Hand Computing: Evaluate $\sum_{k=1}^{2013}f(k/2014)$

In summary, "Hand Computing" is the process of solving mathematical problems using pen and paper. The notation $\sum_{k=1}^{2013}f(k/2014)$ represents a summation where k ranges from 1 to 2013 and f is evaluated at k/2014. The purpose of evaluating this summation is to find its numerical value, which can provide insights into various mathematical problems. To approach hand computing for this summation, one can simplify the expression, substitute a variable for k/2014, and use careful calculations to find the sum.
  • #1
magneto1
102
0
Define:
\[
f(t) := \frac{7^t}{7^t + \sqrt{7}}.
\]
Without the aid of a computer or calculator, evaluate:
\[
\sum_{k=1}^{2013} f \left( \frac{k}{2014} \right).
\]
(Please show the work.)
 
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  • #2
magneto said:
Define:
\[
f(t) := \frac{7^t}{7^t + \sqrt{7}}.
\]
Without the aid of a computer or calculator, evaluate:
\[
\sum_{k=1}^{2013} f \left( \frac{k}{2014} \right).
\]
(Please show the work.)

Notice that
$$f\left(\frac{k}{2014}\right)+f\left(\frac{2014-k}{2014}\right)=1$$
Hence, the sum is:
$$\sum_{k=1}^{2013} f \left( \frac{k}{2014} \right)=1006+f\left(\frac{1007}{2014}\right)=1006+\frac{\sqrt{7}}{2\sqrt{7}}=\boxed{1006.5}$$
 
  • #3
That's correct.
 
  • #4
$$\sum_{k=1}^{2013}f\left ( \frac{k}{2014} \right )= \sum_{k=1}^{1007}\left \{ f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right ) \right \}=^* \sum_{k=1}^{1007}1=1007$$

\[(*). \;\;\; f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right )=\frac{7^{\frac{k}{2014}}}{7^{\frac{k}{2014}}+7^{\frac{1}{2}}}+\frac{7^{\frac{2014-k}{2014}}}{7^{\frac{2014-k}{2014}}+7^{\frac{1}{2}}}\\\\ =\frac{7^{\frac{k}{2014}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{1}{2} }\right )+7^{\frac{2014-k}{2014}}\left ( 7^{\frac{k}{2014}}+7^\frac{1}{2} \right )}{\left ( 7^{\frac{k}{2014}}+7^\frac{1}{2} \right )\left ( 7^{\frac{2014-k}{2014}}+7^\frac{1}{2} \right )} \\\\ =\frac{14 + 7^{\frac{1}{2}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{k}{2014}}\right )}{14 + 7^{\frac{1}{2}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{k}{2014}}\right )}=1\]
 
  • #5
lfdahl said:
$$\sum_{k=1}^{2013}f\left ( \frac{k}{2014} \right )= \sum_{k=1}^{1007}\left \{ f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right ) \right \}=^* \sum_{k=1}^{1007}1=1007$$

\[(*). \;\;\; f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right )=\frac{7^{\frac{k}{2014}}}{7^{\frac{k}{2014}}+7^{\frac{1}{2}}}+\frac{7^{\frac{2014-k}{2014}}}{7^{\frac{2014-k}{2014}}+7^{\frac{1}{2}}}\\\\ =\frac{7^{\frac{k}{2014}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{1}{2} }\right )+7^{\frac{2014-k}{2014}}\left ( 7^{\frac{k}{2014}}+7^\frac{1}{2} \right )}{\left ( 7^{\frac{k}{2014}}+7^\frac{1}{2} \right )\left ( 7^{\frac{2014-k}{2014}}+7^\frac{1}{2} \right )} \\\\ =\frac{14 + 7^{\frac{1}{2}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{k}{2014}}\right )}{14 + 7^{\frac{1}{2}}\left ( 7^{\frac{2014-k}{2014}}+7^{\frac{k}{2014}}\right )}=1\]

Very close.

Consider the pair of $\sum_{k=1}^{1007}\left \{ f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right ) \right \}$ when $k=1007$.
 
  • #6
Oh, my mistake!

\[\sum_{k=1}^{2013}f\left ( \frac{k}{2014} \right )= \sum_{k=1}^{1006}\left \{ f\left ( \frac{k}{2014} \right )+f\left ( \frac{2014-k}{2014} \right ) \right \}+f\left ( \frac{1007}{2014} \right )=^* \sum_{k=1}^{1006} \left \{ 1 \right \}+\frac{1}{2}=\frac{1}{2}2013\]
 

Related to Hand Computing: Evaluate $\sum_{k=1}^{2013}f(k/2014)$

What is "Hand Computing"?

"Hand Computing" refers to the process of solving mathematical problems using pen and paper, rather than using a calculator or computer. It requires careful and meticulous calculations to arrive at the correct answer.

What does the notation $\sum_{k=1}^{2013}f(k/2014)$ mean?

This notation represents a summation, where the variable k takes on values from 1 to 2013, and the function f is evaluated at k/2014 for each value of k. The results of these evaluations are then added together to get the final sum.

Why is the summation range from 1 to 2013?

The range is determined by the upper limit of the summation, which in this case is 2013. This means that the summation will include all values of f(k/2014) where k ranges from 1 to 2013.

What is the purpose of evaluating this summation?

The purpose of evaluating this summation is to find the numerical value of the sum, which can provide insights into various mathematical problems and concepts. It can also be used to approximate the value of an integral or to solve other mathematical equations.

How can one approach hand computing for this summation?

One approach is to first simplify the expression by substituting k/2014 with a variable, such as x. Then, use algebraic manipulations to find the sum in terms of x. Finally, substitute back the original expression for x and evaluate the sum using careful calculations.

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