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.