Show-that-r-1-n-e-rx-e-x-e-nx-1-e-x-1-if-e-rx-cos-irx-isin-irx-and-x-0-Do-not-treat-as-a-GP-to-directly-obtain-the-result-

Question Number 1453 by 112358 last updated on 06/Aug/15

S h o w t h a t

\underset{r = 1}{\sum^{n}} e^{r x} = \frac{e^{x} (e^{n x} - 1)}{e^{x} - 1} (*)

i f e^{r x} = c o s (i r x) - i s i n (i r x) a n d x \neq 0 .

[D o n o t t r e a t (*) a s a G P t o

d i r e c t l y o b t a i n t h e r e s u l t .]

Answered by 123456 last updated on 13/Aug/15

S_{n} = \underset{r = 1}{\sum^{n}} e^{r x}, n \in N^{*}

e^{x} S_{n} = e^{x} \underset{r = 1}{\sum^{n}} e^{r x} = \underset{r = 1}{\sum^{n}} e^{x} e^{r x}

= \underset{r = 1}{\sum^{n}} e^{(r + 1) x} = \underset{p = 2}{\sum^{n + 1}} e^{p x} = \underset{r = 2}{\sum^{n + 1}} e^{r x} (p = r + 1, r = 1 \Rightarrow p = 2, r = n \Rightarrow p = n + 1)

e^{x} S_{n} - S_{n} = \underset{r = 2}{\sum^{n + 1}} e^{r x} - \underset{r = 1}{\sum^{n}} e^{r x}

(e^{x} - 1) S_{n} = e^{(n + 1) x} + \underset{r = 2}{\sum^{n}} e^{r x} - \underset{r = 2}{\sum^{n}} e^{r x} - e^{x}

S_{n} = \frac{e^{(n + 1) x} - e^{x}}{e^{x} - 1} = \frac{e^{x} (e^{n x} - 1)}{e^{x} - 1}, e^{x} \neq 1

e^{x} = 1

S_{n} = \underset{r = 1}{\sum^{n}} 1 = n

Commented by 112358 last updated on 06/Aug/15

I a p p r e c i a t e y o u r a p p r o a c h .

Commented by Rasheed Ahmad last updated on 13/Aug/15

\underset{r = 1}{\sum^{n}} e^{(r + 1) x} = \underset{r = 2}{\sum^{n + 1}} e^{r x}

(\boldsymbol How ? \boldsymbol pl \boldsymbol explain, \boldsymbol I \boldsymbol am \boldsymbol asking

\boldsymbol for \boldsymbol learning . \boldsymbol Question \boldsymbol should \boldsymbol not \boldsymbol be

\boldsymbol treated \boldsymbol as \boldsymbol an \boldsymbol objection .)

Commented by 112358 last updated on 13/Aug/15

\underset{r = 1}{\sum^{n}} e^{(r + 1) x} = e^{2 x} + e^{3 x} + \dots + e^{(n + 1) x} (*)

(*) i s e q u a l t o \underset{r = 2}{\sum^{n + 1}} e^{r x} s i n c e y o u

a c q u i r e t h e s a m e p a t t e r n o f p o w e r s

o f e a s r g o e s u p t o n i f y o u d e n o t e (*) b y \underset{r = 2}{\sum^{n + 1}} e^{r x} .

\underset{r = 2}{\sum^{n + 1}} e^{r x} = e^{2 x} + e^{3 x} + \dots + e^{(n + 1) x} = \underset{r = 1}{\sum^{n}} e^{(r + 1) x}

I h a d h o p e d t h a t s o m e o n e w o u l d

h a v e t r i e d t h e m o r e l e n g t h y r o u t e

w h i c h I h a d f o u n d a s a n a l t e r n a t i v e

w a y t o f i n d i n g \underset{r = 1}{\sum^{n}} e^{r x} .

Commented by Rasheed Ahmad last updated on 13/Aug/15

\boldsymbol THANKS \boldsymbol For \boldsymbol explaining!

Y \boldsymbol our \boldsymbol approach \boldsymbol is \boldsymbol realy \boldsymbol appreciable!

Commented by 123456 last updated on 13/Aug/15

L = \lim_{x \to 0} \frac{e^{(n + 1) x} - e^{x}}{e^{x} - 1} = \lim_{x \to 0} \frac{e^{x} (e^{n x} - 1)}{e^{x} - 1}

L = \lim_{x \to 0} e^{x} \lim_{x \to 0} \frac{e^{n x} - 1}{e^{x} - 1} = \lim_{x \to 0} \frac{e^{n x} - 1}{e^{x} - 1}

\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1

L = \lim_{x \to 0} \frac{e^{n x} - 1}{e^{x} - 1} = \lim_{x \to 0} \frac{\frac{n x}{n x} (e^{n x} - 1)}{\frac{x}{x} (e^{x} - 1)}

L = \lim_{x \to 0} \frac{n x \frac{e^{n x} - 1}{n x}}{x \frac{e^{x} - 1}{x}} = n \lim_{x \to 0} \frac{\frac{e^{n x} - 1}{n x}}{\frac{e^{x} - 1}{x}}

L = n \frac{\lim_{x \to 0} \frac{e^{n x} - 1}{n x}}{\lim_{x \to 0} \frac{e^{x} - 1}{x}} (y = n x, x \to 0 \equiv y \to 0)

L = n \lim_{y \to 0} \frac{e^{y} - 1}{y} = n

Commented by 112358 last updated on 13/Aug/15

L = \lim_{x \to 0} S_{n} = \lim_{x \to 0} \frac{e^{(n + 1) x} - e^{x}}{e^{x} - 1}

B y L^{'} {H o p i t a l}^{'} s r u l e

L = \lim_{x \to 0} \frac{\frac{d}{d x} (e^{(n + 1) x} - e^{x})}{\frac{d}{d x} (e^{x} - 1)} = \lim_{x \to 0} \frac{(n + 1) e^{(n + 1) x} - e^{x}}{e^{x}}

L = \frac{(n + 1) e^{0} - e^{0}}{e^{0}} = n + 1 - 1 = n .

∴ \lim_{x \to 0} S_{n} = n w h i c h s u p p o r t s t h e r e s u l t

S_{n} = n w h e n x = 0 f o r S_{n} = \underset{r = 1}{\sum^{n}} e^{r x} .

Commented by 112358 last updated on 13/Aug/15

I s e e . G o o d a l t e r n a t i v e a p p r o a c h .

H o w m a y o n e s h o w t h a t

\lim_{x \to 0} \frac{e^{x} - 1}{x} = 1 ?

Commented by 123456 last updated on 14/Aug/15

F (a) = \lim_{x \to 0} \frac{a^{x} - 1}{x}, a > 0, a \neq 1

x = \log_{a} (t + 1)

x \to 0 \equiv t \to 0

F (a) = \lim_{t \to 0} \frac{a^{\log_{a} (t + 1)} - 1}{\log_{a} (t + 1)}

F (a) = \lim_{t \to 0} \frac{t}{\log_{a} (t + 1)} [\log_{a} (t + 1) = \frac{\ln (t + 1)}{\ln a}]

F (a) = \lim_{t \to 0} \frac{t \ln a}{\ln (t + 1)} = \lim_{t \to 0} \frac{\ln a}{\frac{1}{t} \ln (1 + t)}

F (a) = \frac{\ln a}{\lim_{t \to 0} \ln {(1 + t)}^{1 / t}} = \frac{\ln a}{\ln \lim_{t \to 0} {(1 + t)}^{1 / t}}

fundamental limit : \lim_{t \to 0} {(1 + t)}^{1 / t} = e

F (a) = \frac{\ln a}{\ln e} = \ln a

a = e

F (e) = 1

Commented by 112358 last updated on 14/Aug/15

A w e s o m e . T h a n k s f o r s h a r i n g

t h a t p r o o f .

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