##\frac {1}{3}##+##\frac {4}{6}=1## is this correct?

 

[andrewkirk]

What do you mean by ##\sum_{0}^1 g(x)##? That string of symbols does not have an unambiguous, accepted meaning.

Perhaps you meant to write ##\sum_{x=0}^{2020} g\left(\frac x{2020}\right)##. If so then no, the sum will be greater than the sum of the first and last terms, because all terms are positive.

 

[chwala]

What do you mean by ##\sum_{0}^1 g(x)##? That string of symbols does not have an unambiguous, accepted meaning.

Perhaps you meant to write ##\sum_{x=0}^{2020} g\left(\frac x{2020}\right)##. If so then no, the sum will be greater than the sum of the first and last terms, because all terms are positive.

noted, i see my mistake...how do i approach this...

 

[chwala]

i still do not get but one thing that i know is,
##g(0)=\frac {1}{3}##
##g(1/2020)=0.333##
##g(1000/2020)=0.498##
##g(1500/2020)=0.58##
##g(2000/2020)=0.663##
##g(2010/2020)=0.665##
##g(2020/2020)= 0.66666##...sum will start from ##{0.333...+...+0.6666...}##

 

[Delta2]

Do you want to find the exact value of that sum or an approximation is good enough for this?
You can approximate the given sum by the integral $$\int_0^{2020} \frac{4^{\frac{x}{2020}}}{4^{\frac{x}{2020}}+2}dx$$ which looks scary but it isn't so hard afterall.

 

[chwala]

Do you want to find the exact value of that sum or an approximation is good enough for this?
You can approximate the given sum by the integral $$\int_0^{2020} \frac{4^{\frac{x}{2020}}}{4^{\frac{x}{2020}}+2}dx$$ which looks scary but it isn't so hard afterall.

I want the exact sum...so we have to integrate? we can't use the series sum approach?

 

[chwala]

using my ti-nspire i am getting sum=##1010##

 

[Delta2]

using my ti-nspire i am getting sum=##1010##

Well that's the value of that integral too!. Maybe i was wrong and the integral is equal to the exact sum.

 

[Delta2]

Wolfram alpha says that the sum is approximately equal to 1010.50 while the integral is exactly 1010.

 

[Delta2]

I want the exact sum...so we have to integrate? we can't use the series sum approach?

Well I ,myself, don't seem to be able to find an exact expression for the sum $$\sum_{n=1}^{2020} \frac{4^{\frac{n}{2020}}}{4^{\frac{n}{2020}}+2}$$ that's why I thought to approximate it by a proper integral.
I think the approximation works because 2020 is quite large number, if we were looking for the sum of $$\sum_{n=1}^{k} \frac{4^{\frac{n}{k}}}{4^{\frac{n}{k}}+2}$$ for ##k=50 or 100## then the approximation would be quite off.

 

[chwala]

wawawawawawawawa maths is not for the faint hearts...you spend hours on a problem, trying to figure out...

 

[Delta2]

wawawawawawawawa maths is not for the faint hearts...you spend hours on a problem, trying to figure out...

I might be wrong but I think we can't find a closed form for this sum (like for example we can find a closed form for the sum ##\sum_{k=1}^n k^2##). Best we can do is approximate it by integral. I believe that is what it was intended by the problem's giver, that's why he gave the sum as a function of x, to inspire us into using integral.

I still might be wrong..

 

[chwala]

could you share your step by step working of the integral? i am ending up with this, as one of my steps...
##\int_0^{2020} \frac {u}{u^2-2u-8},du## where ##u=4^{x/2020}##

 

[Delta2]

Your substitution for ##u=4^{\frac{x}{2020}}## seems correct but the expression you got doesn't seem correct to me. After such substitution you should get that ##du=\frac{\ln4}{2020}udx## so I think the integral transforms to ##\frac{\ln 4}{2020}\int\frac{du}{u+2}##.

 

[chwala]

Your substitution for ##u=4^{\frac{x}{2020}}## seems correct but the expression you got doesn't seem correct to me. After such substitution you should get that ##du=\frac{\ln4}{2020}udx## so I think the integral transforms to ##\frac{\ln 4}{2020}\int\frac{du}{u+2}##.

let me check my working again...

 

[chwala]

a colleague managed to solve it...

 

[Delta2]

Ah very elegant indeed, using a "Gauss method" and first noticing the ##g(x)+g(1-x)=1## equation.

 

[chwala]

Your substitution for ##u=4^{\frac{x}{2020}}## seems correct but the expression you got doesn't seem correct to me. After such substitution you should get that ##du=\frac{\ln4}{2020}udx## so I think the integral transforms to ##\frac{\ln 4}{2020}\int\frac{du}{u+2}##.

something i am not getting here,
now, letting ##u=4^{x/2020}##
##ln u= \frac {x}{2020} ln 4##
##\frac {d}{dx} (ln u)=\frac {d}{dx}[\frac {x}{2020} ln 4]##
it follow that
##\frac{du}{dx}=\frac {ln 4}{2020}##.##u## ...

now from the above step, how did you transform to get the integral term in post 14?...

 

[Delta2]

If you just plug ##u=4^{\frac{x}{2020}}## into the integral, the integral becomes
$$\int \frac{u}{u+2} dx (1)$$

you also have ##du=\frac{\ln 4}{2020}u dx## as you correctly calculate at post #18. We can rewrite this as $$\frac{2020}{\ln 4}du=udx(2)$$

By inserting into (1) the expression for ##udx## by (2) we come to
$$\int \frac{2020}{\ln 4} \frac{du}{u+2}$$

Sorry I was wrong at post #14 about the multiplicative constant, I should have inverse it.

You should always not forget to transform the boundaries of integration when you do change of variable in definite integral. I didnt do this in the above (i omit the boundaries) but you should do it.

 

[chwala]

If you just plug ##u=4^{\frac{x}{2020}}## into the integral, the integral becomes
$$\int \frac{u}{u+2} dx (1)$$

you also have ##du=\frac{\ln 4}{2020}u dx## as you correctly calculate at post #18. We can rewrite this as $$\frac{2020}{\ln 4}du=udx(2)$$

By inserting into (1) the expression for ##udx## by (2) we come to
$$\int \frac{2020}{\ln 4} \frac{du}{u+2}$$

Sorry I was wrong at post #14 about the multiplicative constant, I should have inverse it.

You should always not forget to transform the boundaries of integration when you do change of variable in definite integral. I didnt do this in the above (i omit the boundaries) but you should do it.

yeah, that was a bit confusing...now i think it should be fine...by boundaries you mean the limits of integration?

 

[chwala]

i got it thanks for your time , the limits are from ##u=1## to ##u=4##, plugging on the indefinite integral yields the desired value ##1010##

 

[chwala]

i got it long time ago using the limits ##u=1## and ##u=4## to give us the desired ##1010## value. Thanks for your input. Bingo!

 

[Delta2]

The forum is bugging today, I don't seem to get alerts when someone posts a new message in the threads which I ' ve posted.