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I'm trying to prove that $h(x) > g(x)$ when $0<x<1$.

$$h(x) = (1 + i \cdot x)$$ $$g(x) = (1 + i)^x$$

And $i$ is a real number greater than $0$.

I tried with two differents ways, but i wasn't successful in none of them..

First attempt: $$(1+ i \cdot x) \;\textrm{ ... }\; (1+i)^x$$ $$ \ln{(1 + i \cdot x)} \;\textrm{ ... }\; x \cdot \ln{(1+i)}$$ $$ 1 \;\textrm{ ... }\; \dfrac{x \cdot \ln{(1+i)}}{\ln{(1 + i \cdot x)}}$$ So i have to observe the function $f(x) = \dfrac{x \cdot \ln{(1+i)}}{\ln{(1 + i \cdot x)}}$ in interval $0<x<1$ to determine if it's greater, equal or smaller than 1.

First i have to know if $f(x)$ has a critical point in this interval. $$f '(x) = \dfrac{\ln{(1+i)}}{\ln{(1+i \cdot x)}} - \dfrac{i \cdot\ln{(1+i) \cdot x}}{(i\cdot x +1)\cdot \ln{(i\cdot x +1)^2}}$$ I tried to find the roots (if exit roots, but couldn't resolve..)

So i left this attempt, and tried in a different way:

Second Attempt:

I know that $h(x)$ is a linear function and $g(x)$ is a convex function. If i prove $g(x)$ is convex, and $h(x)$ is a linear combination of $g(x)$ in interval $0<x<1$, then $h(x) > g(x)$ in this interval, alright?

From definition to a function be convex: $$ g(t \cdot a + (1-t)\cdot b)\leq t\cdot g(a) + (1-t) \cdot g(b) \textrm{ , where } t \in [0,1]$$ $$ (1+i)^{(t\cdot a + (1-t)\cdot b)} \leq t\cdot (1+i)^a + (1-t)\cdot (1+i)^b$$ I'm stucked here..

Any idea how can i prove $h(x)>g(x)$??

Chathura Gunasekera
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Luiz Eduardo Zappa
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  • Look up Bernoulli's inequality (and look to see whether this question has been asked here previously, as surely it has been). – Gerry Myerson May 19 '19 at 00:58

2 Answers2

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I think you can define $f(x) = h(x) - g(x) = 1+ix - (1+i)^x$. Then $f^{'} (x) = i - (1+i)^x \ln (1+i)$.

We have $f^{'}(x) = 0$ iff $x = \dfrac{\ln i - \ln \ln (1+i)}{\ln (1+i)}$. Note that $e^i > 1 + i$, hence $x_0 = \dfrac{\ln i - \ln \ln (1+i)}{\ln (1+i)} > 0$.

On the other hand, it is easy seen that $f^{'}(x) < 0$ when $1 > x > x_0$ and $f^{'}(x) > 0$ when $0 < x < x_0$. Since $f(0) = f(1) = 0$, we obtain $f(x) \geq 0$ when $x \in [0,1]$. This mean that $h(x) \geq g(x)$ if $x \in [0,1]$.

ToThichToan
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    Great solution!

    Only a few more steps for those who have difficulty: $e^i > 1+i$, remember $e^i = \sum_{k=0}^\infty {i^k/k! = i^0/0! + i^1/1! + i^2/2! + \cdots = (1 + i) + i^2/2! + \cdots}$, so $e^i > 1+i$

    $\ln{e^i} > \ln{(1+i)}$, so $i>\ln{(1+i)}$ and $\ln{(i)}>\ln{\ln{(1+1)}}$

     – Luiz Eduardo Zappa May 19 '19 at 14:17
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    I couldn't identify why $f′(x)<0$ when $1>x>x_0$ and $f′(x)>0$ when $0<x<x_0$.. I needed to calculate $f''(x_0) = -(1+i)^x \ln{(1+i)^2}$, as $i>0$, then $f''(x)<0$, so $x_0$ is a maximum – Luiz Eduardo Zappa May 19 '19 at 14:27
  • @Luiz Thank you. In fact, we don't need to calculate $f^{''}(x)$. Indeed, it is easy seen that for any $i > 0$ then $(1+i)^{x_1} < (1+i)^{x_2} $ with $0 < x_1 < x_2 < 1$. Since $\ln (1+i ) >0$, we can prove that $(1+i)^{x_1} \ln (1+i) < (1+i)^{x_2} \ln (1+i)$ with $0 < x_1 < x_2 < 1$. Hence $f^{'}(x_1) = 1 - (1+i)^{x_1} \ln (1+i) > 1 - (1+i)^{x_2} \ln (1+i) = f^{'}(x_2)$ with $0 < x_1 < x_2 < 1$. On the other hand, the function $f^{'}(x)$ continues on $(0,1)$ and $f^{'}(0) > 0$, $f^{'}(1) < 0$. Thus we can obtain the claim. – ToThichToan May 23 '19 at 02:27
  • now i got it! thanks – Luiz Eduardo Zappa May 24 '19 at 22:34
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By observation we see that $h(0) = g(0)$ and $h(0) = g(0)$

Let $f(x)=\frac{h(x)}{g(x)}$ then if the postulate $h(x) > g(x)$ is true for $0<x<1$ then $f(x)>1$ for this domain of $x$. As $g(x)>0$ for $x>0$, $f(x)$ is always defined.

Initially we have $f(0) = 1$ and $f(1) = 1$

Then differentiating with respect to x; $$f'(x) = \frac{p(1+p)^x - (1+px)(\ln(1+p))(1+p)^x}{(1+p)^{2x}}$$ $$f'(x) = \frac{(1+p)^x[p - \ln(1+p) - p \ln(p+1)x]}{(1+p)^{2x}}$$

As $f'(x) = 0$ at stationary points; $$ p - \ln(p+1) - p \ln(p+1)x = 0 $$ therefore $$x=\frac{p - \ln(p+1)}{p \ln(p+1)}$$

As $p > \ln(p+1)$ for all $p \in$ $\Re$ therefore $$0 < \frac{p - \ln(p+1)}{p \ln(p+1)} <1$$ checking the gradient of the graph we see that a maxima occurs a the above value.

As $f(0) = 1$ and $f(1) = 1$ and the graph is increasing upto $\frac{p - \ln(p+1)}{p \ln(p+1)}$ we can say for certain that $f(x) > 1$ for $0<x<1$

Thereby for $0<x<1$ , $$\frac{h(x)}{g(x)} > 1$$

Therefore for $0<x<1$ $$ h(x) > g(x)$$

Chathura Gunasekera
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  • Nice! It's better to use $p$ than $i$, because $i$ is the symbol of imaginary number. I forgot about it.. It's because $i$ is used for interest rate – Luiz Eduardo Zappa May 19 '19 at 14:29