Chapter 6.1, Problem 38E

To determine

MRAM_n is always the average of LRAM_n and RRAM_n.

Answer to Problem 38E

The average of LRAM and RRAM is not always equal to MRAM.

Explanation of Solution

Given information:

MRAM_n is always the average of LRAM_n and RRAM_n.

Formula used:

The LRAM and RRAM is used.

Calculation:

Statement is false.

For example,

  $\begin{array}{l}f\left(x\right)=x^3\\ 0\le x\le 1\end{array}$

For

  $\begin{array}{l}n=1\\ {\text{MRAM}}_1=f\left(0,5\right)\\ ={\left(0.5\right)}^3\\ {\text{MRAM}}_1=0.125\\ {\text{LRAM}}_1=f\left(0\right)\\ {\text{LRAM}}_1=0\\ {\text{RRAM}}_1=f\left(1\right)\\ {\text{RRAM}}_1=1\end{array}$

Now take average of LRAM1 and RRAM1

  $\begin{array}{l}\text{average}=\frac{0+1}{2}\\ \text{average}=0.5\end{array}$

It is not equal to MRAM1.

Sometime average Of LRAM and RRAM is equal to MRAM.

Take,

  $\begin{array}{l}f\left(x\right)=x\\ n=1\\ \left[a,b\right]=\left[0,1\right]\\ {\text{MRAM}}_1=f\left(\frac{1}{2}\right)\\ {\text{MRAM}}_1=\frac{1}{2}\\ {\text{LRAM}}_1=f\left(0\right)\\ {\text{LRAM}}_1=0\\ {\text{RRAM}}_1=f\left(1\right)\\ {\text{RRAM}}_1=1\end{array}$

Now take average of LRAM1 and RRAM1

  $\begin{array}{l}\text{average}=\frac{0+1}{2}\\ \text{average}=0.5\end{array}$

It is not equal to MRAM1

Sometime average of LRAM and RRAM is equal to MRAM.

Take,

  $\begin{array}{l}f\left(x\right)=x\\ n=1\\ \left[a,b\right]=\left[0,1\right]\\ {\text{MRAM}}_1=f\left(\frac{1}{2}\right)\\ =\frac{1}{2}\\ {\text{LRAM}}_1=f\left(0\right)\\ =0\\ {\text{RRAM}}_1=f\left(1\right)\\ {\text{RRAM}}_1=1\end{array}$

Now take average of LRAM1 and RRAM1

  $\begin{array}{l}\text{average}=\frac{0+1}{2}\\ \text{average}=0.5\end{array}$

It is equal to MRAM1.

So, above from two examples hence the average of LRAM and RRAM is not always equal to MRAM. Sometimes it is equal, sometime not.

Conclusion:

The average of LRAM and RRAM is not always equal to MRAM.