Thread: a few random questions.

a few random questions :

1. Given :

$\displaystyle 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$

(a) Prove this results from first principles.

2. Show from first principles that,

$\displaystyle \sum^n_{i=1} x^i = x\frac{1-x^n}{1-x}$


Hence give an expression of $\displaystyle \sum^{n-1}_{i=0} x^i$

3. Let

$\displaystyle g(t) = exp$$\displaystyle \{$$\displaystyle \alpha +$ $\displaystyle \beta$$\displaystyle t$$\displaystyle \}$

Given that g(10) = 8.1882 and g(20) = 60.3403, calculate g(15).

4. Consider a function f such that f(11) = 1.234 and f(12)=2.345. Assuming that the function is linear over the interval from 11 to 12, calculate f(11.36).

Someone would please kindly guide me on how to do the questions! Thanks in advance!

Originally Posted by pearlyc

a few random questions :

1. Given :
$\displaystyle 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$

(a) Prove this results from first principles

This depends on what you mean by first principles. Does induction count?

What about this:

Let:

$\displaystyle S(n)=1 + 2 + 3 + ... + n $

Then:

$\displaystyle (\Delta S)(n)=S(n)-S(n-1)=n,\ \ \ n\ge 2$

$\displaystyle
(\Delta^2 S)(n)=(\Delta S)(n)-(\Delta S)(n-1)=1, \ \ \ n\ge 3
$

That is the second differences of $\displaystyle S(n)$ are constant, which means that $\displaystyle S(n)$ is a quadratic in $\displaystyle n$.

So let:

$\displaystyle
S(n)=a n^2 + b n+c
$

Now:

$\displaystyle S(1)=a+b+c=1$
$\displaystyle S(2)=4a+2b+c=3$
$\displaystyle S(3)=9a+3b+c=6$

Thus we have a set of three simultaneous equations in $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ which we solve to find the quadratic for $\displaystyle S(n).$

RonL

Hello, pearlyc!

3. Let $\displaystyle g(t) \:= \:e^{\alpha +\beta\,t}$

Given that: .$\displaystyle g(10) = 8.1882\,\text{ and }\,g(20) = 60.3403$

. . calculate $\displaystyle g(15).$

We have: .$\displaystyle \begin{array}{cccccc}g(10)\:=\:8.1882 & \Rightarrow & e^{(\alpha+10\beta)} &=& 8.1882 & {\color{blue}[1]} \\
g(20)\:=\:60.3402 & \Rightarrow & e^{(\alpha+20\beta)} &=& 60.3402 & {\color{blue}[2]}\end{array}$

Divide [2] by [1]: .$\displaystyle \frac{e^{(\alpha + 20\beta)}}{e^{(\alpha+10\beta)}} \;=\;\frac{60.3402}{8.1882} \quad\Rightarrow\quad e^{10\beta} \:=\:7.369165384$

Then: .$\displaystyle 10\beta \:=\:\ln(7.369165384)$

. . Hence: .$\displaystyle \beta \:=\:0.199730445 \quad\Rightarrow\quad\boxed{\beta \:\approx\:0.2}$

So far: .$\displaystyle g(t) \;=\;e^{(\alpha+0.2t)}$


Then [1] becomes: .$\displaystyle g(10) \;=\;e^{\alpha +2} \;=\;8.1882 \quad\Rightarrow\quad \alpha + 2 \:=\:\ln(8.1882)$

. . Hence: .$\displaystyle \alpha \:=\:0.102694093\quad\Rightarrow\quad \boxed{\alpha \:\approx\:0.1}$


And we have: .$\displaystyle g(t) \;=\;e^{0.1 +0.2t}$

Therefore: .$\displaystyle g(15)\;=\;e^{0.1 + 3} \:=\:e^{3.1} \:=\:22.19795128 \:\approx\:22.2$

Originally Posted by pearlyc

2. Show from first principles that,

$\displaystyle \sum^n_{i=1} x^i = x\frac{1-x^n}{1-x}$


Hence give an expression of $\displaystyle \sum^{n-1}_{i=0} x^i$

$\displaystyle (1-x)\sum^n_{i=1} x^i = \sum_{i=1}^n (x^i-x^{i+1})=(x^1-x^2) +(x^2-x^3) + ... +(x^{n-1}-x^n)+(x^n-x^{n+1})$

Collecting like powers:

$\displaystyle (1-x)\sum^n_{i=1} x^i = x^1-x^{n+1}=x(1-x^n)$

RonL